Binary numbers and arithmetic ADDITION Addition decimal Addition

Binary numbers and arithmetic

ADDITION

Addition (decimal)

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)

Multiplication (binary)

Multiplication (binary) 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 8 16 32 64 bytes 1 2 4 8 C/C++ char short int or long Intel [s]byte [s]word [s]dword [s]qword Sun byte half word xword

Representing numbers (ints) • Fixed, finite number of bits 8 16 32 64 Intel [s]byte [s]word [s]dword [s]qword signed -27. . +27 -1 -215. . +215 -1 -231. . +231 -1 -263. . +263 -1 unsigned 0. . +28 -1 0. . +216 -1 0. . +232 -1 0. . +264 -1 In general, for k bits, the unsigned range is [0. . +2 k-1] and the signed range is [-2 k-1. . +2 k-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 b. D-1

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

1’s complement (diminished radix complement) • Let x be a non-negative number. • Then –x is represented by b. D-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.

1’s complement (diminished radix complement) • Let x be a non-negative number. • Then –x is represented by b. D-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 – 12389 = 100000 – 12389 = 99999 – 12389 = 87610

1’s complement (diminished radix complement) • Let x be a non-negative number. • Then –x is represented by b. D-1+( -x) where b = base D = (total) # of bits (including the sign bit) • Shortcut for base 2? – All combinations used, but 2 zeros!

2’s complement (radix complement) • Let x be a non-negative number. • Then –x is represented by b. D+(-x). – Ex. Let b=2 and D=4. Then -1 is represented by 241 = 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 b. D+(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?

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

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?

excess b. D-1 (biased representation) • For pos, neg, and 0, x is represented by b. D-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 b. D-1 • For pos, neg, and 0, x is represented by b. D-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.

2’s complement vs. excess b. D-1 • In 2’s, positives start with 0; in excess, positives start with 1. • Both have one zero (positive). • Remaining bits are the same.

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

Signed magnitude 1’s complement 2’s complement Excess K (biased) BINARY ARITHMETIC

Signed magnitude BINARY ARITHMETIC

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 else if (|A|>|B|) then else if (|A|==|B|) then else • Complicated? { { R = |A| + |B|; R = |A| - |B|; R = 0; R = |B| - |A|; Rsign = Asign; Rsign = 0; Rsign = Bsign; } }

2’s complement BINARY ARITHMETIC

Representing numbers (ints) using 2’s complement • Fixed, finite number of bits 8 16 32 64 Intel sbyte sword sdword sqword signed -27. . +27 -1 -215. . +215 -1 -231. . +231 -1 -263. . +263 -1 In general, for k bits, the signed range is [-2 k-1. . +2 k-1 -1]. So where does the extra negative value come from?

Representing numbers (ints) • Fixed, finite number of bits 8 16 32 64 Intel sbyte sword sdword sqword signed -27. . +27 -1 -215. . +215 -1 -231. . +231 -1 -263. . +263 -1 In general, for k bits, the signed range is [-2 k-1. . +2 k-1 -1]. So where does the extra negative value come from?

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

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

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 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 #include <stdio. h> 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 ); } } 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.

1’s complement BINARY ARITHMETIC

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

8 -bit 1’s complement addition • Ex. Let X = A 816 and Y = 8616. • Calculate Y - X using 1’s complement.

8 -bit 1’s complement addition • Ex. Let X = A 816 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. ) Y - X = -121 + 87 = -34 (base 10)

8 -bit 1’s complement addition • Ex. Let X = A 816 and Y = 8616. • Calculate X - Y using 1’s complement.

8 -bit 1’s complement addition • Ex. Let X = A 816 and Y = 8616. • Calculate X - Y using 1’s complement. X = 1010 10002 = -8710 Y = 1000 01102 = -12110 ~Y = 0111 10012 end around carry (Note: C=1 out of msb. ) X - Y = -87 + 121 = 34 (base 10)

Excess K (biased) BINARY ARITHMETIC

Binary arithmetic and Excess K (biased) Method: Simply add and then flip the sign bit. -1 +5 -+4 0111 1101 ---0100 -> flip sign -> 1100 +1 -5 --4 1001 0011 ---1100 -> flip sign -> 0100 +1 +5 -+6 1001 1101 ---0110 -> flip sign -> 1110 -1 -5 --6 0111 0011 ---1010 -> toggle sign -> 0010 (Not used for integer arithmetic but employed in IEEE 754 floating point standard. )