在大会的Linux 64的命令行参数 [英] Linux 64 command line parameters in Assembly

问题描述

这说明适用于Linux的32位： 当一个Linux程序开始，所有指向命令行参数都存储在堆栈中。参数的数目被存储为0（％EBP），程序的名称被存储在图4（％EBP），和参数从8（％ebp的）存储

This description is valid for Linux 32 bit: When a Linux program begins, all pointers to command-line arguments are stored on the stack. The number of arguments is stored at 0(%ebp), the name of the program is stored at 4(%ebp), and the arguments are stored from 8(%ebp).

我需要的信息相同的64位。

I need the same information for 64 bit.

编辑： 我有工作code示例展示了如何使用ARGC，ARGV [0]和argv [1]： http://cubbi.com/fibonacci/asm.html

I have working code sample which shows how to use argc, argv[0] and argv[1]: http://cubbi.com/fibonacci/asm.html

.globl _start
_start:
    popq    %rcx        # this is argc, must be 2 for one argument
    cmpq    $2,%rcx
    jne     usage_exit
    addq    $8,%rsp     # skip argv[0]
    popq    %rsi        # get argv[1]
    call ...
...
}

看起来参数是在栈上。由于这code是不明确的，我问这个问题。我的猜测，我可以保持RSP在RBP，然后访问，8（％RBP），16（％RBP）等，这些参数使用0（％RBP）这是正确的？

It looks like parameters are on the stack. Since this code is not clear, I ask this question. My guess that I can keep rsp in rbp, and then access these parameters using 0(%rbp), 8(%rbp), 16(%rbp) etc. It this correct?

推荐答案

它看起来像3.4节的进程初始化的，特别是图3.9中已经提到的System V AMD64 ABI 介绍precisely你想知道什么。

It looks like section 3.4 Process Initialization, and specifically figure 3.9, in the already mentioned System V AMD64 ABI describes precisely what you want to know.

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