我对汇编非常陌生,下面的代码应该通过两个不同的函数交换两个整数:首先使用swap_c
然后使用swap_asm
.
但我怀疑,我是否需要push
(我的意思是保存)汇编代码之前寄存器的每个值和pop
稍后(就在返回之前)main
)。换句话说,如果我返回不同的寄存器内容(而不是像这样的关键内容),CPU会生我的气吗?ebp
or esp
;只是eax
, ebx
, ecx
& edx
)运行后swap_asm
功能?取消装配部分中的行注释是否更好?
这段代码对我来说运行良好,并且我设法减少了 27 行汇编代码C
代码减少到 7 行。
p.s.:系统是Windows 10,VS-2013 Express。
main.c
part
#include <stdio.h>
extern void swap_asm(int *x, int *y);
void swap_c(int *a, int *b) {
int t = *a;
*a = *b;
*b = t;
}
int main(int argc, char *argv[]) {
int x = 3, y = 5;
printf("before swap => x = %d y = %d\n\n", x, y);
swap_c(&x, &y);
printf("after swap_c => x = %d y = %d\n\n", x, y);
swap_asm(&x, &y);
printf("after swap_asm => x = %d y = %d\n\n", x, y);
getchar();
return (0);
}
assembly.asm
part
.686
.model flat, c
.stack 100h
.data
.code
swap_asm proc
; push eax
; push ebx
; push ecx
; push edx
mov eax, [esp] + 4 ; get address of "x" stored in stack into eax
mov ebx, [esp] + 8 ; get address of "y" stored in stack into ebx
mov ecx, [eax] ; get value of "x" from address stored in [eax] into ecx
mov edx, [ebx] ; get value of "y" from address stored in [ebx] into edx
mov [eax], edx ; store value in edx into address stored in [eax]
mov [ebx], ecx ; store value in ecx into address stored in [ebx]
; pop edx
; pop ecx
; pop ebx
; pop eax
ret
swap_asm endp
end