1. CPSC355 Week 3 Hoang Dang

2. Week 3 Assignment 1 Do/while loop Assignment 2 Binary Bitwise operation

3. Assignment 1 Overall very good Problem area: – Conforming to specification – Documenting/Commenting – Removing NOP

4. Conforming to Specification Many missed some parts of the specification – Put maximum in %L0 – Print x, y, and max on each iteration – Print registers at key points

5. Suggestion Read spec carefully Highlight key tasks Check against marking guide

6. Documentation What does the program do? Who coded this? When was this done?

7. Commenting Why, not What – In general, tell me why you are doing it. Not what the code is doing – We assume the reader knows the language to determine what the statements do

8. Commenting… mov 10, %x !move 10 to x mov %x, %o0 !move x to %o0 mov 5, %o1 !move 5 to %o1 call.mul !call multiply mov 10, %x !initialize x mov %x, %o0 !passing x as argument to multiply mov 5, %o1 !passing 5 as argument to multiply(x*5) call.mul

9. NOP Many have problems removing the nop – SPARC instruction processing: F – Fetch the instruction E – Execute the instruction M – Load/Store data to memory W – Write to register SPARC is a pipelined architecture

10. NOP… Linear execution: Pipelined execution: FEMWFEMWF 4 cycle, 1 instruction FEMWFEMWFE FEMWFEMWF FEMWFEMW FEMWFEM 1 cycle, 1 instruction

11. Call and Branch FEMWFEMWFE FEMWFEMWF FEMWFEMW FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 Bge else Mov 3,%l0 Ba next Else: mov 6,%l0 Next:

12. Call and Branch FEMWFEMWFE FEMWFEMWF FEMWFEMW FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 Bge else Mov 3,%l0 Ba next Else: mov 6,%l0 Next: cmp

13. Call and Branch FEMWFEMWFE FEMWFEMWF FEMWFEMW FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 Bge else Mov 3,%l0 Ba next Else: mov 6,%l0 Next: cmp bge

14. Call and Branch FEMWFEMWFE FEMWFEMWF FEMWFEMW FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 Bge else Mov 3,%l0 Ba next Else: mov 6,%l0 Next: cmp bge Mov Doesn’t move to else until end of execution But we’ve already fetched the move instruction belonging to the if, not else! Mov 3

15. Call and Branch FEMWFEMWFE FEMWFEMWF FEMWFEMW FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 Bge else Mov 3,%l0 Ba next Else: mov 6,%l0 Next: cmp bge Mov 3 Mov 6

16. Call and Branch FEMWFEMWFE FEMWFEMWF NOP FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 bge else nop mov 3,%l0 ba next nop else: mov 6,%l0 next:

17. Call and Branch FEMWFEMWFE FEMWFEMWF NOP FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 bge else nop mov 3,%l0 ba next nop else: mov 6,%l0 next: cmp bge

18. Call and Branch FEMWFEMWFE FEMWFEMWF NOP FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 bge else nop mov 3,%l0 ba next nop else: mov 6,%l0 next: cmp bge

19. Call and Branch FEMWFEMWFE FEMWFEMWF NOP FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 bge else nop mov 3,%l0 ba next nop else: mov 6,%l0 next: cmp bge mov 6,%l0 The nop allows us to delay fetching of the next instruction

20. Removing NOP Assembly cmp %l0,5 bge else nop mov 3,%l0 ba next nop else: mov 6,%l0 next: General rule: Replace NOP with next instruction after the branch/call. Next Instruction Assembly cmp %l0,5 bge else nop mov 3,%l0 ba next nop else: mov 6,%l0 next: Assembly cmp %l0,5 bge else mov 6,%l0 mov 3,%l0 ba next nop else: next:

21. Removing NOP FEMWFEMWFE FEMWFEMWF FEMWFEMW FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 bge else mov 6,%l0 mov 3,%l0 ba next nop else: next:

22. Removing NOP FEMWFEMWFE FEMWFEMWF FEMWFEMW FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 bge else mov 6,%l0 mov 3,%l0 ba next nop else: next: cmp %l0,5 bge else

23. Removing NOP FEMWFEMWFE FEMWFEMWF FEMWFEMW FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 bge else mov 6,%l0 mov 3,%l0 ba next nop else: next: cmp %l0,5 bge else mov 6,%l0

24. Removing NOP FEMWFEMWFE FEMWFEMWF FEMWFEMW FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 bge else mov 6,%l0 mov 3,%l0 ba next nop else: next: cmp %l0,5 bge else mov 6,%l0 We have a problem if x <5. We are always going to execute move 6,%l0!

25. Removing NOP FEMWFEMWFE FEMWFEMWF FEMWFEMW FEMWFEM C code If(x<5) x=3; Else x = 6; Assembly cmp %l0,5 Bge,a else mov 6,%l0 mov 3,%l0 ba next nop else: next: cmp %l0,5 bge else mov 3,%l0 We have a problem if x <5. We are always going to execute mov 6,%l0! We can annul it by placing an a after the branch.

26. Removing NOP How would we remove the following NOP set out_string,%o0 mov 6, %o1 call printf,0 nop mov 1, %g1 ta 0

27. Removing NOP How would we remove the following NOP set out_string,%o0 mov 6, %o1 call printf,0 nop mov 1, %g1 ta 0 Next instruction

28. Removing NOP How would we remove the following NOP set out_string,%o0 mov 6, %o1 call printf,0 nop mov 1, %g1 ta 0 Next instruction set out_string,%o0 mov 6, %o1 call printf,0 mov 1, %g1 ta 0

29. Do/While Loop Do first then check loop condition main() { int i=2; do { i--; }while(i>-2); } What is equivalent of this in SPARC?

30. Do/While Loop In SPARC: Mov 2,%i Do: sub i%,1,%i Test: cmp %i,-2 bg Do nop Next: NOP Removed: Mov 2,%i sub i%,1,%1 Test: cmp %i,-2 bg,a Test sub i%,1,%i Next:

31. Assignment 2 2 Parts: – CRC Checksum – Multiplication

32. Binary Base 2 [0,1] 100001111000000 Convert binary to decimal – 1011 = (1*2^0)+(1*2^1)+(0*2^2)+(1*2^3) Signed bit is used to tell negative numbers – 1000011 Low order bitsHigh order bits

33. Bitwise operation NOT – reverses the bits NOT 1001 = 0110 OR AND XOR 101 001 101 101 001 001 101 001 100

34. Bitwise operation C: NOT ~5 And 5 & 5 OR 5 | 5 XOR 5^5 SPARC: NOT not %r,%s And and %r,%s,%rd OR or %r,%s,%rd XOR xor %r,%s,%rd

35. Bit shifting We can shift bits left or right 0101011 1010110 0101011 0010101 Shift left by 1 Shift right by 1 C X << 1 SPARC sll %r, 1, %rd C X >> 1 SPARC srl %r, 1, %rd

36. Bit shifting There are two types of shift – Arithmetic – Keeps the signed bit intact – Logical – ignores the signed bit In SPARC – sll, srl are logical shifts – sra is the arithmetic shift