# B261 Systems Architecture

## Presentation on theme: "B261 Systems Architecture"— Presentation transcript:

B261 Systems Architecture
Representing Numbers B261 Systems Architecture

Previously Computer Architecture Instruction Set (MIPS)
Common misconceptions Performance Instruction Set (MIPS) How data is moved inside a computer How instructions are executed

Outline Numbers How numbers are represented internally Addition
Limitations of computer arithmetic How numbers are represented internally Representing positive integers Representing negative integers Addition Bit-wise operations

Binary Representation
Computers internally represent numbers in binary form The sequence of binary digits dn-1 dn d1 d0 two At its simplest, this represents the decimal number which is the sum of terms 2i for each di = 1. Eg, 10111two = 16ten+4ten+2ten+1ten = 23ten d0 is called the ‘least significant bit’ LSB dn-1 is called the ‘most significant bit’ MSB But we will need to use different meanings for the bits in order to represent negative or floating-point numbers.

Range of Positive Numbers and Word Lengths
For an n-bit word length 2n different numbers can be represented including zero. 0,1,2,..., ((2n) -1) (if only want +ve numbers) A 3-bit word length would support 0,1,2,3,4,5,6,7 MIPS has a 32-bit word length 232 different numbers can be represented 0,1,2,..., ((232)-1) (= 4,294,967,295).

Negative Numbers The range afforded by the word length must simultaneously support both positive and negative numbers, equally. Two requirements: There must be a way, within one word, to tell if it represents a positive or negative number, and its value. A positive number x, and its complement (-x) must clearly add to zero x + (-x) = 0.

Two’s Complement Example of n=3 bit word: 000 = 0 001 = 1 010 = 2
011 = 3 100 = -4 101 = -3 110 = -2 111 = -1 For positive numbers the MSB (‘sign bit’) is always 0 For negative numbers the MSB (‘sign bit’) is always 1

Two’s complement In general for a n-bit word
There will be positive numbers (and zero) 0,1,2, … , ((2n-1) - 1) Negative numbers -(2n-1), … , -1 The highest positive number is ((2n-1) - 1) The lowest negative numbers is -(2n-1 ) Note the slight imbalance of positive and negative.

Conversion For an n-bit word, suppose that dn-1 is the sign bit.
Then the decimal value is: (-dn-1)* (2n-1) + the usual conversion of the remainder of the word. Eg, n=3, then 101 = (-1 * 4) = = -3 011 = (0 * 4) = = 3

Negating a Number There is a simple two-step process for negation (eg, turn 3 = 011 into -3 = 101): invert every bit (e.g. replace 011 by 100) add 1 (e.g = 101). Example, 4-bit word 1101 1101 = = -3 3 = 0011 Invert bits: 1100 add 1: 1101 = -3 as required.

Sign Extension Turning an n-bit representation into one with more bits
Eg, in 3 bits the number -3 is 101 In 4 bits the number -3 is 1101 Replicate the sign bit from the smaller word into all extra slots of larger word For 5-bit word from 3 bits: = -3 For 5-bit word from 4 bits: = -3

Integer Types Some languages (e.g. C++) support two types of int: signed and unsigned. Signed integers have their MSB treated as a sign bit so negative numbers can be represented Unsigned integers have their MSB treated without such an interpretation (no negative numbers).

Signed/Unsigned Int Suppose we had a 4-bit word, then int x;
x could have range 0 to 7 positive and -8 to -1 negative unsigned int x; x has range 0 to 15 only.

Addition Addition is carried out in the same way as decimal arithmetic: 0101 0001 + 0110 Subtractions involve negating the relevant number: = = 0010

Overflow Since there are only a fixed set of bits available for arithmetic things can go wrong: Consider n=4, and (6+5). 0110 0101 + 1011 But 1011 = = -5 !

Overflow Examples Consider (6 - (-5)) -6 - 3 0110 - 1011 = 0110 + 0101
= 1011 = = -5 -6 - 3 = = 0111 = 7

Overflow Situations A + B A - B
where A>0, B>0, A+B too big, result negative where A<0, B<0, A+B too small, result positive A - B where A>0, B<0, A-B too big, result negative where A<0, B>0, A-B too small, result positive.

Bit Operations Shifts shift a word x bits to the right or left:
0110 shift left by 1 is 1100 0110 shift right by 1 is 0011 In C++ shifts are expressed using <<,>> unsigned int y = x<<5, z = x>>3; means y is the value of x shifted 5 to left, and z is x shifted 3 to the right. One application is multiplication/division by powers of 2.

AND/OR AND is a bitwise operation on two words, where for each corresponding bit a and b: a AND b = 1 only when a=1 and b=1, otherwise 0. OR is a bitwise operation, such that a OR b = 0 only if both a=0 and b=0, otherwise 1. For example: 1011 AND 0010 = 0010 1011 OR 0010 = 1011

New MIPS Operations add unsigned subtract unsigned
operands treated as unsigned ints. subtract unsigned subu \$1,\$2,\$3 operands treated as unsigned ints add immediate unsigned addiu \$1,\$2,10

if there is overflow then an ‘exception’ is raised by the hardware control passes to a special procedure to ‘handle’ the exception, and then returns to the next instruction after the one that raised the exception.