1. B261 Systems Architecture
Representing Numbers
B261 Systems Architecture

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

3. 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

4. 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.

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. 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

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) = = -3
011 = (0 * 4) = = 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 = 101).
Example, 4-bit word 1101
1101 = = -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: = -3
For 5-bit word from 4 bits: = -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: =
= 0010

15. 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 !

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

17. 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.

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 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

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.