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| | |} | | |} |
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| − | =Exercise 5= | + | =Exercise 5 (Push and Pop)= |
| | What are the numbers stored in eax and ebx when the loop terminates? If we assume that the default stack is 2 KB long, what is the largest number of times the loop can iterate before the stack overflows (trick question :-)? | | What are the numbers stored in eax and ebx when the loop terminates? If we assume that the default stack is 2 KB long, what is the largest number of times the loop can iterate before the stack overflows (trick question :-)? |
| | {| | | {| |
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| | mov ebx, 0 | | mov ebx, 0 |
| | mov ecx, 10 | | mov ecx, 10 |
| − | for: push ebx | + | for: push ebx |
| | call func1 | | call func1 |
| | pop ebx | | pop ebx |
Revision as of 09:30, 26 October 2012
Back to Weekly Schedule.
Note: Highlight the white section on the right hand-side of each code section to see the solution.
Exercise 1
What is the behavior of this loop?
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mov eax, 0
mov ecx, 10
for: call func1
...
loop for
jmp theEnd
func1: add eax, 1
ret
theEnd: ...
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The loop goes 10 times and the function is called 10 times, adding 1 to eax every time. Eax ends with 10 in it.
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Exercise 2
What is the behavior of the loop below? Could it ever be endless?
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mov eax, 0
mov ecx, 10
for: call func1
...
loop for
jmp theEnd
func1: sub ecx, 1
ret
theEnd: ...
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The function decrements ecx by 1 every time through the loop. So ecx is decremented once by the function, once by the loop instruction. The loop is going to go twice as fast.
The loop instruction stops if ecx is 1 when the instruction is executed. We risk endless looping if ecx is either 2 or 0 when the loop instruction executes. This will happen if we start with an odd number in ecx outside the loop.
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Exercise 3
What is the behavior of this program? Draw the stack as the processor executes this program:
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mov eax, 0
mov ebx, 0
mov ecx, 10
for: call func1
...
loop for
func1: add eax, 1
call func2
ret
func2: add ebx, 1
ret
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Just go through the motion and draw the stack. The return address of the instruction after the call gets pushed in the stack every time the processor executes a call. It is popped out of the stack every time the processor executes a ret.
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Exercise 4
Same question, but now observe that the programmer forgot the ret instruction at the end of the first function.
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mov eax, 0
mov ebx, 0
mov ecx, 10
for: call func1
...
loop for
func1: add eax, 1
call func2
func2: add ebx, 1
ret
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Just go through the motion and draw the stack. Now the problem is that when the main program calls func1 the first time, the return address for the ellipses is pushed in the stack, then the processor starts executing the add eax,1 function. Then it executes the call func2 instruction and pushes the address of the add ebx,1 instruction in the stack. Then, when it executes the ret instruction, the address of add ebx,1 is popped out, and ebx is incremented by 1. Then the ret is executed and we return to the main function. The end result is that func1 is executed once, and func2 twice!
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Exercise 5 (Push and Pop)
What are the numbers stored in eax and ebx when the loop terminates? If we assume that the default stack is 2 KB long, what is the largest number of times the loop can iterate before the stack overflows (trick question :-)?
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mov eax, 0
mov ebx, 0
mov ecx, 10
for: push ebx
call func1
pop ebx
loop for
...
func1: add eax, 1
call func2
ret
func2: add ebx, 1
ret
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The loop does not modify ebx!
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Exercise 6
Same question, but now observe that we pop eax, not ebx...
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mov eax, 0
mov ebx, 0
mov ecx, 10
for: push ebx
call func1
pop eax
loop for
...
func1: add eax, 1
call func2
ret
func2: add ebx, 1
ret
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This one requires the use of the debugger for fully understanding it! :-)
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Exercise 7
What do these 2 code sections do?
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push eax
mov eax, ebx
pop ebx
and
xor eax, ebx
xor ebx, eax
xor eax, ebx
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They both swap the contents of eax and ebx.
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