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Possible Duplicate:
x86 assembly registers — Why do they work the way they do?

I've compiled the following program:

#include <stdio.h>

int square(int x) {
    return x * x;
}

int main() {
    int y = square(9);
    printf("%d\n", y);

    return 0;
}

two times with different options on OSX with GCC 4.2.1:

gcc foo.c -o foo_32.s -S -fverbose-asm -m32 -O1

gcc foo.c -o foo_64.s -S -fverbose-asm -m64 -O1

The result for 32 bit:

_square:                                ## @square
## BB#0:                                ## %entry
    pushl   %ebp
    movl    %esp, %ebp
    movl    8(%ebp), %eax
    imull   %eax, %eax
    popl    %ebp
    ret

And for 64-bit:

_square:                                ## @square
Leh_func_begin1:
## BB#0:                                ## %entry
    pushq   %rbp
Ltmp0:
    movq    %rsp, %rbp
Ltmp1:
    movl    %edi, %eax
    imull   %eax, %eax
    popq    %rbp
    ret

As is evident, the 32-bit version retrieves the parameter from the stack, which is what one would expect with cdecl. The 64-bit version however uses the EDI register to pass the parameter.

Doesn't this violate the System V AMD64 ABI, which specifies that the RDI, RSI, RDX, RCX, R8, R9, XMM0–7 registers should be used? Or is this only the case for true 64-bit values like long?

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1 回答 1

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EDI只是它的下半部分，RDI所以编译器将参数传入RDI，但参数只有 32 位长，所以它只占用了寄存器的一半。

于 2012-09-26T16:09:38.757 回答