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### Representing Numbers

B261 Systems Architecture

Previously

- Computer Architecture
- Common misconceptions
- Performance

- Instruction Set (MIPS)
- How data is moved inside a computer
- How instructions are executed

Outline

- Numbers
- 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-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.

- The sequence of binary digits

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)

- 2n different numbers can be represented
- 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.

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

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.

- There will be positive numbers (and zero)

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

- Then the decimal value is:

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.

- 1101 = -8 + 4 + 0 + 1 = -3

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

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.

- int x;

Addition Subtractions involve negating the relevant number:

- Addition is carried out in the same way as decimal arithmetic:
- 0101
- 0001 +
- 0110

- 0101 - 0011 =
- 0101 + 1101 = 0010

Overflow But 1011 = -8 + 2 + 1 = -5 !

- 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

Overflow Examples

- Consider (6 - (-5))
- 0110 - 1011
- = 0110 + 0101
- = 1011
- = -8 + 3
- = -5

- 0110 - 1011
- -6 - 3
- = 1010 + 1101
- = 0111
- = 7

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.

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.

- shift a word x bits to the right or left:

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
- addu $1,$2,$3
- operands treated as unsigned ints.

- addu $1,$2,$3
- subtract unsigned
- subu $1,$2,$3
- operands treated as unsigned ints

- subu $1,$2,$3
- add immediate unsigned
- addiu $1,$2,10
- operands treated as unsigned ints

- addiu $1,$2,10

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.

- if there is overflow

More MIPS

- Load upper immediate
- lui $1,100
- $1 = 216 * 100
- (100, shifted left by 16 = upper half of the word).

- lui $1,100
- 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).

Summary

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

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