C中的内联汇编

Question

非常自我解释的代码.为什么不起作用!

#include <stdio.h>

int main() {
    __asm__("number dw 0"); // declare number?
    printf("%d",number);
    __asm__("mov %eax,number"
            "inc %eax"
            "mov number,%eax");
    printf("%d",number);
    return 0;
}

cc     ex1.c   -o ex1
ex1.c: In function ‘main’:
ex1.c:22:17: error: ‘number’ undeclared (first use in this function)
ex1.c:22:17: note: each undeclared identifier is reported only once for each function it appears in
make: *** [ex1] Error 1

谢谢.

我有很多知识空白要填补... gcc手册让我对内联汇编问题感到困惑,谷歌搜索教程结果......

在intel i7处理器上工作

Answer 1

使用此语法,您可以访问C内联程序集中声明的变量

#include <stdio.h>

int main() {
    int number = 0;
    printf("%d\n",number);
    asm(
        "mov %[number],%%eax\n"
        "inc %%eax\n"
        "mov %%eax,%[number]\n"
        : [number] "=m" (number) : "m" (number) : "eax", "cc" );
    printf("%d\n",number);
    return 0;
}

您可以通过在输入上指定约束来让编译器为您加载number到eax寄存器中"a"

#include <stdio.h>

int main() {
    int number = 0;
    printf("%d\n",number);
    asm(
        "inc %%eax\n"
        "mov %%eax,%[number]\n"
        : [number] "=m" (number) : "a" (number) : "cc" );
    printf("%d\n",number);
    return 0;
}

由于x86 inc指令可以直接在内存上运行,因此可以将其减少到此

#include <stdio.h>

int main() {
    int number = 0;
    printf("%d\n",number);
    asm(
        "incl %[number]\n" /* incl -> "long" (32-bits) */
        : [number] "=m" (number) : "m" (number) : "cc" );
    printf("%d\n",number);
    return 0;
}

有关更多信息,请参阅gcc文档:

6.41带C表达式操作数的汇编器指令