Question

如果使用Windows x86上的GCC编译器进行编译，我有以下程序（AT＆T汇编语法）可以完美地运行：

LC0:
    .ascii "Hello, world!\0"
.globl  _main
_main:
    pushl   %ebp
    movl    %esp, %ebp
    andl    $-16, %esp
    subl    $16, %esp
    call    ___main
    movl    $LC0, (%esp)
    call    _puts
    movl    $0, %eax
    leave
    ret

这个简单程序可以在C程序中执行吗？我尝试了以下方法：

#include <stdlib.h>

int main()
{

    __asm__ ("LC0:\n\t"
             ".ascii 'Welcome Message\0'\n\t"
             "LC1:\n\t"
             ".ascii 'Hello\0'\n\t"
             "_main:\n\t"
             "LFB11:\n\t"
             "leal  4(%esp), %ecx\n\t"
             "andl  $-16, %esp\n\t"
             "pushl -4(%ecx)\n\t"
             "pushl %ebp\n\t"
             "movl  %esp, %ebp\n\t"
             "pushl %ecx\n\t"
             "subl  $20, %esp\n\t"
             "call  ___main\n\t"
             "movl  $1, 12(%esp)\n\t"
             "movl  $LC0, 8(%esp)\n\t"
             "movl  $LC1, 4(%esp)\n\t"
             "movl  $0, (%esp)\n\t"
             "call  _MessageBoxA@16\n\t"
             "subl  $16, %esp\n\t"
             "movl  $0, %eax\n\t"
             "movl  -4(%ebp), %ecx\n\t"
             "leave\n\t"
             "leal  -4(%ecx), %esp\n\t"
             "ret\n\t");


    return 0;

}

我遇到一个错误：

错误：行尾有垃圾，第一个无法识别的字符是'8'

Answer 1

这很好：

__asm__(
"LC0:\n"
"    .ascii \"Hello, world!\\0\"\n"
".globl  _main\n"
"_main:\n"
"    pushl   %ebp\n"
"    movl    %esp, %ebp\n"
"    andl    $-16, %esp\n"
"    subl    $16, %esp\n"
"    call    ___main\n"
"    movl    $LC0, (%esp)\n"
"    call    _puts\n"
"    movl    $0, %eax\n"
"    leave\n"
"    ret\n"
);

仅用C字符串直义转义了双引号字符串和其他地方的's/^/"/;s/$/\\n"/'。

我的gcc的汇编器不接受单引号中的字符串文字，

LC0:
    .ascii 'Hello, world!\0'
.globl  _main
_main:
    pushl   %ebp
    movl    %esp, %ebp
    andl    $-16, %esp
    subl    $16, %esp
    call    ___main
    movl    $LC0, (%esp)
    call    _puts
    movl    $0, %eax
    leave
    ret

因此，如果您通过__asm__提供它们，我不明白为什么它应该开始接受它们。

C程序（GCC编译器）中的AT＆T汇编语法？

1 个答案: