Java Puzzles
Filipp Shubin

Overall Presentation Goal
Learn some of the quirks of
programming in general and the
Java language in particular;
Have fun!

Learning Objectives
• As a result of this presentation, you
will be able to:
Avoid some common programming
–
pitfalls
Have some fun while you are learning
–

1. “All I Get is Static”
01 class Dog {
02 public static void bark() {
03 System.out.print(\"woof \");
04 }
05 }
06 class Basenji extends Dog {
07 public static void bark() { }
08 }
09 public class Bark {
10 public static void main(String args[]) {
11 Dog woofer = new Dog();
12 Dog nipper = new Basenji();
13 woofer.bark();
14 nipper.bark();
15 }
16 }

What Does It Print?
(a) woof
(b) woof woof
(c) It varies

What Does It Print?
(a) woof
(b) woof woof
(c) It varies
No dynamic dispatch on static methods

Another Look
01 class Dog {
02 public static void bark() {
03 System.out.print(\"woof \");
04 }
05 }
06 class Basenji extends Dog {
07 public static void bark() { }
08 }
09 public class Bark {
10 public static void main(String args[]) {
11 Dog woofer = new Dog();
12 Dog nipper = new Basenji();
13 woofer.bark();
14 nipper.bark();
15 }
16 }

How Do You Fix It?
• Remove static from the bark
method

The Moral
• Static methods can't be overridden
They can only be hidden
–
• Don’t hide static methods
• Never invoke static methods on
instances
Not Instance.staticMethod()
–
But Class.staticMethod()
–

2. “What's in a Name?”
01 public class Name {
02 private String first, last;
03 public Name(String first, String last) {
04 this.first = first;
05 this.last = last;
06 }
07 public boolean equals(Object o) {
08 if (!(o instanceof Name)) return false;
09 Name n = (Name)o;
10 return n.first.equals(first) &&
11 n.last.equals(last);
12 }
13 public static void main(String[] args) {
14 Set s = new HashSet();
15 s.add(new Name(\"Donald\", \"Duck\"));
16 System.out.println(
17 s.contains(new Name(\"Donald\", \"Duck\")));
18 }
19 }

What Does It Print?
(a) True
(b) False
(c) It varies

What Does It Print?
(a) True
(b) False
(c) It varies
Donald is in the set, but the set can’t find
him.
The Name class violates the hashCode contract.

Another Look
01 public class Name {
02 private String first, last;
03 public Name(String first, String last) {
04 this.first = first;
05 this.last = last;
06 }
07 public boolean equals(Object o) {
08 if (!(o instanceof Name)) return false;
09 Name n = (Name)o;
10 return n.first.equals(first) &&
11 n.last.equals(last);
12 }
13 public static void main(String[] args) {
14 Set s = new HashSet();
15 s.add(new Name(\"Donald\", \"Duck\"));
16 System.out.println(
17 s.contains(new Name(\"Donald\", \"Duck\")));
18 }
19 }

How Do You Fix It?
Add a hashCode method:
public int hashCode() {
return 31 * first.hashCode() + last.hashCode();
}

The Moral
• If you override equals, override
hashCode
• Obey general contracts when
overriding
• See Effective Java, Chapter 3

3. “Indecision”
01 class Indecisive {
02 public static void main(String[] args) {
03 System.out.println(waffle());
04 }
05
06 static boolean waffle() {
07 try {
08 return true;
09 } finally {
10 return false;
11 }
12 }
13 }

What Does It Print?
(a) true
(b) false
(c) None of the above

What Does It Print?
(a) true
(b) false
(c) None of the above
The finally is processed after the try.

Another Look
01 class Indecisive {
02 public static void main(String[] args) {
03 System.out.println(waffle());
04 }
05
06 static boolean waffle() {
07 try {
08 return true;
09 } finally {
10 return false;
11 }
12 }
13 }

The Moral
• Avoid abrupt completion of
finally blocks
Wrap unpredictable actions with nested
–
trys
Don't return or throw exceptions
–

4. “The Saga of the Sordid Sort”
01 public class SordidSort {
02 public static void main(String args[]) {
03 Integer big = new Integer( 2000000000);
04 Integer small = new Integer(-2000000000);
05 Integer zero = new Integer(0);
06 Integer[] a = new Integer[] {big, small, zero};
07 Arrays.sort(a, new Comparator() {
08 public int compare(Object o1, Object o2) {
09 return ((Integer)o2).intValue() -
10 ((Integer)o1).intValue();
11 }
12 });
13 System.out.println(Arrays.asList(a));
14 }
15 }

What Does It Print?
(a) [-2000000000, 0, 2000000000]
(b) [2000000000, 0, -2000000000]
(c) [-2000000000, 2000000000, 0]
(d) It varies

What Does It Print?
(a) [-2000000000, 0, 2000000000]
(b) [2000000000, 0, -2000000000]
(c) [-2000000000, 2000000000, 0]
(d) It varies (behavior is undefined)
The comparator is broken!
It relies on int subtraction
•
Int too small to hold difference of 2 arbitrary ints
•

Another Look
01 public class SordidSort {
02 public static void main(String args[]) {
03 Integer big = new Integer( 2000000000);
04 Integer small = new Integer(-2000000000);
05 Integer zero = new Integer(0);
06 Integer[] a = new Integer[] {big,small,zero};
07 Arrays.sort(a, new Comparator() {
08 public int compare(Object o1, Object o2) {
09 return ((Integer)o2).intValue() -
10 ((Integer)o1).intValue();
11 }
12 });
13 System.out.println(Arrays.asList(a));
14 }
15 }

How Do You Fix It?
• Replace comparator with one that
works
01 public int compare(Object o1, Object o2) {
02 int i1 = ((Integer)o1).intValue();
03 int i2 = ((Integer)o2).intValue();
04 return
05 (i2 < i1 ? -1 : (i2 == i1 ? 0 : 1));
06 }

The Moral
• ints aren't integers!
• Think about overflow
• This particular comparison technique
OK only if max - min <= Integer.MAX_VALUE
–
For example: all values positive
–
• Don’t write overly clever code

5. “You're Such a Character”
01 public class Trivial {
02 public static void main(String args[]) {
03 System.out.print(\"H\" + \"a\");
04 System.out.print('H' + 'a');
05 }
06 }

What Does It Print?
(a) HaHa
(b) Ha
(c) None of the above

What Does It Print?
(a) HaHa
(b) Ha
(c) None of the above: It prints Ha169
'H' + 'a' evaluated as int, then converted
to String. Ouch.

The Moral
Use string concatenation (+) with
•
care
At least one operand must be a String
–
If it isn't, cast or convert\" + 'H' +
–
'a');
Be glad operator overloading isn't
•
supported

6. “The Case of the Constructor”
01 public class Confusing {
02 public Confusing(Object o) {
03 System.out.println(\"Object\");
04 }
05 public Confusing(double[] dArray) {
06 System.out.println(\"double array\");
07 }
08 public static void main(String args[]) {
09 new Confusing(null);
10 }
11 }

What Does It Print?
(a) Object
(b) double array
(c) None of the above

What Does It Print?
(a) Object
(b) double array
(c) None of the above
When multiple overloadings
apply, the most specific wins

Another Look
01 public class Confusing {
02 public Confusing(Object o) {
03 System.out.println(\"Object\");
04 }
05 public Confusing(double[] dArray) {
06 System.out.println(\"double array\");
07 }
08 public static void main(String args[]) {
09 new Confusing(null);
10 }
11 }

How Do You Fix It?
• There may be no problem
• If there is, use a cast:
New Confusing((Object)null);

The Moral
• Avoid overloading
• If you overload, avoid ambiguity
• If you do have ambiguous
overloadings, make their behavior
identical
• If you are using a \"broken\" class,
make intentions clear with a cast

7. “A Big Delight in Every Byte”
01 public class ByteMe {
02 public static void main(String[] args) {
03 for (byte b = Byte.MIN_VALUE;
04 b < Byte.MAX_VALUE; b++) {
05 if (b == 0x90)
06 System.out.print(\"Byte me! \");
07 }
08 }
09 }

What Does It Print?
(a) (nothing)
(b) Byte me!
(c) Byte me! Byte me!

What Does It Print?
(a) (nothing)
(b) Byte me!
(c) Byte me! Byte me!
Program compares a byte with an
int
byte is promoted with surprising results
–

Another Look
01 public class ByteMe {
02 public static void main(String[] args) {
03 for (byte b = Byte.MIN_VALUE;
04 b < Byte.MAX_VALUE; b++) {
05 if (b == 0x90) // (b == 144)
06 System.out.print(\"Byte me! \");
07 }
08 }
09 }
10
11 // But (byte)0x90 == -112

How Do You Fix It?
• Cast int to byte
if (b == (byte)0x90)
System.out.println(\"Byte me!\");
• Or convert byte to int, suppressing
sign extension with mask
if ((b & 0xff) == 0x90)
System.out.println(\"Byte me!\");

The Moral
• Bytes aren't ints
• Be careful when mixing primitive
types
• Compare like-typed expressions
Cast or convert one operand as
–
necessary

8. “Time for a Change”
• If you pay $2.00 for a gasket that
costs $1.10, how much change do
you get?
01 public class Change {
02 public static void main(String args[])
03 {
04 System.out.println(2.00 - 1.10);
05 }
06 }

What Does It Print?
(a) 0.9
(b) 0.90
(c) It varies
(d) None of the above

What Does It Print?
(a) 0.9
(b) 0.90
(c) It varies
(d) None of the above:
0.89999999999999
Decimal Values can't be represented
exactly by float or double

How Do You Fix It?
01 import java.math.BigDecimal;
02 public class Change2 {
03 public static void main(String args[]) {
04 System.out.println(
05 new BigDecimal(\"2.00\").subtract(
06 new BigDecimal(\"1.10\")));
07 }
08 }
09
10 public class Change {
11 public static void main(String args[]) {
12 System.out.println(200 - 110);
13 }
14 }

The Moral
• Avoid float and double where
exact answers are required
• Use BigDecimal, int, or long
instead

9. “A Private Matter”
01 class Base {
02 public String name = \"Base\";
03 }
04
05 class Derived extends Base {
06 private String name = \"Derived\";
07 }
08
09 public class PrivateMatter {
10 public static void main(String[] args) {
11 System.out.println(new Derived().name);
12 }
13 }

What Does It Print?
(a) Derived
(b) Base
(c) Compiler error in class Derived:
Can't assign weaker access to name
(d) None of the above

What Does it Print?
(a) Derived
(b) Base
(c) Compiler error in class Derived:
Can't assign weaker access to name
(d) None of the above: Compiler error in
class
PrivateMatter: Can't access name
Private method can't overrides public,
but private field can hide public

Another Look
01 class Base {
02 public String name = \"Base\";
03 }
04
05 class Derived extends Base {
06 private String name = \"Derived\";
07 }
08
09 public class PrivateMatter {
10 public static void main(String[] args) {
11 System.out.println(new Derived().name);
12 }
13 }

How Do You Fix It?
01 class Base {
02 public String getName() { return \"Base\"; }
03 }
04
05 class Derived extends Base {
06 public String getName() { return \"Derived\"; }
07 }
08
09 public class PrivateMatter {
10 public static void main(String[] args) {
11 System.out.println(new Derived().getName());
12 }
13 }

The Moral
• Avoid hiding
Violates subsumption
–
• Avoid public fields
Use accessor methods instead
–

10. “Loopy Behavior”
01 public class Loopy {
02 public static void main(String[] args) {
03 final int start = Integer.MAX_VALUE -
04 100;
05 final int end = Integer.MAX_VALUE;
06 int count = 0;
07 for (int i = start; i <= end; i++)
08 count++;
09 System.out.println(count);
10 }
11 }

What Does It Print?
(a) 100
(b) 101
(c) (nothing)

What Does It Print?
(a) 100
(b) 101
(c) (nothing)
The loop test is broken - infinite loop!

Another Look
01 public class Loopy {
02 public static void main(String[] args) {
03 final int start = Integer.MAX_VALUE -
04 100;
05 final int end = Integer.MAX_VALUE;
06 int count = 0;
07 for (int i = start; i <= end; i++)
08 count++;
09 System.out.println(count);
10 }
11 }

How Do You Fix It?
• Change loop variable from int
to long
for (long i = start; i <= end; i++)
count++;

The Moral
• ints aren't integers!
• Think about overflow
larger type if
• Use
necessary

11. “Random Behavior”
01 public class RandomSet {
02 public static void main(String[] args) {
03 Set s = new HashSet();
04 for (int i = 0; i < 100; i++)
05 s.add(randomInteger());
06 System.out.println(s.size());
07 }
08
09 private static Integer randomInteger() {
10 return new Integer(new Random().nextInt());
11 }
12 }

What Does It Print?
(a) A number close to 1
(b) A number close to 50
(c) A number close to 100
(d) None of the above

What Does It Print?
(a) A number close to 1
(b) A number close to 50
(c) A number close to 100
(d) None of the above
A new random number generator is
created each iteration and the seed
changes rarely if at all.

Another Look
01 public class RandomSet {
02 public static void main(String[] args) {
03 Set s = new HashSet();
04 for (int i=0; i<100; i++)
05 s.add(randomInteger());
06 System.out.println(s.size());
07 }
08
09 private static Integer randomInteger() {
10 return new Integer(new Random().nextInt());
11 }
12 }

How Do You Fix It?
01 public class RandomSet {
02 public static void main(String[] args) {
03 Set s = new HashSet();
04 for (int i=0; i<100; i++)
05 s.add(randomInteger());
06 System.out.println(s.size());
07 }
08
09 private static Random rnd = new Random();
10
11 private static Integer randomInteger() {
12 return new Integer(rnd.nextInt());
13 }
14 }

The Moral
• Use one Random instance for each
sequence
• In most programs, one is all you
need
• In multithreaded programs, you may
want multiple instances for
increased concurrency
─ Seed explicitly or risk identical sequences
─ Generally ok to use one instance to seed
others

12.“Making a Hash of It”
01 public class Name {
02 private String first, last;
03 public Name(String first, String last) {
04 if (first == null || last == null)
05 throw new NullPointerException();
06 this.first = first; this.last = last;
07 }
08 public boolean equals(Name o) {
09 return first.equals(o.first) && last.equals(o.last);
10 }
11 public int hashCode() {
12 return 31 * first.hashCode() + last.hashCode();
13 }
14 public static void main(String[] args) {
15 Set s = new HashSet();
16 s.add(new Name(\"Mickey\", \"Mouse\"));
17 System.out.println(
18 s.contains(new Name(\"Mickey\", \"Mouse\")));
19 }
20 }

What Does It Print?
(a) true
(b) false
(c) It varies

What Does It Print?
(a) true
(b) false
(c) It varies
Name overrides hashCode but not
equals.
The two Name instances are unequal.

Another Look
01 public class Name {
02 private String first, last;
03 public Name(String first, String last) {
04 if (first == null || last == null)
05 throw new NullPointerException();
06 this.first = first; this.last = last;
07 }
08 public boolean equals(Name o) { // Accidental overloading
09 return first.equals(o.first) && last.equals(o.last);
10 }
11 public int hashCode() { // Overriding
12 return 31 * first.hashCode() + last.hashCode();
13 }
14 public static void main(String[] args) {
15 Set s = new HashSet();
16 s.add(new Name(\"Mickey\", \"Mouse\"));
17 System.out.println(
18 s.contains(new Name(\"Mickey\", \"Mouse\")));
19 }
20 }

How Do You Fix It?
• Replace the overloaded equals method
with an overriding equals method
01 public boolean equals(Object o) {
02 if (!(o instanceof Name))
03 return false;
04 Name n = (Name)o;
05 return n.first.equals(first) && n.last.equals(last);
06 }

The Moral
• If you want to override a method:
─ Make sure signatures match
─ The compiler doesn’t check for you
─ Do copy-and-paste declarations!

13. “Ping Pong”
01 class PingPong {
02 public static synchronized void main(String[] a)
{
03 Thread t = new Thread() {
04 public void run() {
05 pong();
06 }
07 };
08
09 t.run();
10 System.out.print(\"Ping\");
11 }
12
13 static synchronized void pong() {
14 System.out.print(\"Pong\");
15 }
16 }

What Does It Print?
(a) PingPong
(b) PongPing
(c) It varies

What Does It Print?
(a) PingPong
(b) PongPing
(c) It varies
Not a multithreaded program!

Another Look
01 class PingPong {
02 public static synchronized void main(String[] a) {
03 Thread t = new Thread() {
04 public void run() {
05 pong();
06 }
07 };
08
09 t.run(); // Common typo!
10 System.out.print(\"Ping\");
11 }
12
13 static synchronized void pong() {
14 System.out.print(\"Pong\");
15 }
16 }

How Do You Fix It?
01 class PingPong {
02 public static synchronized void main(String[] a) {
03 Thread t = new Thread() {
04 public void run() {
05 pong();
06 }
07 };
08
09 t.start();
10 System.out.print(\"Ping\");
11 }
12
13 static synchronized void pong() {
14 System.out.print(\"Pong\");
15 }
16 }

The Moral
• Invoke Thread.start, not
Thread.run
─ Common error
─ Can be very difficult to diagnose
• (Thread shouldn’t implement
Runnable)

14. “Shifty”
01 public class Shifty {
02 public static void main(String[] args) {
03 int distance = 0;
04 while ((-1 << distance) != 0)
05 distance++;
06 System.out.println(distance);
07 }
08 }

What Does It Print?
(a) 31
(b) 32
(c) 33
(d) None of the above

What Does It Print?
(a) 31
(b) 32
(c) 33
(d) None of the above: infinite loop!
Shift distances are calculated mod 32.

Another Look
01 public class Shifty {
02 public static void main(String[] args) {
03 int distance = 0;
04 while ((-1 << distance) != 0)
05 distance++;
06 System.out.println(distance);
07 }
08 }
09
10 // (-1 << 32) == -1

How Do You Fix It?
01 public class Shifty {
02 public static void main(String[] args) {
03 int distance = 0;
04 for (int val = -1; val != 0; val <<= 1)
05 distance++;
06 System.out.println(distance);
07 }
08 }

The Moral
• Shift distances are computed mod
32 (or 64)
• It’s impossible to shift out an entire
int
(or long) using any shift operator or
distance
• Use care when shift distance is
not a literal

15. “Line Printer”
01 public class LinePrinter {
02 public static void main(String[] args) {
03 // Note: \\u000A is Unicode representation for
newline
04 char c = 0x000A;
05 System.out.println(c);
06 }
07 }

What Does It Print?
(a) Two blank lines
(b) 10
(c) Won’t compile
(d) It varies

What Does It Print?
(a) Two blank lines
(b) 10
(c) Won’t compile: Syntax error!
(d) It varies
The Unicode escape in the comment
breaks
it in two. The second half is garbage.

Another Look are processed before comments!
01 // Unicode escapes
02 public class LinePrinter {
03 public static void main(String[] args) {
04 // Note: \\u000A is unicode representation for
newline
05 char c = 0x000A;
06 System.out.println(c);
07 }
08 }
01 // This is what the parser sees
02 public class LinePrinter {
03 public static void main(String[] args) {
04 // Note:
05 is Unicode representation for newline
06 char c = 0x000A;
07 System.out.println(c);
08 }
09 }

How Do You Fix It?
01 public class LinePrinter {
02 public static void main(String[] args) {
03 // Escape sequences (like \\n) are fine in comments
04 char c = '\\n';
05 System.out.println(c);
06 }
07 }

The Moral
• Unicode escapes are dangerous
─Equivalent to the character they
represent!
• Use escape sequences instead, if
possible
• If you must use Unicode escapes, use
with care
─ \\u000A (newline) can break string
literals,
char literals, and single-line comments
─\\u0022 (\") can terminate string literals

16. “All Strung Out”
01 public class Puzzling {
02 public static void main(String[] args) {
03 String s = new String(\"blah\");
04 System.out.println(s);
05 }
06 }
07 class String {
08 java.lang.String s;
09
10 public String(java.lang.String s) {
11 this.s = s;
12 }
13 public java.lang.String toString() {
14 return s;
15 }
16 }

What Does It Print?
(a) Won’t compile
(b) blah
(c) Throws an exception at runtime
(d) Other

What Does It Print?
(a) Won’t compile
(b) blah
(c) Throws an exception at runtime
(d) Other
NoSuchMethodError is thrown
because the Puzzling class is missing
a main method.

Another Look
01 public class Puzzling {
02 public static void main(String[] args) {
03 String s = new String(\"blah\");
04 System.out.println(s);
05 }
06 }
07 class String {
08 java.lang.String s;
09
10 public String(java.lang.String s) {
11 this.s = s;
12 }
13 public java.lang.String toString() {
14 return s;
15 }
16 }

How Do You Fix It?
01 public class Puzzling {
02 public static void main(String[] args) {
03 MyString s = new MyString(\"blah\");
04 System.out.println(s);
05 }
06 }
07 class MyString {
08 String s;
09
10 public MyString(String s) {
11 this.s = s;
12 }
13 public String toString() {
14 return s;
15 }
16 }

The Moral
• Avoid name reuse in all its guises
─ hiding, shadowing, overloading
• Don’t even think about reusing
platform
class names!

17. “Reflection Infection”
01 import java.lang.reflect.*;
02
03 public class Reflector {
04 public static void main(String[] args) throws Exception {
05 Set s = new HashSet();
06 s.add(\"foo\");
07 Iterator i = s.iterator();
08 Method m =
09 i.getClass().getMethod(\"hasNext\", new Class[0]);
10 System.out.println(m.invoke(i, new Object[0]));
11 }
12 }

What Does It Print?
(a) Won’t compile
(b) true
(c) Throws exception
(d) None of the above

What Does It Print?
(a) Won’t compile
(b) true
(c) Throws exception -
IllegalAccessError
(d) None of the above
Attempts to invoke a method on a
private class

Another Look
01 import java.lang.reflect.*;
02
03 public class Reflector {
04 public static void main(String[] args) throws Exception {
05 Set s = new HashSet();
06 s.add(\"foo\");
07 Iterator i = s.iterator();
08 Method m =
09 i.getClass().getMethod(\"hasNext\", new Class[0]);
10 System.out.println(m.invoke(i, new Object[0]));
11 }
12 }

How Do You Fix It?
01 import java.lang.reflect.*;
02
03 public class Reflector {
04 public static void main(String[] args) throws Exception {
05 Set s = new HashSet();
06 s.add(\"foo\");
07 Iterator i = s.iterator();
08 Method m =
09 Iterator.class.getMethod(\"hasNext\",
10 new Class[0]);
11 System.out.println(m.invoke(i, new Object[0]));
12 }
13 }

The Moral
• Reflection has its own access rules
• Avoid reflection when possible
• If you must use reflection
─ Instantiate using reflection
─ Cast to an interface type
─ Access via interface
• Avoid extralinguistic mechanisms

18. “String Cheese”
01 public class StringCheese {
02 public static void main(String args[]) {
03 byte b[] = new byte[256];
04 for(int i = 0; i < 256; i++)
05 b[i] = (byte)i;
06 String str = new String(b);
07 for(int i = 0; i < str.length(); i++)
08 System.out.print((int)str.charAt(i) + \" \");
09 }
10 }

What Does It Print?
(a) The numbers from 0 to 255
(b) The numbers from 0 to 127 then
-128 to -1
(c) It varies
(d) None of the above

What Does It Print?
(a) The numbers from 0 to 255
(b) The numbers from 0 to 127 then
-128 to -1
(c) It varies*
(d) None of the above
The sequence depends on the default
charset,which depends on OS and
locale.
* 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103
104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 8364
65533 8218 402 8222 8230 8224 8225 710 8240 352 8249 338 65533 381 65533 65533 8216 8217 8220 8221
8226 8211 8212 732 8482 353 8250 339 65533 382 376 160 161 162 163 164 165 166 167 168 169 170 171 172

Another Look
01 public class StringCheese {
02 public static void main(String args[]) {
03 byte b[] = new byte[256];
04 for(int i = 0; i < 256; i++)
05 b[i] = (byte)i;
06 String str = new String(b);
07 for(int i = 0; i < str.length(); i++)
08 System.out.print((int)str.charAt(i) + \" \");
09 }
10 }
String(byte[] bytes) -
“Constructs a new String by decoding
the specified array of bytes using
the platform’s default
charset.” [from API Spec.]

How Do You Fix It?
If you want it to print numbers from
0-255
in order:
01 public class StringCheese {
02 public static void main(String args[]) {
03 byte b[] = new byte[256];
04 for(int i = 0; i < 256; i++)
05 b[i] = (byte)i;
06 String str = new String(b, \"ISO-8859-1\");
07 for(int i = 0; i < str.length(); i++)
08 System.out.print((int)str.charAt(i) + \" \");
09 }
10 }
ISO-8859-1 indicates the Latin1 charset.

The Moral
• Converting bytes to chars uses a
charset
• If you don’t specify one, you get
default
─ Depends on OS and locale
• If you need predictability, specify a
charset

19. “Elvis Lives!”
01 public class Elvis {
02 public static final Elvis INSTANCE = new Elvis();
03 private final int beltSize;
04
05 private static final int CURRENT_YEAR =
06 Calendar.getInstance().get(Calendar.YEAR);
07
08 private Elvis() { beltSize = CURRENT_YEAR - 1930; }
09 public int beltSize() { return beltSize; }
10
11 public static void main(String[] args) {
12 System.out.println(\"Elvis wears size \" +
13 INSTANCE.beltSize() + \" belt.\");
14 }
15 }

What Does It Print?
(a) Elvis wears size 0 belt.
(b) Elvis wears size 73 belt.
(c) Elvis wears size -1930 belt.
(d) None of the above.

What Does It Print?
(a) Elvis wears size 0 belt.
(b) Elvis wears size 73 belt.
(c) Elvis wears size -1930 belt.
(d) None of the above.
The value of CURRENT_YEAR is used before it
is
initialized, due to circularity in class
initialization.

Another Look
01 // Static initialization proceeds top to bottom.
02 public class Elvis {
03 // Recursive initialization returns immediately!
04 public static final Elvis INSTANCE = new Elvis();
05 private final int beltSize;
06
07 private static final int CURRENT_YEAR =
08 Calendar.getInstance().get(Calendar.YEAR);
09
10 private Elvis() { beltSize = CURRENT_YEAR - 1930; }
11 public int beltSize() { return beltSize; }
12
13 public static void main(String[] args) {
14 System.out.println(\"Elvis wears size \" +
15 INSTANCE.beltSize() + \" belt.\");
16 }
17 }

How Do You Fix It?
01 public class Elvis {
02 private final int beltSize;
03
04 private static final int CURRENT_YEAR =
05 Calendar.getInstance().get(Calendar.YEAR);
06
07 // Make instance after other initialization complete
08 public static final Elvis INSTANCE = new Elvis();
09
10 private Elvis() { beltSize = CURRENT_YEAR - 1930; }
11 public int beltSize() { return beltSize; }
12
13 public static void main(String[] args) {
14 System.out.println(\"Elvis wears size \" +
15 INSTANCE.beltSize() + \" belt.\");
16 }
17 }

The Moral
• Watch out for circularities in static
initialization
One or more classes may be involved
Circularities aren’t necessarily wrong but…
\" Constructors can run before class fully initialized
\" Static fields can be read before they’re initialized
• Several common patterns are susceptible
Singleton (Effective Java, Item 2)
Typesafe Enum (Effective Java, Item 21)
Service Provider Framework (Effective Java,
Item 1)

20. “What’s the Point?”
01 class Point {
02 protected final int x, y;
03 private final String name; // Cached at construction time
04 protected String makeName() { return \"[\" + x + \",\" + y + \"]\"; }
05 public final String toString() { return name; }
06 Point(int x, int y) {
07 this.x = x; this.y = y;
08 this.name = makeName();
09 }
10 }
11
12 public class ColorPoint extends Point {
13 private final String color;
14 protected String makeName() { return super.makeName() + \":\" +
color; }
15 ColorPoint(int x, int y, String color) {
16 super(x, y);
17 this.color = color;
18 }
19 public static void main(String[] args) {
20 System.out.println(new ColorPoint(4, 2, \"purple\"));
21 }
22 }

What Does It Print?
(a) [4,2]:purple
(b) [4,2]:null
(c) Throws exception at runtime
(d) None of the above

What Does It Print?
(a) [4,2]:purple
(b) [4,2]:null
(c) Throws exception at runtime
(d) None of the above
Superclass constructor runs a subclass
method
before the subclass instance is
initialized.

Another Look
01 class Point {
02 protected final int x, y;
03 private final String name;
04 protected String makeName() { return \"[\" + x + \",\" + y + \"]\"; }
05 public final String toString() { return name; }
06 Point(int x, int y) {
07 this.x = x; this.y = y;
08 this.name = makeName(); // (3) Invokes subclass method
09 }
10 }
11
12 public class ColorPoint extends Point {
13 private final String color;
14 // (4) Subclass method executes before subclass constructor body!
15 protected String makeName() { return super.makeName() + \":\" +
color; }
16 ColorPoint(int x, int y, String color) {
17 super(x, y); // (2) Chains to superclass constructor
18 this.color = color; // (5) Initializes blank final instance
field
19 }
20 public static void main(String[] args) { // (1) Invoke subclass cons.
21 System.out.println(new ColorPoint(4, 2, \"purple\"));
22 }
23 }

How Do You Fix It?
01 class Point {
02 protected final int x, y;
03 private String name; // Lazily initialized (cached on first use)
04 protected String makeName() { return \"[\" + x + \",\" + y + \"]\"; }
05 public final synchronized String toString()
06 { return (name == null ? (name = makeName()) : name); }
07 Point(int x, int y) {
08 this.x = x; this.y = y;
09 // (name initialization removed)
10 }
11 }
12
13 public class ColorPoint extends Point {
14 private final String color;
15 protected String makeName() { return super.makeName() + \":\" +
color; }
16 ColorPoint(int x, int y, String color) {
17 super(x, y);
18 this.color = color;
19 }
20 public static void main(String[] args) {
21 System.out.println(new ColorPoint(4, 2, \"purple\"));
22 }
23 }

The Moral
• Never call overridable methods
from constructors, directly or
indirectly
• Also applies to “pseudo-constructors”
─ readObject()
─ clone()
• See Effective Java, Item 15

21. “Long Division”
01 public class LongDivision {
02 private static final long MILLIS_PER_DAY
03 = 24 * 60 * 60 * 1000;
04 private static final long MICROS_PER_DAY
05 = 24 * 60 * 60 * 1000 * 1000;
06
07 public static void main(String[] args) {
08 System.out.println(MICROS_PER_DAY / MILLIS_PER_DAY);
09 }
10 }

What Does It Print?
(a) 5
(b) 1000
(c) 5000
(d) Throws an exception

What Does It Print?
(a) 5
(b) 1000
(c) 5000
(d) Throws an exception
Computation does overflow

Another Look
01 public class LongDivision {
02 private static final long MILLIS_PER_DAY
03 = 24 * 60 * 60 * 1000;
04 private static final long MICROS_PER_DAY
05 = 24 * 60 * 60 * 1000 * 1000; // >> Integer.MAX_VALUE
06
07 public static void main(String[] args) {
08 System.out.println(MICROS_PER_DAY / MILLIS_PER_DAY);
09 }
10 }

How Do You Fix It?
01 public class LongDivision {
02 private static final long MILLIS_PER_DAY
03 = 24L * 60 * 60 * 1000;
04 private static final long MICROS_PER_DAY
05 = 24L * 60 * 60 * 1000 * 1000;
06
07 public static void main(String[] args) {
08 System.out.println(MICROS_PER_DAY / MILLIS_PER_DAY);
09 }
10 }

The Moral
• When working with large numbers watch
out
for overflow—it’s a silent killer
• Just because variable is big enough to hold
result doesn’t mean computation is of
correct type
• When in doubt, use long

22. “No Pain, No Gain”
01 public class Rhymes {
02 private static Random rnd = new Random();
03 public static void main(String[] args) {
04 StringBuffer word = null;
05 switch(rnd.nextInt(2)) {
06 case 1: word = new StringBuffer('P');
07 case 2: word = new StringBuffer('G');
08 default: word = new StringBuffer('M');
09 }
10 word.append('a');
11 word.append('i');
12 word.append('n');
13 System.out.println(word);
14 }
15 }
Thanks to madbot (also known as Mike McCloskey)

What Does It Print?
(a) Pain, Gain, or Main (varies at
random)
(b) Pain or Main (varies at random)
(c) Main (always)
(d) None of the above

What Does It Print?
(a) Pain, Gain, or Main (varies at
random)
(b) Pain or Main (varies at random)
(c) Main (always)
(d) None of the above: ain (always)
The program has three separate bugs.
One of them is quite subtle.

Another Look
01 public class Rhymes {
02 private static Random rnd = new Random();
03 public static void main(String[] args) {
04 StringBuffer word = null;
05 switch(rnd.nextInt(2)) { // No breaks!
06 case 1: word = new StringBuffer('P');
07 case 2: word = new StringBuffer('G');
08 default: word = new StringBuffer('M');
09 }
10 word.append('a');
11 word.append('i');
12 word.append('n');
13 System.out.println(word);
14 }
15 }

How Do You Fix It?
01 public class Rhymes {
02 private static Random rnd = new Random();
03 public static void main(String[] args) {
04 StringBuffer word = null;
05 switch(rnd.nextInt(3)) {
06 case 1: word = new StringBuffer(\"P\");
break;
07 case 2: word = new StringBuffer(\"G\");
break;
08 default: word = new StringBuffer(\"M\");
break;
09 }
10 word.append('a');
11 word.append('i');
12 word.append('n');
13 System.out.println(word);
14 }
15 }

The Moral
• Use common idioms
─ If you must stray, consult the documentation
• Chars are not strings; they’re more like ints
• Always remember breaks in switch
statement
• Watch out for fence-post errors
• Watch out for sneaky puzzlers

23. “The Name Game”
01 public class NameGame {
02 public static void main(String args[]) {
03 Map m = new IdentityHashMap();
04 m.put(\"Mickey\", \"Mouse\");
05 m.put(\"Mickey\", \"Mantle\");
06 System.out.println(m.size());
07 }
08 }

What Does It Print?
(a) 0
(b) 1
(c) 2
(d) It varies

What Does It Print?
(a) 0
(b) 1
(c) 2
(d) It varies
We’re using an IdentityHashMap, but
string
literals are interned (they cancel each
other out)

Another Look
01 public class NameGame {
02 public static void main(String args[]) {
03 Map m = new IdentityHashMap();
04 m.put(\"Mickey\", \"Mouse\");
05 m.put(\"Mickey\", \"Mantle\");
06 System.out.println(m.size());
07 }
08 }

How Do You Fix It?
01 public class NameGame {
02 public static void main(String args[]) {
03 Map m = new HashMap();
04 m.put(\"Mickey\", \"Mouse\");
05 m.put(\"Mickey\", \"Mantle\");
06 System.out.println(m.size());
07 }
08 }

The Moral
• IdentityHashMap not a general-purpose
Map
─ Don’t use it unless you know it’s what
you want
─ Uses identity in place of equality
─ Useful for topology-preserving
transformations
• (String literals are interned)

24. “More of The Same”
01 public class Names {
02 private Map m = new HashMap();
03 public void Names() {
04 m.put(\"Mickey\", \"Mouse\");
05 m.put(\"Mickey\", \"Mantle\");
06 }
07
08 public int size() { return m.size(); }
09
10 public static void main(String args[]) {
11 Names names = new Names();
12 System.out.println(names.size());
13 }
14 }

What Does It Print?
(a) 0
(b) 1
(c) 2
(d) It varies

What Does It Print?
(a) 0
(b) 1
(c) 2
(d) It varies
No programmer-defined constructor

Another Look
01 public class Names {
02 private Map m = new HashMap();
03 public void Names() { // Not a constructor!
04 m.put(\"Mickey\", \"Mouse\");
05 m.put(\"Mickey\", \"Mantle\");
06 }
07
08 public int size() { return m.size(); }
09
10 public static void main(String args[]) {
11 Names names = new Names(); // Invokes
default!
12 System.out.println(names.size());
13 }
14 }

How Do You Fix It?
01 public class Names {
02 private Map m = new HashMap();
03 public Names() { // No return type
04 m.put(\"Mickey\", \"Mouse\");
05 m.put(\"Mickey\", \"Mantle\");
06 }
07
08 public int size() { return m.size(); }
09
10 public static void main(String args[]) {
11 Names names = new Names();
12 System.out.println(names.size());
13 }
14 }

The Moral
• It is possible for a method to have
the same name as a constructor
• Don’t ever do it
• Obey naming conventions
─ field, method(), Class, CONSTANT

25. “Shades of Gray”
01 public class Gray {
02 public static void main(String[] args){
03 System.out.println(X.Y.Z);
04 }
05 }
06
07 class X {
08 static class Y {
09 static String Z = \"Black\";
10 }
11 static C Y = new C();
12 }
13
14 class C {
15 String Z = \"White\";
16 }
Thanks to Prof. Dominik Gruntz, Fachhochschule Aargau

What Does It Print?
(a) Black
(b) White
(c) Won’t compile
(d) None of the above

What Does It Print?
(a) Black
(b) White
(c) Won’t compile
(d) None of the above
Field Y obscures member class Y (JLS
6.3.2)
The rule: variable > type > package

Another Look
01 public class Gray {
02 public static void main(String[] args){
03 System.out.println(X.Y.Z);
04 }
05 }
06
07 class X {
08 static class Y {
09 static String Z = \"Black\";
10 }
11 static C Y = new C();
12 }
13
14 class C {
15 String Z = \"White\";
16 }
The rule: variable > type > package

How Do You Fix It?
01 public class Gray {
02 public static void main(String[] args){
03 System.out.println(Ex.Why.z);
04 }
05 }
06
07 class Ex {
08 static class Why {
09 static String z = \"Black\";
10 }
11 static See y = new See();
12 }
13
14 class See {
15 String z = \"White\";
16 }

The Moral
• Obey naming conventions
─ field, method(), Class, CONSTANT
─ Single-letter uppercase names reserved
for type variables (new in J2SE 1.5)
• Avoid name reuse, except overriding
─ Overloading, shadowing, hiding, obscuring

26. “It’s Elementary”
01 public class Elementary {
02 public static void main(String[] args) {
03 System.out.println(54321 + 5432l);
04 }
05 }

What Does It Print?
(a) -22430
(b) 59753
(c) 10864
(d) 108642

What Does It Print?
(a) -22430
(b) 59753
(c) 10864
(d) 108642
Program doesn’t say what you think it
does!

Another Look
01 public class Elementary {
02 public static void main(String[] args) {
03 System.out.println(54321 + 5432l);
04 }
05 }
1 - the numeral one
l - the lowercase letter el

How Do You Fix It?
We won’t insult your intelligence

The Moral
• Always use uppercase el (L) for long
literals
─ Lowercase el makes the code
unreadable
─ 5432L is clearly a long, 5432l is
misleading
• Never use lowercase el as a variable
name
─ Not this: List l = new ArrayList();
─ But this: List list = new
ArrayList();

27. “Down For The Count”
01 public class Count {
02 public static void main(String[] args) {
03 final int START = 2000000000;
04 int count = 0;
05 for (float f = START; f < START + 50; f++)
06 count++;
07 System.out.println(count);
08 }
09 }

What Does It Print?
(a) 0
(b) 50
(c) 51
(d) None of the above

What Does It Print?
(a) 0
(b) 50
(c) 51
(d) None of the above
The termination test misbehaves due
to floating point “granularity.”

Another Look
01 public class Count {
02 public static void main(String[] args) {
03 final int START = 2000000000;
04 int count = 0;
05 for (float f = START; f < START + 50; f++)
06 count++;
07 System.out.println(count);
08 }
09 }
// (float) START == (float) (START + 50)

How Do You Fix It?
01 public class Count {
02 public static void main(String[] args) {
03 final int START = 2000000000;
04 int count = 0;
05 for (int f = START; f < START + 50; f++)
06 count++;
07 System.out.println(count);
08 }
09 }

The Moral
• Don’t use floating point for loop
indices
• Not every int can be expressed as a
float
• Not every long can be expressed as a
double
• If you must use floating point, use
double
─ unless you’re certain that float provides
enough precision and you have a
compelling performance need (space or

28. “Classy Fire”
01 public class Classifier {
02 public static void main(String[] args) {
03 System.out.println(
04 classify('n') + classify('+') + classify('2'));
05 }
06 static String classify(char ch) {
07 if (\"0123456789\".indexOf(ch) >= 0)
08 return \"NUMERAL \";
09 if (\"abcdefghijklmnopqrstuvwxyz\".indexOf(ch) >= 0)
10 return \"LETTER \";
11 /* (Operators not supported yet)
12 * if (\"+-*/&|!=\".indexOf(ch) >= 0)
13 * return \"OPERATOR \";
14 */
15 return \"UNKNOWN \";
16 }
17 }

What Does It Print?
(a) LETTER OPERATOR NUMERAL
(b) LETTER UNKNOWN NUMERAL
(c) Throws an exception
(d) None of the above

What Does It Print?
(a) LETTER OPERATOR NUMERAL
(b) LETTER UNKNOWN NUMERAL
(c) Throws an exception
(d) None of the above
As for the intuition, you’ll see in a
moment...

Another Look
01 public class Classifier {
02 public static void main(String[] args) {
03 System.out.println(
04 classify('n') + classify('+') + classify('2'));
05 }
06 static String classify(char ch) {
07 if (\"0123456789\".indexOf(ch) >= 0)
08 return \"NUMERAL \";
09 if (\"abcdefghijklmnopqrstuvwxyz\".indexOf(ch) >= 0)
10 return \"LETTER \";
11 /* (Operators not supported yet)
12 * if (\"+-*/&|!=\".indexOf(ch) >= 0)
13 * return \"OPERATOR \";
14 */
15 return \"UNKNOWN \";
16 }
17 }

How Do You Fix It?
01 public class Classifier {
02 public static void main(String[] args) {
03 System.out.println(
04 classify('n') + classify('+') + classify('2'));
05 }
06 static String classify(char ch) {
07 if (\"0123456789\".indexOf(ch) >= 0)
08 return \"NUMERAL \";
09 if (\"abcdefghijklmnopqrstuvwxyz\".indexOf(ch) >= 0)
10 return \"LETTER \";
11 if (false) { // (Operators not supported yet)
12 if (\"+-*/&|!=\".indexOf(ch) >= 0)
13 return \"OPERATOR \";
14 }
15 return \"UNKNOWN \";
16 }
17 }

The Moral
• You cannot reliably block-comment
out code
─Comments do not nest
• Use “if (false)” idiom or “//”
comments

29. “The Joy of Hex”
01 public class JoyOfHex {
02 public static void main(String[] args) {
03 System.out.println(
04 Long.toHexString(0x100000000L +
0xcafebabe));
05 }
06 }

What Does It Print?
(a) cafebabe
(b) 1cafebabe
(c) ffffffffcafebabe
(d) Throws an exception

What Does It Print?
(a) cafebabe
(b) 1cafebabe
(c) ffffffffcafebabe
(d) Throws an exception
0xcafebabe is a negative number

Another Look
01 public class JoyOfHex {
02 public static void main(String[] args) {
03 System.out.println(
04 Long.toHexString(0x100000000L +
0xcafebabe));
05 }
06 }
1111111
0xffffffffcafebabeL
+ 0x0000000100000000L
0x00000000cafebabeL

How Do You Fix It?
01 public class JoyOfHex {
02 public static void main(String[] args) {
03 System.out.println(
04 Long.toHexString(0x100000000L + 0xcafebabeL));
05 }
06 }

The Moral
• Decimal literals are all positive; not
so for hex
> Negative decimal constants have minus
sign
> Hex literals are negative if the high-order
bit is set
• Widening conversion can cause sign
extension
• Mixed-mode arithmetic is tricky—
avoid it

30. “Animal Farm”
01 public class AnimalFarm {
02 public static void main(String[] args) {
03 final String pig = \"length: 10\";
04 final String dog = \"length: \"+pig.length();
05 System.out.println(\"Animals are equal: \"
06 + pig == dog);
07 }
08 }

What Does It Print?
(a) Animals are equal: true
(b) Animals are equal: false
(c) It varies
(d) None of the above

What Does It Print?
(a) Animals are equal: true
(b) Animals are equal: false
(c) It varies
(d) None of the above: false
The + operator binds tighter than ==

Another Look
01 public class AnimalFarm {
02 public static void main(String[] args) {
03 final String pig = \"length: 10\";
04 final String dog = \"length: \"+pig.length();
05 System.out.println(\"Animals are equal: \"
06 + pig == dog);
07 }
08 }
System.out.println(
(\"Animals are equal: \" + pig) == dog);

How Do You Fix It?
01 public class AnimalFarm {
02 public static void main(String[] args) {
03 final String pig = \"length: 10\";
04 final String dog = \"length: \"+pig.length();
05 System.out.println(\"Animals are equal: \"
06 + (pig == dog));
07 }
08 }

The Moral
• Parenthesize when using string
concatenation
• Spacing can be deceptive; parentheses
never lie
• Don’t depend on interning of string
constants
• Use equals, not ==, for strings

31. “A Tricky Assignment”
01 public class Assignment {
02 public static void main(String[] a) throws Exception {
03 int tricky = 0;
04 for (int i = 0; i < 3; i++)
05 tricky += tricky++;
06 System.out.println(tricky);
07 }
08 }

What Does It Print?
(a) 0
(b) 3
(c) 14
(d) None of the above

What Does It Print?
(a) 0
(b) 3
(c) 14
(d) None of the above
Operands are evaluated left to right.
Postfix increment returns old value.

Another Look
01 public class Assignment {
02 public static void main(String[] a) throws Exception {
03 int tricky = 0;
04 for (int i = 0; i < 3; i++)
05 tricky += tricky++;
06 System.out.println(tricky);
07 }
08 }

Another Look
01 public class Assignment {
02 public static void main(String[] a) throws Exception {
03 int tricky = 0;
04 for (int i = 0; i < 3; i++)
05 tricky += tricky++;
06 System.out.println(tricky);
07 }
08 }

Another Look
05 tricky += tricky++;
(0) (tricky == 0)

Another Look
05 tricky += tricky++;
(0) (tricky == 0)

Another Look
05 tricky += tricky++;
0 0 (tricky == 1)

Another Look
05 tricky += tricky++;
0 0 (tricky == 1)

Another Look
05 tricky += tricky++;
0 0 (tricky == 0)

Another Look
01 public class Assignment {
02 public static void main(String[] a) throws Exception {
03 int tricky = 0;
04 for (int i = 0; i < 3; i++)
05 tricky += tricky++;
06 System.out.println(tricky);
07 }
08 }

How Do You Fix It?
01 public class Assignment {
02 public static void main(String[] a) throws Exception {
03 int tricky = 0;
04 for (int i = 0; i < 3; i++) {
05 tricky++;
06 tricky += tricky; // or tricky *= 2;
07 }
08 System.out.println(tricky);
09 }
10 }

The Moral
• Don’t depend on details of expression
evaluation
• Don’t assign to a variable twice in
one expression
• Postfix increment returns old value
• (Operands are evaluated left to right)

32. “Thrown for a Loop”
01 public class Loop {
02 public static void main(String[] args) {
03 int[][] tests = { { 6, 5, 4, 3, 2, 1 }, { 1, 2 },
04 { 1, 2, 3 }, { 1, 2, 3, 4 },
{ 1 } };
05 int successCount = 0;
06 try {
07 int i = 0;
08 while (true) {
09 if (thirdElementIsThree(tests[i++]))
10 successCount++;
11 }
12 } catch (ArrayIndexOutOfBoundsException e) { }
13 System.out.println(successCount);
14 }
15 private static boolean thirdElementIsThree(int[] a) {
16 return a.length >= 3 & a[2] == 3;
17 }
18 }

What Does It Print?
(a) 0
(b) 1
(c) 2
(d) None of the above

What Does It Print?
(a) 0
(b) 1
(c) 2
(d) None of the above
Not only is the program repulsive, but it has a
bug

Another Look
01 public class Loop {
02 public static void main(String[] args) {
03 int[][] tests = { { 6, 5, 4, 3, 2, 1 }, { 1, 2 },
04 { 1, 2, 3 }, { 1, 2, 3, 4 },
{ 1 } };
05 int successCount = 0;
06 try {
07 int i = 0;
08 while (true) {
09 if (thirdElementIsThree(tests[i++]))
10 successCount++;
11 }
12 } catch (ArrayIndexOutOfBoundsException e) { }
13 System.out.println(successCount);
14 }
15 private static boolean thirdElementIsThree(int[] a) {
16 return a.length >= 3 & a[2] == 3;
17 }
18 }

How Do You Fix It?
01 public class Loop {
02 public static void main(String[] args) {
03 int[][] tests = { { 6, 5, 4, 3, 2, 1 }, { 1, 2 },
04 { 1, 2, 3 }, { 1, 2, 3, 4 }, { 1 } };
05 int successCount = 0;
06 for (int[] test : tests)
07 if (thirdElementIsThree(test))
08 successCount++;
09 System.out.println(successCount);
10 }
11
12 private static boolean thirdElementIsThree(int[] a) {
13 return a.length >= 3 && a[2] == 3;
14 }
15 }

The Moral
• Use exceptions only for exceptional
conditions
> Never use exceptions for normal
control flow
• Beware the logical AND and OR
operators
> Document all intentional uses of & and |
on boolean

33. “Sum Fun”
01 class Cache {
02 static { initIfNecessary(); }
03 private static int sum;
04 public static int getSum() {
05 initIfNecessary();
06 return sum;
07 }
08 private static boolean initialized = false;
09 private static synchronized void initIfNecessary() {
10 if (!initialized) {
11 for (int i = 0; i < 100; i++)
12 sum += i;
13 initialized = true;
14 }
15 }
16 public static void main(String[] args) {
17 System.out.println(getSum());
18 }
19 }

What Does It Print?
(a) 4950
(b) 5050
(c) 9900
(d) None of the above

What Does It Print?
(a) 4950
(b) 5050
(c) 9900
(d) None of the above
Lazy initialization + eager initialization = a
mess

Another Look
01 class Cache {
02 static { initIfNecessary(); }
03 private static int sum;
04 public static int getSum() {
05 initIfNecessary();
06 return sum;
07 }
08 private static boolean initialized = false;
09 private static synchronized void initIfNecessary() {
10 if (!initialized) {
11 for (int i = 0; i < 100; i++)
12 sum += i;
13 initialized = true;
14 }
15 }
16 public static void main(String[] args) {
17 System.out.println(getSum());
18 }
19 }

Another Look
01 class Cache {
02 static { initIfNecessary(); }
03 private static int sum;
04 public static int getSum() {
05 initIfNecessary();
06 return sum;
07 }
08 private static boolean initialized = false; // Ouch!
09 private static synchronized void initIfNecessary() {
10 if (!initialized) {
11 for (int i = 0; i < 100; i++)
12 sum += i;
13 initialized = true;
14 }
15 }
16 public static void main(String[] args) {
17 System.out.println(getSum());
18 }
19 }

How Do You Fix It?
01 class Cache {
02 private static final int SUM = computeSum();
03
04 private static int computeSum() {
05 int result = 0;
06 for (int i = 0; i < 100; i++)
07 result += i;
08 return result;
09 }
10
11 public static int getSum() {
12 return SUM;
13 }
14
15 public static void main(String[] args) {
16 System.out.println(getSum());
17 }
18 }

The Moral
• Use eager or lazy initialization, not
both
> Prefer eager initialization to lazy
• Think about class initialization
• Avoid complex class initialization
sequences

34. “The Mod Squad”
01 public class Mod {
02 public static void main(String[] args) {
03 final int MODULUS = 3;
04 int[] histogram = new int[MODULUS];
05
06 int i = Integer.MIN_VALUE;
07 // This loop iterates over all int values
08 do {
09 histogram[Math.abs(i) % MODULUS]++;
10 } while (i++ != Integer.MAX_VALUE);
11
12 for (int j = 0; j < MODULUS; j++)
13 System.out.print(histogram[j] + \" \");
14 }
15 }

What Does It Print?
(a) 1431655765 1431655765
1431655765
(b) 1431655765 1431655766
1431655765
(c) Throws an exception
(d) None of the above
Hint: 232 / 3 = 1,431,655,765

What Does It Print?
(a) 1431655765 1431655765
1431655765
(b) 1431655765 1431655766
1431655765
(c) Throws an exception: array out of
bounds
(d) None of the above
Math.abs doesn’t always return a nonnegative

Another Look
01 public class Mod {
02 public static void main(String[] args) {
03 final int MODULUS = 3;
04 int[] histogram = new int[MODULUS];
05
06 int i = Integer.MIN_VALUE;
07 // This loop iterates over all int values
08 do {
09 histogram[Math.abs(i) % MODULUS]++;
10 } while (i++ != Integer.MAX_VALUE);
11
12 for (int j = 0; j < MODULUS; j++)
13 System.out.println(histogram[j] + \" \");
14 }
15 }

How Do You Fix It?
Replace:
histogram[Math.abs(i) % MODULUS]++;
With:
histogram[mod(i, MODULUS)]++;
private static int mod(int i, int modulus) {
int result = i % modulus;
return result < 0 ? result + modulus : result;
}

The Moral
• Math.abs can return a negative value
• Two’s-complement integers are
asymmetric
• int arithmetic overflows silently
• i mod m ≠ Math.abs(i) % m

35. “Package Deal”
01 package click;
02 public class CodeTalk {
03 public void doIt() { printMessage(); }
04 void printMessage() { System.out.println(\"Click\"); }
05 }
___________________________________________________________
01 package hack;
02 import click.CodeTalk;
03 public class TypeIt {
04 private static class ClickIt extends CodeTalk {
05 void printMessage()
{ System.out.println(\"Hack\"); }
06 }
07 public static void main(String[] args) {
08 new ClickIt().doIt();
09 }
10 }

What Does It Print?
(a) Click
(b) Hack
(c) Won’t compile
(d) None of the above

What Does It Print?
(a) Click
(b) Hack
(c) Won’t compile
(d) None of the above
There is no overriding in this program

Another Look
01 package click;
02 public class CodeTalk {
03 public void doIt() { printMessage(); }
04 void printMessage() { System.out.println(\"Click\"); }
05 }
___________________________________________________________
01 package hack;
02 import click.CodeTalk;
03 public class TypeIt {
04 private static class ClickIt extends CodeTalk {
05 void printMessage()
{ System.out.println(\"Hack\"); }
06 }
07 public static void main(String[] args) {
08 new ClickIt().doIt();
09 }
10 }

How Do You Fix It?
• If you want overriding
• Make printMessage public or protected
• Use @Override to ensure that you got
overriding

The Moral
• Package-private methods can’t be
overridden by methods outside their
package
• If you can’t see it, you can’t
override it

36. “Lazy Initialization”
01 public class Lazy {
02 private static boolean initialized = false;
03 static {
04 Thread t = new Thread(new Runnable() {
05 public void run() {
06 initialized = true;
07 }
08 });
09 t. start();
10 try {
11 t.join();
12 } catch (InterruptedException e) {
13 throw new AssertionError(e);
14 }
15 }
16 public static void main(String[] args) {
17 System.out.println(initialized);
18 }
19 }

What Does It Print?
(a) true
(b) false
(c) It varies
(d) None of the above

What Does It Print?
(a) true
(b) false
(c) It varies
(d) None of the above: it deadlocks
Intuition: You wouldn’t believe us if we told
you.

Another Look
01 public class Lazy {
02 private static boolean initialized = false;
03 static {
04 Thread t = new Thread(new Runnable() {
05 public void run() {
06 initialized = true; // Deadlocks here!
07 }
08 });
09 t. start();
10 try {
11 t.join();
12 } catch (InterruptedException e) {
13 throw new AssertionError(e);
14 }
15 }
16 public static void main(String[] args) {
17 System.out.println(initialized);
18 }
19 }

How Do You Fix It?
• Don’t use background threads in class
initialization
> If it hurts when you go like that, don’t go
like that!

The Moral
• Never use background threads in class
initialization
• Keep class initialization simple
• Don’t code like my brother

37. “Odd Behavior”
01 public class OddBehavior {
02 public static void main(String[] args) {
03 List<Integer> list = Arrays.asList(-2, -1, 0, 1, 2);
04
05 boolean foundOdd = false;
06 for (Iterator<Integer> it=list.iterator();it.hasNext(); )
07 foundOdd = foundOdd || isOdd(it.next());
08
09 System.out.println(foundOdd);
10 }
11
12 private static boolean isOdd(int i) {
13 return (i & 1) != 0;
14 }
15 }

What Does It Print?
01 public class OddBehavior {
02 public static void main(String[] args) {
03 List<Integer> list = Arrays.asList(-2, -1, 0, 1, 2);
04
05 boolean foundOdd = false;
06 for (Iterator<Integer> it=list.iterator(); it.hasNext(); )
07 foundOdd = foundOdd || isOdd(it.next());
08
09 System.out.println(foundOdd);
10 }
11
12 private static boolean isOdd(int i) {
13 return (i & 1) != 0; (a) true
14 }
(b) false
15 }
(c) Throws exception
(d) None of the above

What Does It Print?
(a) true
(b) false
(c) Throws exception
(d) None of the above: Nothing—
Infinite loop
Conditional OR operator (||) short-
circuits iterator

Another Look
public class OddBehavior {
public static void main(String[] args) {
List<Integer> list = Arrays.asList(-2, -1, 0, 1, 2);
boolean foundOdd = false;
for (Iterator<Integer> it = list.iterator(); it.hasNext();
)
foundOdd = foundOdd || isOdd(it.next());
System.out.println(foundOdd);
}
private static boolean isOdd(int i) {
return (i & 1) != 0;
}
}

You Could Fix it Like This….
public class OddBehavior {
public static void main(String[] args) {
List<Integer> list = Arrays.asList(-2, -1, 0, 1, 2);
boolean foundOdd = false;
for (int i : list)
foundOdd = foundOdd || isOdd(i);
System.out.println(foundOdd);
}
private static boolean isOdd(int i) {
return (i & 1) != 0;
}
}

…But This Is Even Better
public class OddBehavior {
public static void main(String[] args) {
List<Integer> list = Arrays.asList(-2, -1, 0, 1, 2);
System.out.println(containsOdd(list));
}
private static boolean containsOdd(List<Integer> list) {
for (int i : list)
if (isOdd(i))
return true;
return false;
}
private static boolean isOdd(int i) {
return (i & 1) != 0;
}
}

The Moral
• Use for-each wherever possible
> Nicer and safer than explicit iterator or
index usage
• If you must use an iterator, make
sure you call next() exactly once
• Conditional operators evaluate their
right operand only if necessary to
determine result
> This is almost always what you want

38. “Set List”
public class SetList {
public static void main(String[] args) {
Set<Integer> set = new LinkedHashSet<Integer>();
List<Integer> list = new ArrayList<Integer>();
for (int i = -3; i < 3; i++) {
set.add(i);
list.add(i);
}
for (int i = 0; i < 3; i++) {
set.remove(i);
list.remove(i);
}
System.out.println(set + \" \" + list);
}
}

(a) [-3, -2, -1] [-3, -2, -1]
What Does It Print? (b) [-3, -2, -1] [-2, 0, 2]
(c) Throws exception
(d) None of the above
public class SetList {
public static void main(String[] args) {
Set<Integer> set = new LinkedHashSet<Integer>();
List<Integer> list = new ArrayList<Integer>();
for (int i = -3; i < 3; i++) {
set.add(i);
list.add(i);
}
for (int i = 0; i < 3; i++) {
set.remove(i);
list.remove(i);
}
System.out.println(set + \" \" + list);
}
}

What Does It Print?
(a) [-3, -2, -1] [-3, -2, -1]
(b) [-3, -2, -1] [-2, 0, 2]
(c) Throws exception
(d) None of the above
Autoboxing + overloading = confusion

Another Look
public class SetList {
public static void main(String[] args) {
Set<Integer> set = new LinkedHashSet<Integer>();
List<Integer> list = new ArrayList<Integer>();
for (int i = -3; i < 3; i++) {
set.add(i);
list.add(i);
}
for (int i = 0; i < 3; i++) {
set.remove(i);
list.remove(i); // List.remove(int)
}
System.out.println(set + \" \" + list);
}
}

How Do You Fix It?
public class SetList {
public static void main(String[] args) {
Set<Integer> set = new LinkedHashSet<Integer>();
List<Integer> list = new ArrayList<Integer>();
for (int i = -3; i < 3; i++) {
set.add(i);
list.add(i);
}
for (int i = 0; i < 3; i++) {
set.remove(i);
list.remove((Integer) i);
}
System.out.println(set + \" \" + list);
}
}

The Moral
• Avoid ambiguous overloadings
• Harder to avoid in release 5.0
> Autoboxing, varargs, generics
• Design new APIs with this in mind
> Old rules no longer suffice
• Luckily, few existing APIs were
compromised
> Beware List<Integer>

39. “Powers of Ten”
public enum PowerOfTen {
ONE(1), TEN(10),
HUNDRED(100) {
@Override public String toString() {
return Integer.toString(val);
}
};
private final int val;
PowerOfTen(int val) { this.val = val; }
@Override public String toString() {
return name().toLowerCase();
}
public static void main(String[] args) {
System.out.println(ONE + \" \" + TEN + \" \" + HUNDRED);
}
}

What Does It Print? (a) ONE TEN HUNDRED
(b) one ten hundred
public enum PowerOfTen { (c) one ten 100
ONE(1), TEN(10),
(d) None of the above
HUNDRED(100) {
@Override public String toString() {
return Integer.toString(val);
}
};
private final int val;
PowerOfTen(int val) { this.val = val; }
@Override public String toString() {
return name().toLowerCase();
}
public static void main(String[] args) {
System.out.println(ONE + \" \" + TEN + \" \" + HUNDRED);
}
}

What Does It Print?
(a) ONE TEN HUNDRED
(b) one ten hundred
(c) one ten 100
(d) None of the above: Won’t compile
Non-static variable val can’t be referenced from static context
return Integer.toString(val);
^
Private members are never inherited

Another Look
public enum PowerOfTen {
ONE(1), TEN(10),
HUNDRED(100) { // Creates static anonymous class
@Override public String toString() {
return Integer.toString(val);
}
};
private final int val;
PowerOfTen(int val) { this.val = val; }
@Override public String toString() {
return name().toLowerCase();
}
public static void main(String[] args) {
System.out.println(ONE + \" \" + TEN + \" \" + HUNDRED);
}
}

How Do You Fix It?
public enum PowerOfTen {
ONE(1), TEN(10),
HUNDRED(100) {
@Override public String toString() {
return Integer.toString(super.val);
}
};
private final int val;
PowerOfTen(int val) { this.val = val; }
@Override public String toString() {
return name().toLowerCase();
}
public static void main(String[] args) {
System.out.println(ONE + \" \" + TEN + \" \" + HUNDRED);
}
}

The Moral
• Nest-mates can use each others’ private
members
• But private members are never inherited
• Constant-specific enum bodies define static
anonymous classes
• Compiler diagnostics can be confusing

40. “Testy Behavior”
import java.lang.reflect.*;
@interface Test { }
public class Testy {
@Test public static void test() { return; }
@Test public static void test2() { new RuntimeException(); }
public static void main(String[] args) throws Exception {
for (Method m : Testy.class.getDeclaredMethods()) {
if (m.isAnnotationPresent(Test.class)) {
try {
m.invoke(null);
System.out.print(\"Pass \");
} catch (Throwable ex) {
System.out.print(\"Fail \");
}
}
}
}
}

What Does It Print? (a) Pass Fail
(b) Pass Pass
import java.lang.reflect.*; (c) It varies
(d) None of the above
@interface Test { }
public class Testy {
@Test public static void test() { return; }
@Test public static void test2() { new RuntimeException(); }
public static void main(String[] args) throws Exception {
for (Method m : Testy.class.getDeclaredMethods()) {
if (m.isAnnotationPresent(Test.class)) {
try {
m.invoke(null);
System.out.print(\"Pass \");
} catch (Throwable ex) {
System.out.print(\"Fail \");
}
}
}
}
}

What Does It Print?
(a) Pass Fail
(b) Pass Pass
(c) It varies
(d) None of the above: In fact,
nothing!
The program contains two bugs, both
subtle

Another Look
import java.lang.reflect.*;
@interface Test { } // By default, annotations are discarded at runtime
public class Testy {
@Test public static void test() { return; }
@Test public static void test2() { new RuntimeException(); } // Oops !
public static void main(String[] args) throws Exception {
for (Method m : Testy.class.getDeclaredMethods()) {
if (m.isAnnotationPresent(Test.class)) {
try {
m.invoke(null);
System.out.print(\"Pass\");
} catch (Throwable ex) {
System.out.print(\"Fail \");
}
}
}
}
}

How Do You Fix It?
import java.lang.reflect.*;
import java.lang.annotation.*;
@Retention(RetentionPolicy.RUNTIME) @interface Test { }
public class Testy {
@Test public static void test() { return; }
@Test public static void test2() { throw new RuntimeException(); }
public static void main(String[] args) throws Exception {
for (Method m : Testy.class.getDeclaredMethods()) {
if (m.isAnnotationPresent(Test.class)) {
try {
m.invoke(null);
System.out.print(\"Pass \");
} catch (Throwable ex) {
System.out.print(\"Fail \");
}
}
}
}
}

The Moral
• By default, annotations are discarded
at runtime
> If you need annotations at runtime, use
@Retention(RetentionPolicy.RUNTIME
)
> If you want them omitted from class file,
use
@Retention(RetentionPolicy.SOURCE)
• No guarantee on order of reflected
entities

41. “What the Bleep?”
public class Bleep {
String name = \"Bleep\";
void setName(String name) {
this.name = name;
}
void backgroundSetName() throws InterruptedException {
Thread t = new Thread() {
@Override public void run() { setName(\"Blat\"); }
};
t.start();
t.join();
System.out.println(name);
}
public static void main(String[] args) throws InterruptedException
{
new Bleep().backgroundSetName();
}
}

What Does It Print? (a) Bleep
(b) Blat
public class Bleep {
String name = \"Bleep\";
(c) It varies
void setName(String name) {
(d) None of the above
this.name = name;
}
void backgroundSetName() throws InterruptedException {
Thread t = new Thread() {
@Override public void run() { setName(\"Blat\"); }
};
t.start();
t.join();
System.out.println(name);
}
public static void main(String[] args) throws InterruptedException
{
new Bleep().backgroundSetName();
}
}

What Does It Print?
(a) Bleep
(b) Blat
(c) It varies
(d) None of the above
Bleep.setName isn’t getting called

Another Look
public class Bleep {
String name = \"Bleep\";
void setName(String name) { // Does this look familiar?
this.name = name;
}
void backgroundSetName() throws InterruptedException {
Thread t = new Thread() {
// Invokes Thread.setName (shadowing)
@Override public void run() { setName(\"Blat\"); }
};
t.start();
t.join();
System.out.println(name);
}
public static void main(String[] args) throws InterruptedException
{
new Bleep().backgroundSetName();
}
}

How Do You Fix It?
public class Bleep {
String name = \"Bleep\";
void setName(String name) {
this.name = name;
}
void backgroundSetName() throws InterruptedException {
Thread t = new Thread(new Runnable() {
public void run() { setName(\"Blat\"); }
});
t.start();
t.join();
System.out.println(name);
}
public static void main(String[] args) throws InterruptedException
{
new Bleep().backgroundSetName();
}
}

The Moral
• Don’t extend Thread
> Use new Thread(Runnable) instead
• Often the Executor Framework is
better still
> Much more flexible
> See java.util.concurrent for more
information
• Beware of shadowing

42. “Beyond Compare”
public class BeyondCompare {
public static void main(String[] args) {
Object o = new Integer(3);
System.out.println(new Double(3).compareTo(o) == 0);
}
}

What Does It Print?
public class BeyondCompare {
public static void main(String[] args) {
Object o = new Integer(3);
System.out.println(new Double(3).compareTo(o) == 0);
}
}
(a) true
(b) false
(c) Throws exception
(d) None of the above

What Does It Print?
(a) true
(b) false
(c) Throws exception
(d) None of the above: Won’t compile (it did in 1.4)
compareTo(Double) in Double cannot be applied to (Object)
System.out.println(new Double(3).compareTo(o) == 0);
^
The Comparable interface was generified in 5.0

Another Look
public class BeyondCompare {
public static void main(String[] args) {
Object o = new Integer(3);
System.out.println(new Double(3).compareTo(o) == 0);
}
}
// Interface Comparable was generified in release 5.0
public interface Comparable<T> {
int compareTo(T t); // Was Object
}
public class Double extends Number
implements Comparable<Double>

How Do You Fix It?
// Preserves 1.4 semantics
public class BeyondCompare {
public static void main(String[] args) {
Object o = new Integer(3);
System.out.println(
new Double(3).compareTo((Double) o) == 0);
}
}
// Fixes the underlying problem
public class BeyondCompare {
public static void main(String[] args) {
Double d = 3.0;
System.out.println(Double.valueOf(3).compareTo(d) == 0);
}
}

The Moral
• Binary compatibility is preserved at all costs
• Source compatibility broken for good cause
(rare)
• Comparable<T> alerts you to errors at compile
time
• Take compiler diagnostics seriously
> Often there is an underlying problem

43. “Fib O’Nacci”
public class Fibonacci {
private static final int LENGTH = 7;
public static void main(String[] args) {
int[] fib = new int[LENGTH];
fib[0] = fib[1] = 1; // First 2 Fibonacci numbers
for (int i = 2; i < LENGTH; i++)
fib[i] = fib[i - 2] + fib[i - 1];
System.out.println(Arrays.asList(fib));
}
}

What Does It Print?
public class Fibonacci {
private static final int LENGTH = 7;
public static void main(String[] args) {
int[] fib = new int[LENGTH];
fib[0] = fib[1] = 1; // First 2 Fibonacci numbers
for (int i = 2; i < LENGTH; i++)
fib[i] = fib[i - 2] + fib[i - 1];
System.out.println(Arrays.asList(fib));
}
}
(a) [1, 1, 2, 3, 5, 8, 13]
(b) Throws exception
(c) It varies
(d) None of the above

What Does It Print?
(a) [1, 1, 2, 3, 5, 8, 13]
(b) Throws exception
(c) It varies: Depends on hashcode
[[I@ad3ba4]
(d) None of the above
Arrays.asList only works on arrays of
object refs

Another Look
public class Fibonacci {
private static final int LENGTH = 7;
public static void main(String[] args) {
int[] fib = new int[LENGTH];
fib[0] = fib[1] = 1; // First 2 Fibonacci numbers
for (int i = 2; i < LENGTH; i++)
fib[i] = fib[i - 2] + fib[i - 1];
// Idiom only works for arrays of object references
System.out.println(Arrays.asList(fib));
}
}

How Do You Fix It?
public class Fibonacci {
private static final int LENGTH = 7;
public static void main(String[] args) {
int[] fib = new int[LENGTH];
fib[0] = fib[1] = 1; // First 2 Fibonacci numbers
for (int i = 2; i < LENGTH; i++)
fib[i] = fib[i - 2] + fib[i - 1];
System.out.println(Arrays.toString(fib));
}
}

The Moral
• Use varargs sparingly in your APIs
> It can hide errors and cause confusion
> This program wouldn't compile under 1.4
• Arrays.asList printing idiom is
obsolete
> use Arrays.toString instead
> Prettier, safer, and more powerful
• A full complement of array utilities
added in 5.0
• equals, hashCode, toString for all array

44. “Parsing Is Such Sweet Sorrow”
public class Parsing {
/**
* Returns Integer corresponding to s, or null if s is null.
* @throws NumberFormatException if s is nonnull and
* doesn't represent a valid integer
*/
public static Integer parseInt(String s) {
return (s == null) ?
(Integer) null : Integer.parseInt(s);
}
public static void main(String[] args) {
System.out.println(parseInt(\"-1\") + \" \" +
parseInt(null) + \" \" +
parseInt(\"1\"));
}
}

(a) -1 null 1
What Does It Print? (b) -1 0 1
(c) Throws exception
public class Parsing {
(d) None of the above
/**
* Returns Integer corresponding to s, or null if s is null.
* @throws NumberFormatException if s is nonnull and
* doesn't represent a valid integer
*/
public static Integer parseInt(String s) {
return (s == null) ?
(Integer) null : Integer.parseInt(s);
}
public static void main(String[] args) {
System.out.println(parseInt(\"-1\") + \" \" +
parseInt(null) + \" \" +
parseInt(\"1\"));
}
}

What Does It Print?
(a) -1 null 1
(b) -1 0 1
(c) Throws exception:
NullPointerException
(d) None of the above
Program attempts to auto-unbox null

Another Look
public class Parsing {
/**
* Returns Integer corresponding to s, or null if s is null.
* @throws NumberFormatException if s is nonnull and
* doesn't represent a valid integer.
*/
public static Integer parseInt(String s) {
return (s == null) ? // Mixed-type computation: Integer and int
(Integer) null : Integer.parseInt(s);
}
public static void main(String[] args) {
System.out.println(parseInt(\"-1\") + \" \" +
parseInt(null) + \" \" +
parseInt(\"1\"));
}
}

How Do You Fix It?
public class Parsing {
/**
* Returns Integer corresponding to s, or null if s is null.
* @throws NumberFormatException if s is nonnull and
* doesn't represent a valid integer.
*/
public static Integer parseInt(String s) {
return (s == null) ? null : Integer.valueOf(s);
}
public static void main(String[] args) {
System.out.println(parseInt(\"-1\") + \" \" +
parseInt(null) + \" \" +
parseInt(\"1\"));
}
}

The Moral
• Mixed-type computations are
confusing
• Especially true for ?: expressions
• Avoid null where possible
• Auto-unboxing and null are a
dangerous mix

Resources and Summary

Resources
• Send more puzzles
puzzlers@javapuzzles.com
–

Conclusion
Java platform is simple and elegant
•
But it has a few sharp corners — avoid
–
them!
Keep programs simple
•
Avoid name reuse: overloading, hiding,
–
shadowing
If you aren't sure what a program
•
does,
it probably doesn't do what you
want it to