在哪里可以找到在GCC汇编输出中调用的过程的汇编代码? [英] Where do I find the assembly code for a procedure called in the GCC assembly output?

问题描述

我一直在玩GCC的Assembly Output开关:

I've been playing around with the assembly output switch for GCC:

gcc -S -c helloworld.c

helloworld.c :

#include <stdio.h>

int main(void){
    printf("Hello World!\n");
    return 0;
}

helloworld.s :

    .file   "helloworld.c"
    .section    .rodata
.LC0:
    .string "Hello World!"
    .text
    .globl  main
    .type   main, @function
main:
.LFB0:
    .cfi_startproc
    pushq   %rbp
    .cfi_def_cfa_offset 16
    .cfi_offset 6, -16
    movq    %rsp, %rbp
    .cfi_def_cfa_register 6
    movl    $.LC0, %edi
    call    puts
    movl    $0, %eax
    popq    %rbp
    .cfi_def_cfa 7, 8
    ret
    .cfi_endproc
.LFE0:
    .size   main, .-main
    .ident  "GCC: (Debian 4.7.2-5) 4.7.2"
    .section    .note.GNU-stack,"",@progbits

在输出的 helloworld.s 文件中，我注意到输出"Hello World！"的汇编命令.文字很简单:

In the output helloworld.s file, I notice that the assembly command to output the "Hello World!" text is simply:

call puts

但是， helloworld.s1 文件中没有 put 过程程序集代码.我在哪里可以查看此汇编代码?

However, the helloworld.s1 file doesn't have the puts procedure assembly code within it. Where would I be able to view this assembly code?

推荐答案

作为用户展开:

这是因为您的C程序期望与标准C链接运行时库，其中包含 puts().您当然可以挖掘图书馆如果需要的话，请在系统上将其分解为文件.

This is because your C program is expecting to be linked with the standard C runtime library, which has puts() in it. You can of course dig up the library file on your system as disassemble it, if you feel like it.

因此，我使用 objdump 来反汇编 libc.a ，并在< puts> 部分下找到了汇编代码: objdump-d/usr/lib/x86_64-linux-gnu/libc.a

So I used objdump to disassemble libc.a and found the assembly code under the <puts> section: objdump -d /usr/lib/x86_64-linux-gnu/libc.a

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