1. 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
One
001
Zero + one = one.
Makes sense!

2. Add one repeatedly, use up all possible patterns:
Zero
Called the
Unsigned Integer System
No negative integers!

3. Two additions: 5 9

4. Two additions: 2 010 + 3 + 011 5 101 Yes! 5 = 101 4 100 + 5 + 101
Yes! 5 = 101
9  001; 001 represents one.
is = 1???

5. Addition of unsigned integers
Error
detected by presence of
"carry"

6. How do we subtract unsigned integers?
We need the concept of the
"Two's complement"

7. One's complement Take any string Invert every bit
This is One's complement
"NOT"

8. Two's complement Given a string One's complement then add one
This is called
two's complement

9. To subtract unsigned A-B
Perform
A + 2's comp (B)
= A + Not (B) + 1

10. Example:
010
110
=2; Good!
Carry!
=6; BAD!
No Carry!

11. 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

12. 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!

13. 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

14. Repeatedly add -1: Zero 000 - 1 111 - 2 110 Less than - 3 101 zero
Less than
zero
No!
High order bit called "sign bit"

15. Signed 3-bit integers 3 011 2 010 1 001 0 000 -1 111 -2 110 -3 101
Not symmetrical
around zero!!!

16. Sign bit The high order bit in a number Also called "N"-bit
Value is negative when
this bit is "1"

17. Let's try A + B
(-1) 111
*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

18. Some additions A
+(-3) (-4) 1 0 0
+(-2)

19. Some additions B
+(-3) (-4) 1 0 0
-4 C C
+(-2)

20. Some additions C
OK BAD
+(-3) (-4) 1 0 0
-4 C C
+(-2)
OK OK

21. Some additions D
OK BAD
+(-3) (-4) 1 0 0
-4 C C
+(-2)
OK OK

22. 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)

23. Some subtractions A 3 0 1 1 3 0 1 1 - 1 + 1 1 1 -(-1) + 0 0 1 2 4
(-1)
- (-1) (-1)

24. Some subtractions B 3 0 1 1 3 0 1 1 - 1 + 1 1 1 -(-1) + 0 0 1
(-1)
2 C
- (-1) (-1)
0 C
-4 C C

25. Some subtractions C 3 0 1 1 3 0 1 1 - 1 + 1 1 1 -(-1) + 0 0 1
(-1)
2 C
OK BAD
- (-1) (-1)
0 C
OK OK
-4 C C
OK BAD

26. Some subtractions D 3 0 1 1 3 0 1 1 - 1 + 1 1 1 -(-1) + 0 0 1
(-1)
2 C
OK BAD
- (-1) (-1)
0 C
OK OK
-4 C C
OK BAD

27. Error during signed subtraction:
R = A - B
A, B different sign
and
B, R same sign
called overflow

28. Arithmetic- logic unit (ALU)
Condition codes
C, V, N, Z 32 A 32 C 32 B
Operation
C = carry
V = overflow
N = sign bit of R
Z = 1 if R = 0

29. Compare two unsigned numbers?
is A < B ?
Easy! Compute A - B and examine carry
But – to compare two signed numbers?
Most common mistake:
Compute R = A - B, then look at sign of R.
If R < 0 then A < B (N-bit)
Not good enough!

30. To compare two signed numbers:
What about A = - 4 B = 3 (- 4) c 001
Assumption: “If R neg then A < B”
We conclude A  B, that is
- 4  3
Wrong!

31. Some examples A 3 0 1 1 3 0 1 1 - 1 + 1 1 1 -(-1) + 0 0 1
(-1)
- (-1) (-1)

32. Some examples B 3 0 1 1 3 0 1 1 - 1 + 1 1 1 -(-1) + 0 0 1
(-1)
2 C
3 < +1? No! < -1? No!
- (-1) (-1)
0 C
-1 < -1? No! < -1? No!
-4 C C
-3 < +1? Yes! < 1? Yes!

33. Some examples C 3 0 1 1 3 0 1 1 - 1 + 1 1 1 -(-1) + 0 0 1
(-1)
2 C
3 < +1? No! < -1? No!
N = 0, V = N = 1, V = 1
- (-1) (-1)
0 C
-1 < -1? No! < -1? No!
N = 0, V = N = 0, V = 0
-4 C C
-3 < +1? Yes! < 1? Yes!
N = 1, V = N = 0, V = 1

34. To compare signed numbers:
Compute R = A - B
A < B true if N and V are different
A<B = exor(N,V) after computation