# binary numbers and arithmetic

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Binary numbers and arithmetic. addition. Addition (decimal). Addition (binary). Addition (binary). Addition (binary). So can we count in binary?. Counting in binary (4 bits). 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15. 0000 0001 …. multiplication. Multiplication (decimal). - PowerPoint PPT Presentation

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• Binary numbers and arithmetic

So can we count in binary?

• Counting in binary (4 bits)012345678910111213141500000001

• MULTIPLICATION

• Multiplication (decimal)

• Multiplication (binary)

• Multiplication (binary)Its 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.

bitsbytesC/C++IntelSun81char[s]bytebyte162short[s]wordhalf324int or long[s]dwordword648long long[s]qwordxword

• Representing numbers (ints)Fixed, finite number of bits.

bitsIntelsignedunsigned8[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.signed magnitude

excess bD-1

• Signed magnitudeEx. 4-bit signed magnitude1 bit for sign3 bits for magnitude

• Signed magnitudeEx. 4-bit signed magnitude1 bit for sign3 bits for magnitude

• 1s 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)

Ex. Let b=2 and D=4.Then -1 is represented by 24-1-1 = 1410 or 11102.

• 1s complement(diminished radix complement)Let x be a non-negative number.Then x is represented by bD-1+(-x) whereb = base & D = (total) # of bits (including the sign bit)Ex. What is the 9s complement of 1238910?Given b=10 and D=5. Then the 9s complement of 12389= 105 1 12389= 100000 1 12389= 99999 12389= 87610

• 1s 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!

• 2s 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 10s complement of 12389 = 105 12389 = 100000 12389 = 87611.

• 2s 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?

• 2s complement(radix complement)Shortcut for base 2?Yes! Flip the bits and add 1.

• 2s 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 2s complement?

• 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-1For 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.

Both have one zero (positive).

Remaining bits are the same.

• Summary of methods for representing signed ints.1000=-8| 0000 unused

• BINARY ARITHMETICSigned magnitude1s complement2s complementExcess K (biased)

• BINARY ARITHMETICSigned magnitude

• Addition w/ signed magnitude algorithmFor 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 ARITHMETIC2s complement

• Representing numbers (ints) using 2s complementFixed, finite number of bits.

bitsIntelsigned8sbyte-27..+27-116sword-215..+215-132sdword-231..+231-164sqword-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.

bitsIntelsigned8sbyte-27..+27-116sword-215..+215-132sdword-231..+231-164sqword-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?

• Addition of 2s complement binary numbersConsider 8-bit 2s 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

• Addition of 2s complement binary numbersConsider 8-bit 2s complement binary numbers.Addition of a negative to a positive.

What are the values of these 2 terms?-88 and 122-88 + 122 = 34

• So how can we perform subtraction?

• Addition of 2s complement binary numbersConsider 8-bit 2s complement binary numbers.Subtraction is nothing but addition of the 2s complement.Ex. 58 40 = 58 + (-40) = 18discard carry

• Carry vs. overflow

• Addition of 2s complement binary numbersCarry vs. overflow when adding A + BIf 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 2s 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

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 2s 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?Whats 127 as a 2s complement binary number?

What is 111111102?Flip the bits: 00000001.Then add 1: 00000010.This is -2.

• BINARY ARITHMETIC1s complement

• Addition with 1s complementNote: 1s complement has two 0s!1s complement addition is tricky (end-around-carry).

• 8-bit 1s complement additionEx. Let X = A816 and Y = 8616.Calculate Y - X using 1s complement.

• 8-bit 1s complement additionEx. Let X = A816 and Y = 8616.Calculate Y - X using 1s complement. Y = 1000 01102 = -12110

X = 1010 10002 = -8710~X = 0101 01112

(Note: C=0 out of msb.)Y - X = -121 + 87 = -34 (base 10)

• 8-bit 1s complement additionEx. Let X = A816 and Y = 8616.Calculate X - Y using 1s complement.

• 8-bit 1s complement additionEx. Let X = A816 and Y = 8616.Calculate X - Y using 1s complement.X = 1010 10002 = -8710

Y = 1000 01102 = -12110~Y = 0111 10012

(Note: C=1 out of msb.)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 -> 0010(Not used for integer arithmetic but employed in IEEE 754 floating point standard.)