Representing Numbers

1. Representing Numbers B261 Systems Architecture

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

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

4. Binary Representation • Computers internally represent numbers in binary form • The sequence of binary digits • dn-1 dn-2 ... 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.

5. 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).

6. 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.

7. For positive numbers the MSB (‘sign bit’) is always 0 For negative numbers the MSB (‘sign bit’) is always 1 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

8. 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.

9. 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) + 0 + 1 = -4 + 1 = -3 • 011 = (0 * 4) + 2 + 1 = 0 + 3 = 3

10. 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. 100+1 = 101). • Example, 4-bit word 1101 • 1101 = -8 + 4 + 0 + 1 = -3 • 3 = 0011 • Invert bits: 1100 • add 1: 1101 = -3 as required.

11. 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: 11101 = -3 • For 5-bit word from 4 bits: 11101 = -3

12. 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).

13. 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.

14. Addition • Addition is carried out in the same way as decimal arithmetic: • 0101 • 0001 + • 0110 • Subtractions involve negating the relevant number: • 0101 - 0011 = • 0101 + 1101 = 0010

15. Overflow • Since there are only a fixed set of bits available for arithmetic things can go wrong: • Consider n=4, and 0110 + 0101 (6+5). • 0110 • 0101 + • 1011 • But 1011 = -8 + 2 + 1 = -5 !

16. Overflow Examples • Consider (6 - (-5)) • 0110 - 1011 • = 0110 + 0101 • = 1011 • = -8 + 3 • = -5 • -6 - 3 • = 1010 + 1101 • = 0111 • = 7

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

18. 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.

19. 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

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

21. Exceptions • Where the signed versions (add, addi, etc) are used • 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.

22. More MIPS • Load upper immediate • lui $1,100 • $1 = 216 * 100 • (100, shifted left by 16 = upper half of the word). • and, or, and immediate (andi), or immediate (ori), shift left logical (sll), shift right logical (srl) • all of form $1,$2,$3 or $1,$2,10 (for immediate).

23. Summary • Basic number representation (integers) • Representation of negatives • Basic arithmetic (+ and -) • Logical operations • Next time - building an ALU.