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Lecture 09- Computer Arithmetic

EE

Computer Architecture

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Limits of Computer Numbers

 Bits can represent anything!

 Characters?

26 letter => 5 bits

upper/lower case + punctuation => 7 bits

(in 8)

rest of the world’s languages => 16 bits

(unicode)

 Logical values?

0 -> False, 1 => True

 colors?

 locations / addresses? commands?

 but N bits => only 2N^ things

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Positive & Negative Numbers

So far, unsigned numbers

Obvious solution: define leftmost bit to be

sign!

Representation called sign and magnitude

MIPS uses 32-bit integers. +

ten

would be:

And - 1

ten

in sign and magnitude would be:

Arithmetic circuit more complicated

Special steps depending whether signs are

2’s Complement Number line

 2 N-1^ non-

negatives

 2 N-1^ negatives

 one zero

 how many

positives?

 comparison?

 overflow?

Sign extension

Convert 2’s complement number using n bits

to more than n bits

Simply replicate the most significant bit (sign

bit) of smaller to fill new bits

2’s comp. positive number has infinite 0s

2’s comp. negative number has infinite 1s

16-bit -

ten

to 32-bit:

two

two

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Signed v. Unsigned Comparisons

X = 1111 1111 1111 1111 1111 1111 1111

two

Y = 0011 1011 1001 1010 1000 1010 0000

two

Is X > Y?

unsigned: YES

signed: NO

Converting to decimal to check

Signed comparison:

ten

ten?

Unsigned comparison:

ten

ten?

slt, slti and sltu, sltiu instructions

Determining Overflow

Overflow occurs when the signs of the values are the same,

and the sign of the result is different.

Overflow if Cin to MSB is different from Cout of MSB

Arithmetic and Logic Unit

The ALU is at the heart of the CPU

Does math and logic

• The ALU is primarily involved in R-type

instructions

– Perform an operation on two registers and

produce a result

• Where is the operation specified?

– The instruction type specifies the operation

– The ALU will have to be controlled by the

instruction opcode

ALU - Logic Operations

0 1 A B Operation Result 2-to-1 Mux If Operation = 0, Result = A • B If Operation = 1, Result = A Ú B Start out by supporting AND and OR operations Two operands, two results. We need only one result... The Operation input comes from logic that looks at the opcode

Arithmetic Unit

a b cin s cout 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

ab

cin

Adding to our ALU

CarryIn CarryOut ALU A B Cout Cin Result Cin Cout Op (2 bits) Operation Function 00 A • B 01 A Ú B 10 A + B Operation Function 00 A • B 01 A Ú B 10 A + B

(Op is now 2 bits) Add an Adder Connect CarryIn (from previous bit) and CarryOut (to next bit) Expand Mux to 3-to-1 (Op is now 2 bits) 0 1 Operation Result A B 2 0 1

1-bit & 32-bit ALU

• Connect to common

Operation controls

• (^) Now we can do 32-bit

AND and OR operations

• Stack 32 of our 1-bit

ALU’s together

• (^) Each one gets one bit

from A and one from B

• Connect C

out

’s to C

in

’s

• (^) Now, 32-bit adds will

work

• (^) Note: Carry will ripple

through the stages, one

at a time

• (^) Ripple-Carry Adder

Subtraction

a input b input Cin input output A 0 0 A 0 B 0 B A 0 1 A+ A 0 0 A – 1* A B 1 A+B+ A B* 1 A – B A B 0 A + B A B* 0 A – B -