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Y86 Assembly Programming Example: Program Structure

This example showcases the program structure using Y86 assembly language, starting at address 0x0 with stack at 0x100. Learn about initializing arrays, function calls, stack setup, and more.

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Y86 Assembly Programming Example: Program Structure

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  1. asum.ys A Y86 Programming Example

  2. Y86 Sample Program Structure Program starts at address 0x0 Stack starts at address 0x100 Initialize the array (data) .pos 0 init: # Initialization ... call Main halt .align 4 # Program data array: ... Main: # Main function ... call Sum ... ret Sum: # Length function ... ret .pos 0x100 # Place stack Stack:

  3. 3.init: irmovl Stack, %esp# Set up stack pointer 4. irmovl Stack, %ebp# Set up base pointer • %esp • %ebp Because line 46 and line 47 makes the label “Stack” at address 0x100, So the irmovls make %esp == 0x100 and %ebp == 0x100 now.

  4. 5. call main • %esp • %esp • %ebp • %esp 0x11 0x11 is the address of “6: halt”.

  5. 15. Main: pushl %ebp • %ebp • %esp • %esp 0x11 • %esp 0x100

  6. 16. rrmovl %esp,%ebp • %ebp • %ebp 0x11 • %ebp • %esp 0x100

  7. 17.irmovl $4,%eax 18.pushl %eax# Push 4 0x11 • %ebp • %esp • %esp 0x100 • %esp 4 4 is the value of count (4 elements in the array).

  8. 19.irmovl array,%edx 20.pushl %edx • %ebp • %esp • %esp • %esp 0x14 0x14 is the first element’s address of the array. Here we finished storing the arguments to be passed.

  9. 21.call Sum • %ebp • %esp • %esp • %esp 0x3d 0x3d is the address of “22: rrmovl %ebp,%esp”.

  10. 27.Sum: pushl %ebp 28. rrmovl %esp,%ebp • %ebp • %ebp • %esp • %esp • %esp • %ebp 0xf8

  11. 29.mrmovl 8(%ebp),%ecx # ecx = Start 30. mrmovl 12(%ebp),%edx # edx = Count

  12. 31.xorl %eax,%eax# sum = 0

  13. 32.andl %edx,%edx# Set condition codes

  14. 33.je End It’s like a if statement for checking. Why need line 32 and 33? How to improve them?

  15. 34.Loop: mrmovl (%ecx),%esi# get *Start Note that the parenthesis of(%ecx) is necessary.

  16. 35. addl %esi,%eax# add to sum 36. irmovl $4,%ebx #

  17. 37. addl %ebx,%ecx# Start++ 38. irmovl $-1,%ebx #

  18. 39. addl %ebx,%edx# Count-- 40. jne Loop # Stop when 0

  19. 34. Loop: mrmovl (%ecx),%esi# get *Start 35. addl %esi,%eax# add to sum

  20. 36. irmovl $4,%ebx # 37. addl %ebx,%ecx# Start++

  21. 38. irmovl $-1,%ebx # 39. addl %ebx,%edx# Count--

  22. 40. jne Loop # Stop when 0

  23. 34. Loop: mrmovl (%ecx),%esi# get *Start 35. addl %esi,%eax# add to sum

  24. 36. irmovl $4,%ebx # 37. addl %ebx,%ecx # Start++

  25. 38. irmovl $-1,%ebx # 39. addl %ebx,%edx # Count--

  26. 30. jne Loop # Stop when 0

  27. 34. Loop: mrmovl (%ecx),%esi# get *Start 35. addl %esi,%eax# add to sum

  28. 36. irmovl $4,%ebx # 37. addl %ebx,%ecx# Start++

  29. 38. irmovl $-1,%ebx # 39. addl %ebx,%edx# Count--

  30. 40. jne Loop # Stop when 0

  31. 41. End: rrmovl %ebp,%esp

  32. 42. popl %ebp • %ebp • %esp • %esp • %esp • %ebp • %ebp

  33. 0x7c 43. ret • %ebp • %esp • %esp • %esp 0x3d 0x3d

  34. 22. rrmovl %ebp,%esp • %esp • %esp • %esp

  35. 23. popl %ebp • %ebp • %esp • %esp • %esp • %ebp • %ebp

  36. 0x41 24. ret • %esp • %esp • %esp 0x11 0x11 0x11

  37. 6. halt

  38. A small question How to change line 32 and line 33 so that if count <= 0 the loop will not execute? 32: andl %edx, %edx 33: je End rrmovl %edx, %ebx # use %ebx as temporary place subl %eax, %ebx # here %eax == 0, so calculate %ebx - 0 jle End

  39. Some Takeaways In the called function: Fun: pushl %ebp rrmovl %esp,%ebp # Set up the stack space Before ret operation: rrmovl %ebp,%esp popl %ebp Use conditional jumps to implement if statement and loops call operation: push the address of next instruction onto the stack ret operation: pop stack top value to PC (program counter)

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