Why is the compiler generating a push/pop instruction pair?

Question

I compiled the code below with the VC++ 2010 compiler:

__declspec(dllexport)
unsigned int __cdecl __mm_getcsr(void) { return _mm_getcsr(); }

and the generated code was:

push ECX
    stmxcsr [ESP]
    mov EAX, [ESP]
pop ECX
retn

Why is there a push ECX/pop ECX instruction pair?

score 12 · Accepted Answer · answered Jan 14 '12 at 15:24

12

The compiler is making room on the stack to store the MXCSR. It could have equally well done this:

sub esp,4
stmxcsr [ESP]
mov EAX, [ESP]
add esp,4
retn

But "push ecx" is probably shorter or faster.

answered Jan 14 '12 at 15:24

Robᵩ

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D'oh... totally missed that. :) Thanks a lot. – user541686 Jan 14 '12 at 15:25
And how does that explain the pop? – CodesInChaos Jan 14 '12 at 15:26
@CodeInChaos - the `pop` restores the stack pointer, just as `add esp,4` would. – Robᵩ Jan 14 '12 at 15:29
3

but wouldn't `pop EAX` instead of `mov EAX, [ESP]; pop ECX` be even better? – CodesInChaos Jan 14 '12 at 15:32

score 3 · Answer 2 · answered Jan 14 '12 at 15:26

The push here is used to allocate 4 bytes of temporary space. [ESP] would normally point to the pushed return address, which we cannot overwrite.

ECX will be overwritten here, however, ECX is a probably a volatile register in the ABI you're targeting, so functions don't have to preserve ECX.

The reason a push/pop is used here is a space (and possibly speed) optimization.

score 0 · Answer 3 · answered Jan 14 '12 at 15:30

0

It creates an top-of-stack entry that ESP now refers to as the target for the stmxcsr instruction. Then the result is stored in EAX for the return.

answered Jan 14 '12 at 15:30

Mark Tolonen

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Why is the compiler generating a push/pop instruction pair?

3 Answers3