gcc 给出了奇怪的 Intel 语法

发布于 2024-10-11 21:29:00 字数 738 浏览 2 评论 0原文

我一直试图通过使用编译器生成不同优化级别的各种 C 程序的汇编程序来更好地了解幕后发生的情况。有件事困扰我一段时间了。

当我按如下方式编译 tc 时，

gcc -S t.c

我得到 AT&T 语法的程序集，如下所示。

function:
    pushl   %ebp
    movl    %esp, %ebp
    movl    12(%ebp), %eax
    addl    8(%ebp), %eax
    popl    %ebp
    ret
    .size   function, .-function

当我使用 masm 参数进行编译时，如下所示：-

gcc -S t.c -masm=intel

我得到以下输出。

function:
    push    %ebp
    mov %ebp, %esp
    mov %eax, DWORD PTR [%ebp+12]
    add %eax, DWORD PTR [%ebp+8]
    pop %ebp
    ret
    .size   function, .-function

语法发生了变化，但寄存器符号之前仍然有“%”（这就是为什么我一开始就不喜欢 AT&T 语法）。

有人可以解释为什么会发生这种情况吗？我该如何解决这个问题？

原文

I have been trying to get a better idea of what happens under the hood by using the compiler to generate the assembly programs of various C programs at different optimization levels. There is something that has been bothering me for a while.

When I compile t.c as follows,

gcc -S t.c

I get the assembly in AT&T syntax as follows.

function:
    pushl   %ebp
    movl    %esp, %ebp
    movl    12(%ebp), %eax
    addl    8(%ebp), %eax
    popl    %ebp
    ret
    .size   function, .-function

When I compile using the masm argument as follows:-

gcc -S t.c -masm=intel

I get the following output.

function:
    push    %ebp
    mov %ebp, %esp
    mov %eax, DWORD PTR [%ebp+12]
    add %eax, DWORD PTR [%ebp+8]
    pop %ebp
    ret
    .size   function, .-function

There is a change in syntax but there are still "%"s before the notation of registers(this is why I don't prefer AT&T syntax in the first place).

Can someone shed some light on why this is happening? How do I solve this issue?

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