问题描述
此说明适用于Linux 32位:
当Linux程序开始时,所有指向命令行参数的指针都存储在堆栈中。参数的数量存储在0(%ebp),程序的名称存储在4(%ebp),参数存储在8(%ebp)。
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64位的相同信息。
I need the same information for 64 bit.
编辑:
我有工作代码示例演示如何使用argc,argv [0]和argv [1]:
.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 ...
...
}
看起来参数在堆栈上。因为这段代码不清楚,我问这个问题。我猜想我可以保留rsp在rbp,然后访问这些参数使用0(%rbp),8(%rbp),16(%rbp)等这是正确的?
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
它看起来像第3.4节进程初始化,特别是图3.9,在已经提到的正确描述了您想要知道的内容。
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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