datorteknik integeraddsub bild 1 integer arithmetic depends what you mean by "integer"...

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Integer arithmetic

Depends what you mean by "integer"

Assume at 3-bit string.– Then we define

zero = 000

one = 001

Use zero, one and binary addition:

Zero 000

One + 001

001

Zero + one = one. Makes sense!

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Add one repeatedly, use up all possible patterns:

Zero 000

1 001

2 010

3 011

4 100

5 101

6 110

7 111

Called the

Unsigned Integer System

No negative integers!

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Two additions:

2 010

+ 3 + 011

5

4 100

+ 5 + 101

9

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Two additions:

2 010

+ 3 + 011

5 101

Yes! 5 = 101

4 100

+ 5 + 101

9 001

9 001; 001 represents one.

is 4 + 5 = 1???

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Addition of unsigned integers

Error detected by presence of

"carry"

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How do we subtract unsigned integers?

We need the concept of the

"Two's complement"

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One's complement

Take any string

Invert every bit

0 1

This is One's complement "NOT"

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Two's complement

Given a string

One's complement

then add one

This is called

two's complement

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To subtract unsigned A-B

Perform

A + 2's comp (B)

= A + Not (B) + 1

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Example:

5 101 101

- 3 - 011 + 100

2 + 1

010

3 011 011

- 5 - 101 + 010

- 2 + 1

110

Carry!

=2; Good!

No Carry!

=6; BAD!

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Subraction of unsigned integers

Error

detected by

absence of carry!

– Warning: Some machines invert the carry bit on subtraction

– So that "carry" => Error for both add and sub

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Conclusion

For unsigned arithmetic we are interested in carry

Pay attention!

I never used the word "overflow"thats something completely different.

Also notice:– 3-bit operands gave 3-bit results.

– Don't be tempted to write that 4'th bit down!

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How about negative numbers?

How should we represent -1?

How would we compute 0 - 1?0 + 2's compl (1)

We choose this as our "-1"

1 = 001

- 1 = 110

+ 1

111

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Repeatedly add -1:

Zero 000

- 1 111

- 2 110 Less than

- 3 101 zero

- 4 110

- 5 011 No!

High order bit called "sign bit"

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Signed 3-bit integers

3 011

2 010

1 001

0 000

-1 111

-2 110

-3 101

-4 100Not symmetrical around zero!!!

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Sign bit

The high order bit in a number

Also called "N"-bit

Value is negative when

this bit is "1"

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Let's try A + B

1 001 1 001

+2 010 +(-1) 111

3 011 0 *000

Both results is OK

But: Left case: no carry

Right case: carry

Conclusion: For signed addition carry is worthless

Same conclusion for signed subtraction

* carry

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Some additions A

1 0 0 1 2 0 1 0

+2 0 1 0 + 2 0 1 0

3 4

-1 1 1 1 -1 1 1 1

+(-3) 1 0 1 +(-4) 1 0 0

-4 -5

1 0 0 1 -2 1 1 0

+(-2) 1 1 0 +1 0 0 1

-1 -1

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Some additions B

1 0 0 1 2 0 1 0

+2 0 1 0 + 2 0 1 0

3 0 1 1 4 1 0 0

-1 1 1 1 -1 1 1 1

+(-3) 1 0 1 +(-4) 1 0 0

-4 C 1 0 0 -5 C 0 1 1

1 0 0 1 -2 1 1 0

+(-2) 1 1 0 +1 0 0 1

-1 1 1 1 -1 1 1 1

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Some additions C

1 0 0 1 2 0 1 0

+2 0 1 0 + 2 0 1 0

3 0 1 1 4 1 0 0

3 -4

OK BAD

-1 1 1 1 -1 1 1 1

+(-3) 1 0 1 +(-4) 1 0 0

-4 C 1 0 0 -5 C 0 1 1

-4 3

OK BAD

1 0 0 1 -2 1 1 0

+(-2) 1 1 0 +1 0 0 1

-1 1 1 1 -1 1 1 1

-1 -1

OK OK

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Some additions D

1 0 0 1 2 0 1 0

+2 0 1 0 + 2 0 1 0

3 0 1 1 4 1 0 0

3 -4

OK BAD

-1 1 1 1 -1 1 1 1

+(-3) 1 0 1 +(-4) 1 0 0

-4 C 1 0 0 -5 C 0 1 1

-4 3

OK BAD

1 0 0 1 -2 1 1 0

+(-2) 1 1 0 +1 0 0 1

-1 1 1 1 -1 1 1 1

-1 -1

OK OK

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Error during signed addition:

R = A + B

A, B same sign

and

R opposite sign

called overflow

Notice: Matematically, signed addition is the same as unsigned addition

The same is true for signed subtraction and unsigned subtraction

A - B –> A + (-B) –> A + 2's compl (B)

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Some subtractions A

3 0 1 1 3 0 1 1

- 1 + 1 1 1 -(-1) + 0 0 1

2 4

-1 1 1 1 1 0 0 1

- (-1) + 0 0 1 - (-1) + 0 0 1

0 2

- 3 1 0 1 - 4 1 0 0

- 1 + 1 1 1 - 1 + 1 1 1

-4 - 5

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Some subtractions B

3 0 1 1 3 0 1 1

- 1 + 1 1 1 -(-1) + 0 0 1

2 C 0 1 0 4 1 0 0

-1 1 1 1 1 0 0 1

- (-1) + 0 0 1 - (-1) + 0 0 1

0 C 0 0 0 2 0 1 0

- 3 1 0 1 - 4 1 0 0

- 1 + 1 1 1 - 1 + 1 1 1

-4 C 1 0 0 - 5 C 0 1 1

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Some subtractions C

3 0 1 1 3 0 1 1

- 1 + 1 1 1 -(-1) + 0 0 1

2 C 0 1 0 4 1 0 0

2 -4

OK BAD

-1 1 1 1 1 0 0 1

- (-1) + 0 0 1 - (-1) + 0 0 1

0 C 0 0 0 2 0 1 0

0 2

OK OK

- 3 1 0 1 - 4 1 0 0

- 1 + 1 1 1 - 1 + 1 1 1

-4 C 1 0 0 - 5 C 0 1 1

-4 3

OK BAD

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Some subtractions D

3 0 1 1 3 0 1 1

- 1 + 1 1 1 -(-1) + 0 0 1

2 C 0 1 0 4 1 0 0

2 -4

OK BAD

-1 1 1 1 1 0 0 1

- (-1) + 0 0 1 - (-1) + 0 0 1

0 C 0 0 0 2 0 1 0

0 2

OK OK

- 3 1 0 1 - 4 1 0 0

- 1 + 1 1 1 - 1 + 1 1 1

-4 C 1 0 0 - 5 C 0 1 1

-4 3

OK BAD

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Error during signed subtraction:

R = A - B

A, B different sign

and

B, R same sign

called overflow

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Arithmetic- logic unit (ALU)

C = carry

V = overflow

N = sign bit of R

Z = 1 if R = 0

32

32

32

A

B

C

Operation

Condition codesC, V, N, Z

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Compare two unsigned numbers?

is A < B ?

Easy! Compute A - B and examine carry

But – to compare two signed numbers?

is A < B ?

Most common mistake:– Compute R = A - B, then look at sign

of R.

– If R < 0 then A < B (N-bit)

Not good enough!

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To compare two signed numbers:

What about

A = - 4B = 3

(- 4) 100 - 3 + 101 c 001

Assumption:“If R neg then A < B”

We conclude A B, that is - 4 3

Wrong!

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Some examples A

3 0 1 1 3 0 1 1

- 1 + 1 1 1 -(-1) + 0 0 1

-1 1 1 1 1 0 0 1

- (-1) + 0 0 1 - (-1) + 0 0 1

- 3 1 0 1 - 4 1 0 0

- 1 + 1 1 1 - 1 + 1 1 1

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Some examples B

3 0 1 1 3 0 1 1

- 1 + 1 1 1 -(-1) + 0 0 1

2 C 0 1 0 4 1 0 0

3 < +1? No! 3 < -1? No!

-1 1 1 1 1 0 0 1

- (-1) + 0 0 1 - (-1) + 0 0 1

0 C 0 0 0 2 0 1 0

-1 < -1? No! 1 < -1? No!

- 3 1 0 1 - 4 1 0 0

- 1 + 1 1 1 - 1 + 1 1 1

-4 C 1 0 0 - 5 C 0 1 1

-3 < +1? Yes! -4 < 1? Yes!

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Some examples C

3 0 1 1 3 0 1 1

- 1 + 1 1 1 -(-1) + 0 0 1

2 C 0 1 0 4 1 0 0

3 < +1? No! 3 < -1? No!

N = 0, V = 0 N = 1, V = 1

-1 1 1 1 1 0 0 1

- (-1) + 0 0 1 - (-1) + 0 0 1

0 C 0 0 0 2 0 1 0

-1 < -1? No! 1 < -1? No!

N = 0, V = 0 N = 0, V = 0

- 3 1 0 1 - 4 1 0 0

- 1 + 1 1 1 - 1 + 1 1 1

-4 C 1 0 0 - 5 C 0 1 1

-3 < +1? Yes! -4 < 1? Yes!

N = 1, V = 0 N = 0, V = 1

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To compare signed numbers:

Compute R = A - B

A < B true if N and V are different

A<B = exor(N,V) after computation