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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