Question

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).

I need the same information for 64 bit.

Edit:
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 ...
...
}

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?

Answer 1

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.