CS 33. Data Representation, Part 1. CS33 Intro to Computer Systems VII 1 Copyright 2017 Thomas W. Doeppner. All rights reserved.

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CS 33 Data Representation, Part 1 CS33 Intro to Computer Systems VII 1 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Number Representation Hindu-Arabic numerals developed by Hindus starting in 5 th century» positional notation» symbol for 0 adopted and modified somewhat later by Arabs» known by them as Rakam Al-Hind (Hindu numeral system) 1999 rather than MCMXCIX» (try doing long division with Roman numerals!) CS33 Intro to Computer Systems VII 2 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Which Base? 1999 base 10» 9 10 0 +9 10 1 +9 10 2 +1 10 3 base 2» 11111001111 1 2 0 +1 2 1 +1 2 2 +1 2 3 +0 2 4 +0 2 5 +1 2 6 +1 2 7 +1 2 8 +1 2 9 +1 2 10 base 8» 3717 7 8 0 +1 8 1 +7 8 2 +3 8 3» why are we interested? base 16» 7CF 15 16 0 +12 16 1 +7 16 2» why are we interested? CS33 Intro to Computer Systems VII 3 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Words 0 1 1 1 1 1 0 0 1 1 1 1 12-bit computer word 3 7 1 7 0 0 0 0 0 1 1 1 1 1 0 0 1 1 1 1 16-bit computer word 0 7 C F CS33 Intro to Computer Systems VII 4 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Algorithm void basex(unsigned int num, unsigned int base) { char digits[] = {'0', '1', '2', '3', '4', '5', '6', }; char buf[8*sizeof(unsigned int)+1]; int i; for (i = sizeof(buf) - 2; i >= 0; i--) { buf[i] = digits[num%base]; } num /= base; if (num == 0) break; } buf[sizeof(buf) - 1] = '\0'; printf("%s\n", &buf[i]); CS33 Intro to Computer Systems VII 5 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Or $ bc obase=16 1999 7CF $ CS33 Intro to Computer Systems VII 6 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Quiz 1 What s the decimal (base 10) equivalent of 23 16? a) 19 b) 33 c) 35 d) 37 CS33 Intro to Computer Systems VII 7 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Encoding Byte Values Byte = 8 bits binary 000000002 to 111111112 decimal: 010 to 25510 hexadecimal 0016 to FF16» base 16 number representation» use characters 0 to 9 and A to F» write FA1D37B16 in C as 0xFA1D37B 0xfa1d37b 0 0 0000 1 1 0001 2 2 0010 3 3 0011 4 4 0100 5 5 0101 6 6 0110 7 7 0111 8 8 1000 9 9 1001 A 10 1010 B 11 1011 C 12 1100 D 13 1101 E 14 1110 F 15 1111 CS33 Intro to Computer Systems VII 8 Copyright 2017 Thomas W. Doeppner. All rights reserved.

And Boolean Algebra Developed by George Boole in 19th Century algebraic representation of logic» encode true as 1 and false as 0 n A&B = 1 when both A=1 and B=1 Or n A B = 1 when either A=1 or B=1 Not n ~A = 1 when A=0 Exclusive-Or (Xor) n A^B = 1 when either A=1 or B=1, but not both CS33 Intro to Computer Systems VII 9 Copyright 2017 Thomas W. Doeppner. All rights reserved.

General Boolean Algebras Operate on bit vectors operations applied bitwise 01101001 01010101 01101001 & 01010101 01101001 ^ 01010101 ~ 01010101 01000001 01111101 00111100 10101010 All of the properties of boolean algebra apply CS33 Intro to Computer Systems VII 10 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Example: Representing & Manipulating Sets Representation width-w bit vector represents subsets of {0,, w 1} a j = 1 iff j A 01101001 { 0, 3, 5, 6 } 76543210 01010101 { 0, 2, 4, 6 } 76543210 Operations & intersection 01000001 { 0, 6 } union 01111101 { 0, 2, 3, 4, 5, 6 } ^ symmetric difference 00111100 { 2, 3, 4, 5 } ~ complement 10101010 { 1, 3, 5, 7 } CS33 Intro to Computer Systems VII 11 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Bit-Level Operations in C Operations &,, ~, ^ available in C apply to any integral data type» long, int, short, char view arguments as bit vectors arguments applied bit-wise Examples (char datatype) ~0x41 0xBE ~010000012 101111102 ~0x00 0xFF ~000000002 111111112 0x69 & 0x55 0x41 011010012 & 010101012 010000012 0x69 0x55 0x7D 011010012 010101012 011111012 CS33 Intro to Computer Systems VII 12 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Contrast: Logic Operations in C Contrast to Logical Operators &&,,!» view 0 as false» anything nonzero as true» always return 0 or 1» early termination/short-circuited execution Examples (char datatype)!0x41 0x00!0x00 0x01!!0x41 0x01 0x69 && 0x55 0x01 0x69 0x55 0x01 p && *p (avoids null pointer access) CS33 Intro to Computer Systems VII 13 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Contrast: Logic Operations in C Contrast to Logical Operators &&,,!» view 0 as false» anything nonzero as true Watch» always out return for 0 && or 1 vs. & (and vs. )» early termination/short-circuited execution One of the more common oopsies in Examples C programming (char datatype)!0x41 0x00!0x00 0x01!!0x41 0x01 0x69 && 0x55 0x01 0x69 0x55 0x01 p && *p (avoids null pointer access) CS33 Intro to Computer Systems VII 14 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Shift Operations Left 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 left Undefined Behavior shift amount < 0 or word size Argument x << 3 Log. >> 2 Arith. >> 2 Argument x << 3 Log. >> 2 Arith. >> 2 01100010 00010000 00011000 00011000 10100010 00010000 00101000 11101000 CS33 Intro to Computer Systems VII 15 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Unsigned Integers b w-1 b w-2 b w-3 b 2 b 1 b 0 value = w'1! b i $ 2 i i)0 CS33 Intro to Computer Systems VII 16 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Signed Integers Sign-magnitude b w-1 b w-2 b w-3 b 2 b 1 b 0 sign magnitude value = two representations of zero! computer must have two sets of instructions one for signed arithmetic, one for unsigned CS33 Intro to Computer Systems VII 17 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Signed Integers Ones complement negate a number by forming its bitwise complement» e.g., (-1) 01101011 = 10010100 value = = if b w-1 = 0 two zeroes! = if b w-1 = 1 CS33 Intro to Computer Systems VII 18 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Signed Integers Two s complement b w-1 = 0 Þ non-negative number value = b w-1 = 1 Þ negative number one zero! value = CS33 Intro to Computer Systems VII 19 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Signed Integers Negating two s complement value = b w 1 2 w 1 + how to compute value? (~value)+1 w 2 i=0 b i 2 i CS33 Intro to Computer Systems VII 20 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Signed Integers Negating two s complement (continued) value + (~value + 1) = (value + ~value) + 1 = (2 w 1) + 1 = 2 w 0 0 0 0 0 0 w = 1 CS33 Intro to Computer Systems VII 21 Copyright 2017 Thomas W. Doeppner. All rights reserved.

Quiz 2 We have a computer with 4-bit words that uses two s complement to represent negative numbers. What is the result of subtracting 0010 (2) from 0001 (1)? a) 0111 b) 1001 c) 1110 d) 1111 CS33 Intro to Computer Systems VII 22 Copyright 2017 Thomas W. Doeppner. All rights reserved.

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