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![Representing numbers (ints)
Fixed, finite number of bits.
bits bytes C/C++ Intel Sun
8 1 char [s]byte byte
16 2 short [s]word half
32 4 int or long [s]dword word
64 8 long long [s]qword xword](https://image.slidesharecdn.com/mbaadmissioninindia-140909020154-phpapp02/75/Mba-admission-in-india-14-2048.jpg)
![Representing numbers (ints)
Fixed, finite number of bits.
bits Intel signed unsigned
8 [s]byte -27..+27-1 0..+28-1
16 [s]word -215..+215-1 0..+216-1
32 [s]dword -231..+231-1 0..+232-1
64 [s]qword -263..+263-1 0..+264-1
In general, for k bits, the unsigned range is [0..+2k-1] and
the signed range is [-2k-1..+2k-1-1].](https://image.slidesharecdn.com/mbaadmissioninindia-140909020154-phpapp02/75/Mba-admission-in-india-15-2048.jpg)
![Representing numbers (ints)
using 2’s complement
Fixed, finite number of bits.
bits Intel signed
8 sbyte -27..+27-1
16 sword -215..+215-1
32 sdword -231..+231-1
64 sqword -263..+263-1
In general, for k bits, the signed range is [-2k-1..+2k-1-1].
So where does the extra negative value come from?](https://image.slidesharecdn.com/mbaadmissioninindia-140909020154-phpapp02/75/Mba-admission-in-india-34-2048.jpg)
![Representing numbers (ints)
Fixed, finite number of bits.
bits Intel signed
8 sbyte -27..+27-1
16 sword -215..+215-1
32 sdword -231..+231-1
64 sqword -263..+263-1
In general, for k bits, the signed range
is
[-2k-1..+2k-1-1].
So where does the extra negative value
come from?
n n
0 0000 0000
1 0001 1111
2 0010 1110
3 0011 1101
4 0100 1100
5 0101 1011
6 0110 1010
7 0111 1001
8 1000](https://image.slidesharecdn.com/mbaadmissioninindia-140909020154-phpapp02/75/Mba-admission-in-india-35-2048.jpg)
![Addition of 2’s complement
binary numbers
class test {
public static void main ( String
args[] )
{
byte A = 127;
byte B = 127;
byte result = (byte)(A + B);
System.out.println( "A + B = "
+ result );
}
}
#include <stdio.h>
int main ( int argc, char* argv[] )
{
char A = 127;
char B = 127;
char result = (char)(A + B);
printf( "A + B = %d n", result );
return 0;
} Result = -2 in both
Java (left) and C++
(right). Why?](https://image.slidesharecdn.com/mbaadmissioninindia-140909020154-phpapp02/75/Mba-admission-in-india-42-2048.jpg)
![Addition of 2’s complement
binary numbers
class test {
public static void main ( String
args[] )
{
byte A = 127;
byte B = 127;
byte result = (byte)(A + B);
System.out.println( "A + B = "
+ result );
}
}
Result = -2 in both Java and
C++.
Why?
What’s 127 as a 2’s
complement binary number?
01111111
01111111
11111110
What is 111111102?
Flip the bits: 00000001.
Then add 1: 00000010.](https://image.slidesharecdn.com/mbaadmissioninindia-140909020154-phpapp02/75/Mba-admission-in-india-43-2048.jpg)
Uploaded byEdhole.com
Mba admission in india
The document discusses binary numbers and arithmetic. It covers topics like addition, subtraction, multiplication in binary, and different methods for representing signed integers like two's complement. It explains how two's complement works by using bitwise operations to represent negative numbers. For example, it shows that adding two positive 8-bit binary numbers in two's complement is simply the bitwise addition, while subtraction can be performed by adding the number and the two's complement of the subtrahend. The document also discusses issues like carry vs overflow that can occur during binary arithmetic operations.
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- 1. MBA admission inIndia By: Admission.edhole.com
- 2. Binary numbers andarithmetic Admission.edhole.com
- 3.
- 4. Addition (decimal) 1 4 5 5 14 19 5 1 5 10 6 1 5 11 1 1 12 99 111 Admission.edhole.com
- 5. Addition (binary) 0 0 0 1 0 1 0 1 1 1 1 1 10 Admission.edhole.com
- 6. Addition (binary) 11 1 1 01101 01011 11000 Admission.edhole.com
- 7. Addition (binary) 0 0 1 0 So ca0n we count in 1 binary? 0 1 1 1 1 1 10 Admission.edhole.com
- 8. Counting in binary(4 bits) 0 0000 1 0001 2 … 3 4 5 6 7 8 9 10 11 12 13 14 15 Admission.edhole.com
- 9.
- 10. Multiplication (decimal) 13 11 13 130 143 Admission.edhole.com
- 11. Multiplication (binary) 1101 1011 1101 11010 1101000 10001111 Admission.edhole.com
- 12. Multiplication (binary) 1101 1011 1101 11010 1101000 10001111 It’s interesting to note that binary multiplication is a sequence of shifts and adds of the first term (depending on the bits in the second term. 110100 is missing here because the corresponding bit in the second terms is 0. Admission.edhole.com
- 13. Representing signed (positive and negative) numbers
- 14. Representing numbers (ints) Fixed, finite number of bits. bits bytes C/C++ Intel Sun 8 1 char [s]byte byte 16 2 short [s]word half 32 4 int or long [s]dword word 64 8 long long [s]qword xword
- 15. Representing numbers (ints) Fixed, finite number of bits. bits Intel signed unsigned 8 [s]byte -27..+27-1 0..+28-1 16 [s]word -215..+215-1 0..+216-1 32 [s]dword -231..+231-1 0..+232-1 64 [s]qword -263..+263-1 0..+264-1 In general, for k bits, the unsigned range is [0..+2k-1] and the signed range is [-2k-1..+2k-1-1].
- 16. Methods for representingsigned ints. 1. signed magnitude 2. 1’s complement (diminished radix complement) 3. 2’s complement (radix complement) 4. excess bD-1
- 17. Signed magnitude Ex. 4-bit signed magnitude 1 bit for sign 3 bits for magnitude N N 0 0000 1000 1 0001 1001 2 0010 1010 3 0011 1011 4 0100 1100 5 0101 1101 6 0110 1110 7 0111 1111
- 18. Signed magnitude Ex. 4-bit signed magnitude 1 bit for sign 3 bits for magnitude N N 0 0000 1000 1 0001 1001 2 0010 1010 3 0011 1011 4 0100 1100 5 0101 1101 6 0110 1110 7 0111 1111
- 19. 1’s complement (diminishedradix complement) Let x be a non-negative number. Then –x is represented by bD-1+(-x) where b = base D = (total) # of bits (including the sign bit) Ex. Let b=2 and D=4. Then -1 is represented by 24-1-1 = 1410 or 11102.
- 20. 1’s complement (diminishedradix complement) Let x be a non-negative number. Then –x is represented by bD-1+(-x) where b = base & D = (total) # of bits (including the sign bit) Ex. What is the 9’s complement of 1238910? Given b=10 and D=5. Then the 9’s complement of 12389 = 105 – 1 – 12389 = 100000 – 1 – 12389 = 99999 – 12389 = 87610
- 21. 1’s complement (diminishedradix complement) Let x be a non-negative number. Then –x is represented by bD-1+(-x) where b = base D = (total) # of bits (including the sign bit) Shortcut for base 2? All combinations used, but 2 zeros! N N 0 0000 1111 1 0001 1110 2 0010 1101 3 0011 1100 4 0100 1011 5 0101 1010 6 0110 1001 7 0111 1000
- 22. 2’s complement (radixcomplement) Let x be a non-negative number. Then –x is represented by bD+(-x). Ex. Let b=2 and D=4. Then -1 is represented by 24- 1 = 15 or 11112. Ex. Let b=2 and D=4. Then -5 is represented by 24 – 5 = 11 or 10112. Ex. Let b=10 and D=5. Then the 10’s complement of 12389 = 105 – 12389 = 100000 – 12389 = 87611.
- 23. 2’s complement (radixcomplement) Let x be a non-negative number. Then –x is represented by bD+(-x). Ex. Let b=2 and D=4. Then -1 is represented by 24-1 = 15 or 11112. Ex. Let b=2 and D=4. Then -5 is represented by 24 – 5 = 11 or 10112. Shortcut for base 2? N N 0 0000 0000 1 0001 1111 2 0010 1110 3 0011 1101 4 0100 1100 5 0101 1011 6 0110 1010 7 0111 1001
- 24. 2’s complement (radixcomplement) Shortcut for base 2? Yes! Flip the bits and add 1. N N 0 0000 0000 1 0001 1111 2 0010 1110 3 0011 1101 4 0100 1100 5 0101 1011 6 0110 1010 7 0111 1001
- 25. 2’s complement (radixcomplement) Are all combinations of 4 bits used? No. (Now we only have one zero.) 1000 is missing! What is 1000? Is it positive or negative? Does -8 + 1 = -7 work in 2’s complement? N N 0 0000 0000 1 0001 1111 2 0010 1110 3 0011 1101 4 0100 1100 5 0101 1011 6 0110 1010 7 0111 1001
- 26. excess bD-1 (biased representation) For pos, neg, and 0, x is represented by bD-1 + x Ex. Let b=2 and D=4. Then the excess 8 (24-1) representation for 0 is 8+0 = 8 or 10002. Ex. Let b=2 and D=4. Then excess 8 for -1 is 8 – 1 = 7 or 01112.
- 27. excess bD-1 For pos, neg, and 0, x is represented by bD-1 + x. Ex. Let b=2 and D=4. Then the excess 8 (24-1) representation for 0 is 8+0 = 8 or 10002. Ex. Let b=2 and D=4. Then excess 8 for -1 is 8 – 1 = 7 or 01112. N N 0 1000 1000 1 1001 0111 2 1010 0110 3 1011 0101 4 1100 0100 5 1101 0011 6 1110 0010 7 1111 0001
- 28. 2’s complement vs.excess bD-1 In 2’s, positives start with 0; in excess, positives start with 1. Both have one zero (positive). Remaining bits are the same. N N 0 1000 1000 1 1001 0111 2 1010 0110 3 1011 0101 4 1100 0100 5 1101 0011 6 1110 0010 7 1111 0001
- 29. Summary of methodsfor representing signed ints. signedMag sComp sComp excess 1 2 8 N n n n n n n 0 0000 1000 1111 0000 1000 1000 1 0001 1001 1110 1111 0111 1001 2 0010 1010 1101 1110 0110 1010 3 0011 1011 1100 1101 0101 1011 4 0100 1100 1011 1100 0100 1100 5 0101 1101 1010 1011 0011 1101 6 0110 1110 1001 1010 0010 1110 7 0111 1111 1000 1001 0001 1111 1000=-8| 0000 unused
- 30. Binary arithmetic Signedmagnitude 1’s complement 2’s complement Excess K (biased)
- 31. Binary Arithmetic Signedmagnitude
- 32. Addition w/ signedmagnitude algorithm For A - B, change the sign of B and perform addition of A + (-B) (as in the next step) For A + B: if (Asign==Bsign) then { R = |A| + |B|; Rsign = Asign; } else if (|A|>|B|) then { R = |A| - |B|; Rsign = Asign; } else if (|A|==|B|) then { R = 0; Rsign = 0; } else { R = |B| - |A|; Rsign = Bsign; } Complicated?
- 33. Binary Arithmetic 2’scomplement
- 34. Representing numbers (ints) using 2’s complement Fixed, finite number of bits. bits Intel signed 8 sbyte -27..+27-1 16 sword -215..+215-1 32 sdword -231..+231-1 64 sqword -263..+263-1 In general, for k bits, the signed range is [-2k-1..+2k-1-1]. So where does the extra negative value come from?
- 35. Representing numbers (ints) Fixed, finite number of bits. bits Intel signed 8 sbyte -27..+27-1 16 sword -215..+215-1 32 sdword -231..+231-1 64 sqword -263..+263-1 In general, for k bits, the signed range is [-2k-1..+2k-1-1]. So where does the extra negative value come from? n n 0 0000 0000 1 0001 1111 2 0010 1110 3 0011 1101 4 0100 1100 5 0101 1011 6 0110 1010 7 0111 1001 8 1000
- 36. Addition of 2’scomplement binary numbers Consider 8-bit 2’s complement binary numbers. Then the msb (bit 7) is the sign bit. If this bit is 0, then this is a positive number; if this bit is 1, then this is a negative number. Addition of 2 positive numbers. Ex. 40 + 58 = 98 1 1 1 00101000 00111010 01100010
- 37. Addition of 2’scomplement binary numbers Consider 8-bit 2’s complement binary numbers. Addition of a negative to a positive. What are the values of these 2 terms? -88 and 122 -88 + 122 = 34 1 1 1 1 10101000 01111010 1 00100010
- 38. So how canwe perform subtraction?
- 39. Addition of 2’scomplement binary numbers Consider 8-bit 2’s complement binary numbers. Subtraction is nothing but addition of the 2’s complement. Ex. 58 – 40 = 58 + (-40) = 18 1 1 1 1 00111010 11011000 1 00010010 discard carry
- 40.
- 41. Addition of 2’scomplement binary numbers Carry vs. overflow when adding A + B If A and B are of opposite sign, then overflow cannot occur. If A and B are of the same sign but the result is of the opposite sign, then overflow has occurred (and the answer is therefore incorrect). Overflow occurs iff the carry into the sign bit differs from the carry out of the sign bit.
- 42. Addition of 2’scomplement binary numbers class test { public static void main ( String args[] ) { byte A = 127; byte B = 127; byte result = (byte)(A + B); System.out.println( "A + B = " + result ); } } #include <stdio.h> int main ( int argc, char* argv[] ) { char A = 127; char B = 127; char result = (char)(A + B); printf( "A + B = %d n", result ); return 0; } Result = -2 in both Java (left) and C++ (right). Why?
- 43. Addition of 2’scomplement binary numbers class test { public static void main ( String args[] ) { byte A = 127; byte B = 127; byte result = (byte)(A + B); System.out.println( "A + B = " + result ); } } Result = -2 in both Java and C++. Why? What’s 127 as a 2’s complement binary number? 01111111 01111111 11111110 What is 111111102? Flip the bits: 00000001. Then add 1: 00000010.
- 44. Binary Arithmetic 1’scomplement
- 45. Addition with 1’scomplement Note: 1’s complement has two 0’s! 1’s complement addition is tricky (end-around-carry). N N 0 0000 1111 1 0001 1110 2 0010 1101 3 0011 1100 4 0100 1011 5 0101 1010 6 0110 1001 7 0111 1000
- 46. 8-bit 1’s complementaddition Ex. Let X = A816 and Y = 8616. Calculate Y - X using 1’s complement.
- 47. 8-bit 1’s complementaddition Ex. Let X = A816 and Y = 8616. Calculate Y - X using 1’s complement. Y = 1000 01102 = -12110 X = 1010 10002 = -8710 ~X = 0101 01112 (Note: C=0 out of msb.) 1000 0110 0101 0111 1101 1101 Y - X = -121 + 87 = -34 (base 10)
- 48. 8-bit 1’s complementaddition Ex. Let X = A816 and Y = 8616. Calculate X - Y using 1’s complement.
- 49. 8-bit 1’s complementaddition Ex. Let X = A816 and Y = 8616. Calculate X - Y using 1’s complement. X = 1010 10002 = -8710 Y = 1000 01102 = -12110 ~Y = 0111 10012 (Note: C=1 out of msb.) 1010 1000 0111 1001 1 0010 0001 1 end around carry 0010 0010 X - Y = -87 + 121 = 34 (base 10)
- 50. Binary Arithmetic ExcessK (biased)
- 51. Binary arithmetic andExcess K (biased) Method: Simply add and then flip the sign bit. -1 0111 +5 1101 -- ---- +4 0100 -> flip sign -> 1100 +1 1001 -5 0011 -- ---- -4 1100 -> flip sign -> 0100 +1 1001 +5 1101 -- ---- +6 0110 -> flip sign -> 1110 -1 0111 -5 0011 -- ---- -6 1010 -> toggle sign -> 0010 N N 0 1000 1000 1 1001 0111 2 1010 0110 3 1011 0101 4 1100 0100 5 1101 0011 6 1110 0010 7 1111 0001 (Not used for integer arithmetic but employed in IEEE 754 floating point standard.)