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