mov 0, %eax 上带有 x86 程序集的 Segfault [英] Segfault with x86 assembly on mov 0, %eax

问题描述

我正在尝试组装一小段 x86 代码.我在一台 32 位机器上，我编写了以下代码.它应该只是将值添加到 eax 然后返回.我意识到不会有任何输出.当我使用

I'm trying to assemble a small piece of x86 code. I'm on a 32 bit machine and I have written the following code. It should just add values into eax and then return. I realize there will not be any output. When I compile this using

gcc main.S -o main

它编译没有错误.但是当我运行它时会出现段错误(gdb 声称它在第一条 movl 指令上出现段错误).main.S 中包含以下代码.我做错了什么?

It compiles with no errors. But when I run it seg faults (gdb claims that it segfaults on the first movl instruction). main.S has the following code in it. What am I doing wrong?

.text  
.globl main  
main:  
pushl    %ebp  
movl     %esp, %ebp  
movl 0,  %eax  
addl $3, %eax  
addl $3, %eax  
leave 
ret

推荐答案

不是你的第一个，而是你的第二个 movl

Not your first, but your second movl

movl  0,%eax

这是来自绝对地址 0 的内存源操作数的加载，这当然是段错误.

That's a load from a memory source operand with absolute address 0 which of course segfaults.

使用 mov $0, %eax 将 mov-immediate 移入寄存器.(或者对于零，xor %eax, %eax 更有效地将寄存器归零.)

Use mov $0, %eax for mov-immediate into a register. (Or for zero specifically, xor %eax, %eax to more efficiently zero a register.)

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