This is a possible assembly code of int x = plus_10(40);
push 40 ; push argument
call plus_10 ; call function
retadd: add esp, 4 ; clean up stack (dummy pop)
; result of the function call is in EAX, per the calling convention
; if compiled without optimization, the caller might just store it:
mov DWORD PTR [ebp-x], eax ; store return value
; (in eax) in x
Now when you call plus_10
, the address retadd
is pushed on the stack by the call
instruction. It's effectively a push
+jmp
, and ret
is effectively pop eip
.
So your stack looks like this in the plus_10
function:
| ... |
+--------+
| 40 | <- ESP+4 points here (the function argument)
+--------+
| retadd | <- ESP points here
+--------+
ESP
points to a memory location that contains the return address.
Now if you use pop edx
the return address goes into edx
and the stack looks like this:
| ... |
+--------+
| 40 | <- ESP points here
+--------+
Now if you execute ret
at this point, the program will actually jump to address 40 and most likely segfault or behave in some other unpredictable way.
The actual assembly code generated by the compiler may be different, but this illustrates the problem.
BTW, a more efficient way to write your function is this: it's what most compilers would do with optimization enabled, for a non-inline version of this tiny function.
global plus_10
plus_10:
mov eax, [esp+4] ; retval = first arg
add eax, 10 ; retval += 10
ret
This is smaller and slightly more efficient than
mov eax, 10
add eax, [esp+4] ; decode to a load + add.
ret
pop edx
moves the stack pointer,mov edx, [esp+4]
doesn't. Normally in C it's up to the caller to clean the stack. – Antagonistpop edx
actually pops the return adress from the stack and when theret
is executed the processor jumps to whatever address is on the stack – Antagonist