1. Computer Engineering page 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 000 One + 001 001 Zero + one = one. Makes sense!

2. Computer Engineering page 2 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!

3. Computer Engineering page 3 Two additions: 2 010 + 3 + 011 5 4 100 + 5 + 101 9

4. Computer Engineering page 4 Two additions: 2 010 + 3 + 011 5 101 Yes! 5 = 101 4 100 + 5 + 101 9 001 But 001 represents one. is 4 + 5 = 1???

5. Computer Engineering page 5 Addition of unsigned integers Error detected by presence of “carry”

6. Computer Engineering page 6 How do we subtract unsigned integers? We need the concept of the; “ Two’s complement”

7. Computer Engineering page 7 One’s complement Take any string; Invert every bit; 0 1 This is One’s complement. "NOT”.

8. Computer Engineering page 8 Two’s complement Given a string; One's complement; then add one. This is called; two’s complement

9. Computer Engineering page 9 To subtract unsigned A- B Perform: A + 2’s compl (B) = A + Not (B) + 1

10. Computer Engineering page 10 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!

11. Computer Engineering page 11 Subtraction 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. Computer Engineering page 12 Conclusion For unsigned arithmetic we are interested in carry Pay attention! I never used the word "overflow" that's something completely different. Also notice: –3-bit operands gave 3-bit results. –Don't be tempted to write that 4'th bit down!

13. Computer Engineering page 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. Computer Engineering page 14 Repeatedly add -1: Zero000 - 1 111 - 2 110Less than - 3 101zero - 4 110 - 5 011No! High order bit called "sign bit"

15. Computer Engineering page 15 Signed 3-bit integers 3011 2010 1001 0000 -1111 -2110 -3101 -4100 Not symmetrical around zero!!!

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

17. Computer Engineering page 17 Let's try A + B 10011001 +2010 +(-1)111 30110 *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. Computer Engineering page 18 Some additions A 1 0 0 120 1 0 +2 0 1 0 + 20 1 0 3 4 -1 1 1 1 -11 1 1 +(-3) 1 0 1 +(-4)1 0 0 -4 -5 1 0 0 1 -21 1 0 +(-2) 1 0 0 +10 0 1 -1 -1

19. Computer Engineering page 19 Some additions B 1 0 0 120 1 0 +2 0 1 0 + 20 1 0 3 0 1 1 4 1 0 0 -1 1 1 1 -11 1 1 +(-3) 1 0 1 +(-4)1 0 0 -4 C 1 0 0 -5 C 0 1 1 1 0 0 1 -21 1 0 +(-2) 1 0 0 +10 0 1 -1 1 1 1 -11 1 1

20. Computer Engineering page 20 Some additions C 1 0 0 120 1 0 +2 0 1 0 + 20 1 0 3 0 1 1 4 1 0 0 3 -4 OK BAD -1 1 1 1 -11 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 -21 1 0 +(-2) 1 0 0 +10 0 1 -1 1 1 1 -11 1 1 -1 -1 OK OK

21. Computer Engineering page 21 Some additions D 1 0 0 120 1 0 +2 0 1 0 + 20 1 0 3 0 1 1 4 1 0 0 3 -4 OK BAD -1 1 1 1 -11 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 -21 1 0 +(-2) 1 0 0 +10 0 1 -1 1 1 1 -11 1 1 -1 -1 OK OK

22. Computer Engineering page 22 Error during signed addition: R = A + B A, B same sign and R opposite sign called overflow Notice: Mathematically, 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. Computer Engineering page 23 Some subtractions A 3 0 1 130 1 1 - 1 + 1 1 1 -(-1) + 0 0 1 2 4 -1 1 1 1 10 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

24. Computer Engineering page 24 Some subtractions B 3 0 1 130 1 1 - 1 + 1 1 1 -(-1) + 0 0 1 2 C 0 1 0 4 1 0 0 -1 1 1 1 10 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

25. Computer Engineering page 25 Some subtractions C 3 0 1 130 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 10 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

26. Computer Engineering page 26 Some subtractions D 3 0 1 130 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 10 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

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

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

29. Computer Engineering page 29 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!

30. Computer Engineering page 30 To compare two signed numbers: What about –A = - 4 –B = 3 – - 4 100 – - 3 + 101 – c 001 “If R neg then A < B” – We conclude A ≥ B, that is - 4 ≥ 3 Wrong!

31. Computer Engineering page 31 Some examples A 3 0 1 130 1 1 - 1 + 1 1 1 -(-1) + 0 0 1 -1 1 1 1 10 0 1 - (-1) + 0 0 1 - 3 1 0 1 - 4 1 0 0 - 1 + 1 1 1 - 1 + 1 1 1

32. Computer Engineering page 32 Some examples B 3 0 1 130 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 10 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!

33. Computer Engineering page 33 Some examples C 3 0 1 130 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 10 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

34. Computer Engineering page 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