binary numbers and arithmetic. addition addition (decimal)

Binary numbers and arithmetic

ADDITION

Addition (decimal)

5

4

1

19

14

5

10

5

51

11

5

61

111

99

21 11

Addition (binary)

0

0

0

1

0

1

1

1

0

10

1

11

Addition (binary)

11000

01011

101101111

Addition (binary)

So can we count in binary?

0

0

0

1

0

1

1

1

0

10

1

11

Counting in binary (4 bits)0123456789101112131415

00000001…

MULTIPLICATION

Multiplication (decimal)

143

130

13

11

13

Multiplication (binary)

10001111

1101000

11010

1101

1011

1101

Multiplication (binary)

10001111

1101000

11010

1101

1011

1101

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.

REPRESENTING SIGNED (POSITIVE AND NEGATIVE) NUMBERS

Representing numbers (ints)

• Fixed, finite number of bits.

bits bytes C/C++ Intel Sun8 1 char [s]byte byte16 2 short [s]word half32 4 int or long [s]dword word64 8 long long [s]qword xword

Representing numbers (ints)

• Fixed, finite number of bits.

bits Intel signed unsigned8 [s]byte -27..+27-1 0..+28-116 [s]word -215..+215-1 0..+216-132 [s]dword -231..+231-1 0..+232-164 [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].

Methods for representing signed ints.

1. signed magnitude

2. 1’s complement (diminished radix complement)

3. 2’s complement (radix complement)

4. excess bD-1

Signed magnitude

• Ex. 4-bit signed magnitude– 1 bit for sign– 3 bits for magnitude

111101117

111001106

110101015

110001004

101100113

101000102

100100011

100000000

NN

Signed magnitude

• Ex. 4-bit signed magnitude– 1 bit for sign– 3 bits for magnitude

111101117

111001106

110101015

110001004

101100113

101000102

100100011

100000000

NN

1’s complement(diminished radix complement)

• Let x be a non-negative number.• Then –x is represented by bD-1+(-x) where

b = baseD = (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.

1’s complement(diminished radix 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

1’s complement(diminished radix complement)

• Let x be a non-negative number.• Then –x is represented by bD-1+

(-x) whereb = baseD = (total) # of bits (including the

sign bit)

• Shortcut for base 2?– All combinations used, but 2

zeros!100001117

100101106

101001015

101101004

110000113

110100102

111000011

111100000

NN

2’s complement(radix complement)

• 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.

2’s complement(radix complement)

• 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?

100101117

101001106

101101015

110001004

110100113

111000102

111100011

000000000

NN

2’s complement(radix complement)

• Shortcut for base 2?– Yes! Flip the bits and add 1.

100101117

101001106

101101015

110001004

110100113

111000102

111100011

000000000

NN

2’s complement(radix complement)

• 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?

100101117

101001106

101101015

110001004

110100113

111000102

111100011

000000000

NN

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.

excess bD-1

• For pos, neg, and 0, x is represented bybD-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.

000111117

001011106

001111015

010011004

010110113

011010102

011110011

100010000

NN

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.

000111117

001011106

001111015

010011004

010110113

011010102

011110011

100010000

NN

Summary of methods for representing signed ints.

1111000110011000111101117

1110001010101001111001106

1101001110111010110101015

1100010011001011110001004

1011010111011100101100113

1010011011101101101000102

1001011111111110100100011

1000100000001111100000000

821

nnnnnnN

excesssCompsCompsignedMag

1000=-8| 0000 unused

BINARY ARITHMETIC

Signed magnitude1’s complement2’s complementExcess K (biased)

BINARY ARITHMETICSigned magnitude

Addition w/ signed magnitude 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?

BINARY ARITHMETIC2’s complement

Representing numbers (ints) using 2’s complement

• Fixed, finite number of bits.

bits Intel signed8 sbyte -27..+27-116 sword -215..+215-132 sdword -231..+231-164 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?

Representing numbers (ints)

• Fixed, finite number of bits.

bits Intel signed8 sbyte -27..+27-116 sword -215..+215-132 sdword -231..+231-164 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?

10008

100101117

101001106

101101015

110001004

110100113

111000102

111100011

000000000

nn

Addition of 2’s complement 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

10001010

11101000

10000100111

Addition of 2’s complement 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

00100010 1

01111010

100001011111

So how can we perform subtraction?

Addition of 2’s complement 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

00010010 1

11011000

101011001111

discard carry

Carry vs. overflow

Addition of 2’s complement 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.

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?

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?

What is 111111102?

Flip the bits: 00000001.Then add 1: 00000010.This is -2.

11111110

01111111

01111111

BINARY ARITHMETIC1’s complement

Addition with 1’s complement

• Note: 1’s complement has two 0’s!

• 1’s complement addition is tricky (end-around-carry).

100001117

100101106

101001015

101101004

110000113

110100102

111000011

111100000

NN

8-bit 1’s complement addition

• Ex. Let X = A816 and Y = 8616.

• Calculate Y - X using 1’s complement.

8-bit 1’s complement addition

• 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.)

1101 1101

0111 0101

0110 1000

Y - X = -121 + 87 = -34 (base 10)

8-bit 1’s complement addition

• Ex. Let X = A816 and Y = 8616.

• Calculate X - Y using 1’s complement.

8-bit 1’s complement addition

• 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.) 0010 0010

1

0001 0010 1

1001 0111

1000 1010

X - Y = -87 + 121 = 34 (base 10)

end around carry

BINARY ARITHMETICExcess K (biased)

Binary arithmetic and Excess 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 -> 0010000111117

001011106

001111015

010011004

010110113

011010102

011110011

100010000

NN

(Not used for integer arithmetic but employed in IEEE 754 floating point standard.)