Integers Topics Numeric Encodings (2.2) Programming Implications Unsigned & Two’s complement Programming Implications C promotion rules Basic operations (2.3) Addition, negation, multiplication Consequences of overflow Using shifts to perform power-of-2 multiply/divide

Encoding Integers Sign Bit Unsigned Two’s Complement Sign Bit short int x = 15213; short int y = -15213; Sign Bit C short 2 bytes long Sign Bit For 2’s complement, most significant bit indicates sign 0 for nonnegative 1 for negative

Numeric Ranges Unsigned Values UMin = 0 UMax = 2w – 1 000…0 UMax = 2w – 1 111…1 Two’s Complement Values TMin = –2w–1 100…0 TMax = 2w–1 – 1 011…1 Other Values Minus 1 111…1 Values for W = 16

Unsigned & Signed Numeric Values Given n bits, all a matter of mapping (rule) X B2U(X) B2S(X) B2O(X) B2T(X) 000 001 1 010 2 011 3 100 4 -0 -3 -4 101 5 -1 -2 110 6 111 7

Unsigned & Signed Numeric Values X B2T(X) B2U(X) 0000 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 –8 8 –7 9 –6 10 –5 11 –4 12 –3 13 –2 14 –1 15 1000 1001 1010 1011 1100 1101 1110 1111 Equivalence Same encodings for nonnegative values Uniqueness Every bit pattern represents unique integer value Each representable integer has unique bit encoding  Can Invert Mappings U2B(x) = B2U-1(x) Bit pattern for unsigned integer T2B(x) = B2T-1(x) Bit pattern for two’s comp integer

Values for Different Word Sizes C Programming  #include <limits.h> K&R App. B11 Declares constants, e.g.,  ULONG_MAX  LONG_MAX  LONG_MIN Values platform-specific Observations |TMin | = TMax + 1 Asymmetric range Umax = 2 * TMax + 1 -Tmin ?

Prob. 2.19 X T2B4(X) -8 -3 -2 -1 5

Decimal Numbers Digits: 0,1,…9 49 – 2 49 + (100 – 2) = 49 + 98 = 147 = 47 2 – 49 2 + (100 – 49) = 2 + 51 = 53 = -47 Pat. 10’s comp. 0 0 1 1 2 2 3 3 4 4 5 -5 6 -4 7 -3 8 -2 9 -1 Pat. 10’s comp 00 0 01 1 .. 49 49 50 -50 51 -49 52 -48 98 -2 99 -1

2.2. Casting Signed to Unsigned C Allows Conversions from Signed to Unsigned Resulting Value No change in bit representation Nonnegative values unchanged ux = 15213 Negative values change into (large) positive values uy = 50323 short int x = 15213; unsigned short int ux = (unsigned short) x; short int y = -15213; unsigned short int uy = (unsigned short) y;

Signed vs. Unsigned in C Constants Casting By default are considered to be signed integers Unsigned if have “U” as suffix 0U, 4294967259U Casting Explicit casting between signed & unsigned same as U2T and T2U int tx, ty; unsigned ux, uy; tx = (int) ux; uy = (unsigned) ty; Implicit casting also occurs via assignments and procedure calls tx = ux; uy = ty;

Casting Surprises Expression Evaluation 0 == 0U Constant1 Constant2 If mix unsigned and signed in single expression, signed values implicitly cast to unsigned Including comparison operations <, >, ==, <=, >= Examples for W = 8 Constant1 Constant2 0 == 0U -1 < 0 -1 > 0U 127 > -128 127U < -128 (unsigned) -1 > -2 127 < 128U 127 > (int) 128U

Prob. 2.44 int x = foo(); /* arbitrary value */ int y = bar(); /* arbitrary value */ unsigned us = x; unsigned uy = y; For any x and y, are the following always true ? (x>0) || (x-1<0) (x & 7) !=7 || (x<<29 < 0) (x*x) >=0 x < 0 || -x <= 0 x > 0 || -x >= 0 x+y == uy+ux x * ~y + uy * ux == -x

Why Casting Surprises ? 2’s Comp.  Unsigned Ordering Inversion Negative  Big Positive TMax TMin –1 –2 UMax UMax – 1 TMax + 1 2’s Comp. Range Unsigned

Sign Extension Task: Rule: Given w-bit signed integer x Convert it to w+k-bit integer with same value Rule: Make k copies of sign bit: X  = xw–1 ,…, xw–1 , xw–1 , xw–2 ,…, x0 • • • X X  w k k copies of MSB

Sign Extension Example short int x = 15213; int ix = (int) x; short int y = -15213; int iy = (int) y; Decimal Hex Binary x 15213 3B 6D 00111011 01101101 ix 00 00 3B 6D 00000000 00000000 00111011 01101101 y -15213 C4 93 11000100 10010011 iy FF FF C4 93 11111111 11111111 11000100 10010011 Converting from smaller to larger integer data type C automatically performs sign extension

Justification For Sign Extension Prove Correctness by Induction on k Induction Step: extending by single bit maintains value Key observation: –2w–1 = –2w +2w–1 Look at weight of upper bits: X –2w–1 xw–1 X  –2w xw–1 + 2w–1 xw–1 = –2w–1 xw–1 - • • • X X  + w+1 w

Shift Operations Left Shift: x << y Right Shift: x >> y Shift bit-vector x left y positions Throw away extra bits on left Fill with 0’s on right Right Shift: x >> y Shift bit-vector x right y positions Throw away extra bits on right Logical shift Fill with 0’s on left Arithmetic shift Replicate most significant bit on right Useful with two’s complement integer representation Argument x 01100010 << 3 00010000 00010000 00010000 Log. >> 2 00011000 0001100 00011000 Arith. >>2 00011000 00011000 0001100 Argument x 10100010 << 3 00010000 00010000 00010000 Log. >> 2 0010100 00101000 00101000 Arith. >> 2 11101000 11101000 1110100

Prob. 2.23 Given a 32-bit machine, int func1(unsigned word){ return (int) ((word <<24) >> 24); } int func2(unsigned word){ return ((int) word <<24) >> 24; w func1(w) func2(w) 0x00000076 0x87654321 0x000000C9 0xEDCBA987