c - 内联汇编中的 Printf

Question

我正在尝试编写一个交换两个值的内联 assebly 函数。(并且我正在使用扩展的 ASM 格式)

此代码有效:

#include <stdio.h>
void Exchange(int *x, int *y)
{
printf("In Exchange function: Before exchange x is: %d\n",*x);
printf("In Exchange function: Before exchange y is: %d\n",*y);

asm("xchg %%eax,%%edx\n\t" \
:"+a"(*x),"+d"(*y));

printf("In Exchange function: After exchange x is: %d\n",*x);
printf("In Exchange function: After exchange y is: %d\n",*y);


}


int main()
{
int x=20;
int y=30;
printf("In main: Before exchange x is: %d\n",x);
printf("In main: Before exchange y is: %d\n",y);
Exchange(&x,&y);
printf("In main: After exchange x is: %d\n",x);
printf("In main: After exchange y is: %d\n",y);


return 0;
}

但是当我尝试像下面这样在完整的汇编中编写它时，我得到了段错误(核心转储)错误。

void Exchange(int *x, int *y)
{
asm("subl $8,%%esp\n\t" \
    "movl %%eax,4(%%esp)\n\t" \
    "movl %%edx,(%%esp)\n\t" \
    "call printf\n\t" \
    "addl $8,%%esp\n\t" \   
    "xchg %%eax,%%edx\n\t" \
    "subl $8,%%esp\n\t" \
    "movl %%eax,4(%%esp)\n\t" \
    "movl %%edx,(%%esp)\n\t" \
    "call printf\n\t" \
    "addl $8,%%esp\n\t" \
:"+a"(*x),"+d"(*y));

}
 int main() 
{   
int x=20;
int y=30; 
printf("In main: Before exchange x is: %d\n",x);
printf("In main: Before exchange y is: %d\n",y);
Exchange(&x,&y);
printf("In main: After exchange x is: %d\n",x);
printf("In main: After exchange y is: %d\n",y);


return 0;
}

我们不允许在汇编部分使用 printf 函数吗？

Answer 1

您的 asm 代码使用两个整数参数调用 printf —— 没有格式字符串。因此它尝试将第一个整数取消引用为指向格式字符串的指针并崩溃。

此外，调用 printf 会破坏 %eax 和 %edx 中的值，因为它们不会在标准 x86 调用约定中跨调用保留。