# Bits, Data Types, and Operations Chapter 2 Bits, Data Types, and Operations How do we represent data in a computer? At the lowest level, a computer is an electronic machine. works by controlling the flow of electrons Easy to recognize two conditions: 1. presence of a voltage well call this state 1 2. absence of a voltage well call this state 0 Could base state on value of voltage, but control and detection circuits more complex. compare turning on a light switch to measuring or regulating voltage intermediate version, i.e., 4 or 8 states represented by different voltages used to increase capacity/throughput in special cases TLC, QLC, PAM4 Computer is a binary digital system. Digital system: finite number of symbols Binary (base two) system: has two states: 0 and 1 Basic unit of information is the binary digit, or bit. Values with more than two states require multiple bits. A collection of two bits has four possible states: 00, 01, 10, 11 A collection of three bits has eight possible states: 000, 001, 010, 011, 100, 101, 110, 111 A collection of n bits has 2n possible states.

What kinds of data do we need to represent? Numbers signed, unsigned, integers, floating point, complex, rational, irrational, Logical true, false Text characters, strings, Instructions (binary) LC-3, x-86 .. Images jpeg, gif, bmp, png ... Sound mp3, wav.. Data type: representation and operations within the computer Well start with numbers Unsigned Integers Non-positional notation could represent a number (5) with a string of ones (11111) problems? Weighted positional notation like decimal numbers: 329 3 is worth 300, because of its position, while 9 is only worth 9 329 102 101 100 3x100 + 2x10 + 9x1 = 329 most significant 22

101 21 least significant 20 1x4 + 0x2 + 1x1 = 5 Unsigned Integers (cont.) An n-bit unsigned integer represents 2n values: from 0 to 2n-1. 22 21 20 0 0 0 0 0 0 1

1 0 1 0 2 0 1 1 3 1 0 0 4 1 0 1 5

1 1 0 6 1 1 1 7 Unsigned Binary Arithmetic Base-2 addition just like base-10! add from right to left, propagating carry carry 10010 + 1001 11011 10010 + 1011 11101 10111 + 111 Subtraction, multiplication, division, 1111

+ 1 10000 Try it out 17 + 37 = ? 1 + 63 = ? A. B. C. D. E. A. B. C. D. E. 110111 010110 110110 101101 110011 000000 010110 1101100 1000000 D or possibly A Signed Integers

With n bits, we have 2n distinct values. assign about half to positive integers (1 through 2n-1) and about half to negative (- 2n-1 through -1) that leaves two values: one for 0, and one extra Positive integers just like unsigned zero in most significant (MS) bit 00101 = 5 Negative integers: formats sign-magnitude set MS bit to show negative, other bits are the same as unsigned 10101 = -5 ones complement flip every bit to represent negative 11010 = -5 in either case, MS bit indicates sign: 0=positive, 1=negative Twos Complement Problems with sign-magnitude and 1s complement two representations of zero (+0 and 0) arithmetic circuits are complex How to add two sign-magnitude numbers? e.g., try 2 + (-3) How to add to ones complement numbers? e.g., try 4 + (-3) Solution for each positive number (X), assign value to its negative (-X), such that X + (-X) = 0 with normal addition, ignoring carry out Twos Complement Twos complement representation developed to make circuits easy for arithmetic.

for each positive number (X), assign value to its negative (-X), such that X + (-X) = 0 with normal addition, ignoring carry out 00101 (5) + 11011 (-5) 00000 (0) 01001 (9) + (-9) 00000 (0) Twos Complement Representation If number is positive or zero, normal binary representation, zeroes in upper bit(s) If number is negative, start with positive number flip every bit (i.e., take the ones complement) then add one 00101 (5) 01001 (9) 11010 (1s comp) (1s comp) + 1 + 1 11011 (-5) (-9) Twos Complement Shortcut

To take the twos complement of a number: copy bits from right to left until (and including) the first 1 flip remaining bits to the left 011010000 100101111 + 1 100110000 011010000 (1s comp) (flip) (copy) 100110000 Twos Complement Signed Integers MS bit is sign bit it has weight 2n-1. Range of an n-bit number: -2n-1 through 2n-1 1. The most negative number (-2n-1) has no positive counterpart. -23 22 21 20 -23 22

21 20 0 0 0 0 0 1 0 0 0 -8 0 0 0 1 1

1 0 0 1 -7 0 0 1 0 2 1 0 1 0 -6 0 0

1 1 3 1 0 1 1 -5 0 1 0 0 4 1 1 0 0

-4 0 1 0 1 5 1 1 0 1 -3 0 1 1 0 6 1

1 1 0 -2 0 1 1 1 7 1 1 1 1 -1 Converting Binary (2s C) to Decimal 1. If leading bit is one, take twos complement to get a positive number. 2. Add powers of 2 that have 1 in the corresponding bit positions.

3. If original number was negative, add a minus sign. X = 01101000two = 26+25+23 = 64+32+8 = 104ten Assuming 8-bit 2s complement numbers. n 2n 0 1 2 3 4 5 6 7 8 9 1 0 1 2 4 8 16 32 64 128 256 512 102 4

More Examples X = 00100111two = 25+22+21+20 = 32+4+2+1 = 39ten X = -X = = = 11100110two 00011010 24+23+21 = 16+8+2 26ten X = -26ten Assuming 8-bit 2s complement numbers. n 2n 0 1 2 3 4 5 6 7 8 9 10 1 2 4

8 16 32 64 128 256 512 1024 Converting Decimal to Binary (2s C) First Method: Division 1. Find magnitude of decimal number. (Always positive.) 2. Divide by two remainder is least significant bit. 3. Keep dividing by two until answer is zero, writing remainders from right to left. 4. Append a zero as the MS bit; if original number was negative, take twos complement. X = 104ten 104/2 = 52 r0 52/2 26/2 13/2 6/2 3/2 X = 01101000two = = = = =

26 r0 13 r0 6 r1 3 r0 1 r1 1/2 = 0 r1 bit 0 bit 1 bit 2 bit 3 bit 4 bit 5 bit 6 Converting Decimal to Binary (2s C) n 2n Second Method: Subtract Powers of Two 1. Find magnitude of decimal number. 2. Subtract largest power of two less than or equal to number. 3. Put a one in the corresponding bit position. 4. Keep subtracting until result is zero. 5. Append a zero as MS bit; if original was negative, take twos complement. X = 104ten X = 01101000two 104 - 64 = 40

bit 6 40 - 32 = 8 8-8 = 0 bit 5 bit 3 0 1 2 3 4 5 6 7 8 9 10 1 2 4 8 16 32 64 128 256 512 1024 Operations: Arithmetic and Logical Recall:

a data type includes representation and operations. We now have a good representation for signed integers, so lets look at some arithmetic operations: Addition Subtraction Sign Extension Well also look at overflow conditions for addition. Multiplication, division, etc., can be built from these basic operations. Logical operations are also useful: AND OR NOT Addition As weve discussed, 2s comp. addition is just binary addition. assume all integers have the same number of bits ignore carry out for now, assume that sum fits in n-bit 2s comp. representation 01101000 (104) 11110110 (-10) + 11110000 (-16) + (-9) 01011000 (98) (-19) Assuming 8-bit 2s complement numbers. Subtraction Negate subtrahend (2nd no.) and add. assume all integers have the same number of bits ignore carry out for now, assume that difference fits in n-bit 2s comp.

representation 01101000 - 00010000 01101000 + 11110000 01011000 (104) 11110110 (-10) (16) (-9) (104) 11110110 (-10) (-16) + (9) (88) (-1) Assuming 8-bit 2s complement numbers. Sign Extension To add two numbers, we must represent them with the same number of bits. If we just pad with zeroes on the left: 4-bit 8-bit 0100 (4) 00000100 (still 4) 1100 (-4) 00001100 (12, not -4) Instead, replicate the MS bit -- the sign bit: 4-bit 0100 (4) 1100 (-4)

8-bit 00000100 (still 4) 11111100 (still -4) Overflow If operands are too big, then sum cannot be represented as an n-bit 2s comp number. 01000 (8) + 01001 (9) 10001 (-15) 11000 (-8) + 10111 (-9) 01111 (+15) We have overflow if: signs of both operands are the same, and sign of sum is different. Another test -- easy for hardware: carry into MS bit does not equal carry out Logical Operations Operations on logical TRUE or FALSE two states -- takes one bit to represent: TRUE=1, FALSE=0 A 0 0 1 1

B A AND B 0 0 1 0 0 0 1 1 A 0 0 1 1 B A OR B 0 0 1 1 0 1 1 1 A NOT A 0 1 1 0 View n-bit number as a collection of n logical values operation applied to each bit independently

Examples of Logical Operations AND useful for clearing bits AND with zero = 0 AND with one = no change AND 11000101 00001111 00000101 OR 11000101 00001111 11001111 OR useful for setting bits OR with zero = no change OR with one = 1 NOT unary operation -- one argument flips every bit NOT 11000101 00111010 XOR

Result is 1 if inputs are different 0 otherwise Useful for checking if two numbers are equal among other things A B A XOR B 0 0 0 0 1 1 1 0 1 1 1 0 XOR 11000101 11000101 00000000 Hexadecimal Notation It is often convenient to write binary (base-2) numbers as hexadecimal (base-16) numbers instead. fewer digits -- four bits per hex digit less error prone -- easy for humans to corrupt long string of 1s and 0s Binary Hex Decimal

Binary Hex Decimal 0000 0001 0010 0011 0100 0101 0110 0111 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7

1000 1001 1010 1011 1100 1101 1110 1111 8 9 A B C D E F 8 9 10 11 12 13 14 15 Converting from Binary to Hexadecimal Every four bits is a hex digit. start grouping from right-hand side 011101010001111010011010111 3

A 8 F 4 This is not a new machine representation, just a convenient way to write the number. D 7 Fractions: Fixed-Point How can we represent fractions? Use a binary point to separate positive from negative powers of two -- just like decimal point. 2s comp addition and subtraction still work. if binary points are aligned 2-1 = 0.5 2-2 = 0.25 2-3 = 0.125 00101000.101 (40.625) + 11111110.110 (-1.25) 00100111.011 (39.375) No new operations -- same as integer arithmetic. Very Large and Very Small: Floating-Point Large values: 6.023 x 1023 -- requires 79 bits Small values: 6.626 x 10-34 -- requires >110 bits Use equivalent of scientific notation: F x 2E

Need to represent F (fraction), E (exponent), and sign. IEEE 754 Floating-Point Standard (32-bits): 1b 8b S Exponent 23b Fraction N ( 1)S 1.fraction 2exponent 127 , 1 exponent 254 N ( 1)S 0.fraction 2 126 , exponent 0 Special Values Zero: exponent = 0, fraction/mantissa = 0, sign bit +0 or -0 Infinity: exponent = all 1s, fraction/mantissa = 0, sign bit + or NaN: exponent = all 1s, fraction/mantissa = non-zero Question: how many NaN representations? Floating Point Example Single-precision IEEE floating point number: 10111111010000000000000000000000 sign exponent fraction Sign is 1 number is negative. Exponent field is 01111110 = 126 (decimal). Fraction is 0.100000000000 = 0.5 (decimal). Value = -1.5 x 2(126-127) = -1.5 x 2-1 = -0.75.

https://www.h-schmidt.net/FloatConverter/IEEE754.html Decimal to 32 bit floating point 1. Change absolute value of the decimal number to binary 2. Move radix point so there is only a single 1 bit to the left of the radix point. Every position moved to the left increases the exponent size by one. Every position moved to the right decreases the exponent size by one. The initial exponent is 0. 3. Remove leading 1 from resulting binary number and store this number in bits 0-22. 4. Add 127 to exponent and store binary representation of exponent in bits 23-30 5. Store sign in bit 31, 1 for negative, 0 for positive. 32 bit floating point to decimal 1. Check bit MSB (31) for sign, 1 negative, 0 positive 2. Extract bits 30 23, and find their value in binary then subtract 127 to get the exponent 3. Extract bits 22 0 and add implicit bit with value 1 to location 23 to get the fractional part 4. Change value of exponent to 0 by shifting radix point of fractional part right to reduce exponent and left to increase exponent 5. Convert resulting binary number to decimal and add correct sign

Floating-Point Operations Will regular 2s complement arithmetic work for Floating Point numbers? (Hint: In decimal, how do we compute 3.07 x 1012 + 9.11 x 108? Need to work with exponents ) Text: ASCII Characters ASCII: Maps 128 characters to 7-bit code. both printable and non-printable (ESC, DEL, ) characters 00 01 02 03 04 05 06 07 08 09 0a 0b 0c 0d 0e 0f nul 10 soh 11 stx 12 etx 13 eot 14 enq 15 ack 16 bel 17

bs 18 ht 19 nl 1a vt 1b np 1c cr 1d so 1e si 1f dle 20 dc1 21 dc2 22 dc3 23 dc4 24 nak 25 syn 26 etb 27 can 28 em 29 sub 2a esc 2b fs 2c gs 2d rs 2e us 2f sp ! " # \$ % & '

( ) * + , . / 30 31 32 33 34 35 36 37 38 39 3a 3b 3c 3d 3e 3f 0 1 2 3 4 5 6 7 8

9 : ; < = > ? 40 41 42 43 44 45 46 47 48 49 4a 4b 4c 4d 4e 4f @ A B C D E F G H

I J K L M N O 50 51 52 53 54 55 56 57 58 59 5a 5b 5c 5d 5e 5f P Q R S T U V W X

Y Z [ \ ] ^ _ 60 61 62 63 64 65 66 67 68 69 6a 6b 6c 6d 6e 6f ` a b c d e f g h

i j k l m n o 70 71 72 73 74 75 76 77 78 79 7a 7b 7c 7d 7e 7f p q r s t u v w x

y z { | } ~ del Interesting Properties of ASCII Code What is relationship between a decimal digit ('0', '1', ) and its ASCII code? What is the difference between an upper-case letter ('A', 'B', ) and its lower-case equivalent ('a', 'b', )? Given two ASCII characters, how do we tell which comes first in alphabetical order? Unicode: 128 characters are not enough. 1990s Unicode was standardized, Java used Unicode. No new operations -- integer arithmetic and logic. Other Data Types Text strings sequence of characters, terminated with NULL (0) typically, no hardware support Image array of pixels monochrome: one bit (1/0 = black/white) color: red, green, blue (RGB) components (e.g., 8 bits each) other properties: transparency hardware support: typically none, in general-purpose processors MMX -- multiple 8-bit operations on 32-bit word

Sound sequence of fixed-point numbers LC-3 Data Types Some data types are supported directly by the instruction set architecture. For LC-3, there is only one hardware-supported data type: 16-bit 2s complement signed integer Operations: ADD, AND, NOT Other data types are supported by interpreting 16-bit values as logical, text, fixed-point, etc., in the software that we write.