Programming Methodology if Statements

Programming Methodology &
Abstractions
1
Lecture 8
Decision and Control Statements:
if
CS106
Programming Methodology
& Abstractions
FALL 2005
Balochistan University
of I.T & M.S
Faculty of System Sciences
Sadique Ahmed Bugti

Abstractions
2
Conditional Expressions
 As we have seen, the following logical,
relational and equality operators exist in C:
 These operators are used in conjunction with
decision and control statements.

Abstractions
3
 For example:
heart_rate > 75
performs the necessary comparison and
evaluates to 1 (true) when heart_rate
is over 75; and evaluates to 0 (false)
when heart_rate is not greater than
75.

Abstractions
4
 For example:
marks is in the range 40 to 100, (40  marks 
b) inclusive
(marks >= 40) && (marks <= 100)
performs the necessary comparison and
evaluates to 1 (true) when marks is greater than
or equal to 40 AND marks is less than or equal to
100; and evaluates to 0 (false) when the either
expression is false.

Abstractions
5
Decision and Control Statements
 Decision or control-flow statements specify the
order in which computations are performed.
• they are branching statements
 There are two principle control statements:
•if-else
•switch

Abstractions
6
if-else statement
• The if-else statement is used to carry out
a logical test and then take one of two
possible actions, depending on whether the
outcome of the test is true or false.
• The else portion of the statement is
optional.

Abstractions
7
if statement
if structure (Single Selection)
F
T

Abstractions
8
if statement
 The simplest possible if-else statement takes the
form:
if(expression)
statement;
 The expression is evaluated:
• If expression has a nonzero value (i.e., if
expression if true), statement is executed.
• If expression has a value of zero (i.e., if
expression is false) then the statement will be
ignored.
 The expression must be placed in
parenthesis, as shown.

Abstractions
9
if statement
 For example:
if (heart_rate > 75)
printf(“Heart rate is
normaln”);
if (y != 0.0)
x = x / y;

Abstractions
10
if statement
 Be careful! What happens here?
a = -12;
if (a > 0)
printf(“a is positiven”);
b = sqrt(a);

Abstractions
11
if statement
 Where appropriate, compound statements can
be used to group a series of statements under
the control of a single if expression.
• For example:
if (j < k) if (j < k) {
min = j; min = j;
if (j < k) max =
k;
max = k; }

Abstractions
12
Compound if statement
 Computes the growth rate of a population from
time t1 to time t2.
if (pop_t2 > pop_t1)
{
growth = pop_t2 – pop_t1;
growth_pct = (growth/pop_t1) *
100;
printf(“The growth % is %.2fn”,
growth_pct);
}

Abstractions
13
if-else statement
 The if-else statement is an extension of if used in
situations where there are two alternatives:
if(expression)
statement1;
else
statement2;
 The expression is evaluated:
• If expression has a nonzero value (i.e., if expression
is true), statement1 is executed.
• If the expression has a value of zero (i.e., if
expression is false) and there is an else part,
statement2 is executed.

Abstractions
14
if-else statement
if else structure
(double selection)
TF

Abstractions
15
if-else statement
 For example:
if (marks >=40)
printf(“Pass);
else
printf(“Fail”);
 If marks >= y is true, then “Pass” will be
printed; if it is false, “Fail” will be printed.

Abstractions
16
if-else statement
 For example:
if (ch >= ‘a’ && ch <= ‘z’)
++lower_char;
else
++other_char;
 If ch >= ‘a’ && ch <= ‘z’ (c is a
lowercase character) is true, the variable
lower_char is incremented; if it is false,
other_char is incremented.

Abstractions
17
Example: Finding the Minimum Value
/* Find the minimum of two values */
#include <stdio.h>
void main(void)
{
int x, y, min;
printf(“Input two integers: “);
scanf(“%d%d”, &x, &y);
if (x < y)
min = x;
else
min = y;
printf(“The minimum value is %dn”, min);
}

Abstractions
18
if-else statement
 An example of a syntax error:
if (a != b){
a = a + 1;
b = b + 1;
};
else
c = a + b;
 The syntax error occurs because the semicolon
following the right brace creates an empty
statement, and consequently the else has
nowhere to attach.

Abstractions
19
The “dangling-else” Problem
 There is an ambiguity when an else is
omitted from a nested if statement:
“dangling else” problem
if (marks>= 40)
if (marks > 90)
printf(“Excellent”);
else
printf(“Failed”);
 This is resolved by associating the else with
the closest previous else-less if.

Abstractions
20
The “dangling-else” Problem
 If that’s not what you want, use braces to
force the proper association.
if (marks>=40{
if (marks >90)
printf(“Excellent”);
}
else
printf(“Fail”);
 Use braces where there are nested ifs.

Abstractions
21
Nested if statement
 Sometimes the if-else statement is
used for a multi-way decision.
 The expressions are evaluated in order;
if any expression is true, the statement
associated with it is executed, and this
terminates the whole chain.
 The last else can be used to handle the
default case where none of the other
conditions is satisfied.

Abstractions
22
Nested if statement
if (expression1)
statement1;
else if (expression2)
statement2;
:
:
else if (expressionN)
statementN;
else
default_statement;

Abstractions
23
Nested if statement
 For example
if (marks >= 80)
printf(“Grade A”);
else if (marks >=70)
printf(“Grade B”);
else if(marks>=50)
printf(“Grade C”)
else /* marks less than 50 */
printf(“Fail”);

Abstractions
24
Example: Quadratic Equation
 A simple C program that calculates the
real roots of a quadratic equation;
ax+bx+c=0
using the quadratic formula.

Abstractions
25
Example: Quadratic Equation (quadratic.c)
#include <stdio.h>
#include <math.h>
void main()
{
double a, b, c, d, x1, x2;
/* Read input data */
printf("na = ");
scanf("%lf", &a);
printf("b = ");
scanf("%lf", &b);
printf("c = ");
scanf("%lf", &c);
/* Perform calculation */
d = sqrt(b * b - 4. * a * c);
x1 = (-b + d) / (2 * a);
x2 = (-b - d) / (2 * a);
/* Display output */
printf("nx1 = %12.3e n x2 = %12.3en", x1, x2);
}

Abstractions
26
 The program quadratic.c, listed
previously, is incapable of dealing
correctly with cases where the
 roots are complex (i.e., b2< 4ac),
 or cases where a = 0.

Abstractions
27
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
void main()
{
double a, b, c, d, e, x1, x2;
/* Read input data */
printf("na = ");
scanf("%lf", &a);
printf("b = ");
scanf("%lf", &b);
printf("c = ");
scanf("%lf", &c);

Abstractions
28
/* Test for complex roots */
e = b * b - 4. * a * c;
if (e < 0.)
{
printf("nError: roots are complexn");
exit(1);
}
/* Test for a = 0. */
if (a == 0.)
{
printf("nError: a = 0.n");
exit(1);
}
d = sqrt(e);
x1 = (-b + d) / (2. * a);
x2 = (-b - d) / (2. * a);
/* Display output */
printf("nx1 = %12.3e x2 = %12.3en", x1, x2);
}

Abstractions
29
 The standard library function call exit(1)
(header file: stdlib.h) causes the program
to abort with an error status.
 Execution of the above program for the case
of complex roots yields the following output:
a = 4
b = 2
c = 6
Error: roots are complex %

Abstractions
30
e = b * b - 4. * a * c;
if (e > 0.) // Test for real roots
{
/* Case of real roots */
d = sqrt(e);
x1 = (-b + d) / (2. * a);
x2 = (-b - d) / (2. * a);
printf("nx1 = %12.3e x2 = %12.3en", x1, x2);
}
else
{
/* Case of complex roots */
d = sqrt(-e);
x1 = -b / (2 * a);
x2 = d / (2 * a);
printf("nx1 = (%12.3e, %12.3e) nx2 = (%12.3e, %12.3e)n",
x1, x2, x1, -x2);
}

Abstractions
31
 The output from the above program for
the case of complex roots looks like:
a = 9
b = 2
c = 2
x1 = ( -1.111e-01, 4.581e-01)
x2 = ( -1.111e-01, -4.581e-01)

Abstractions
32
The End

Programming Methodology if Statements

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