汇编代码理解求助：C代码生成的x86汇编指令疑问解析

Hey there! Let's break down each of those marked ?? instructions and clear up your follow-up confusion, linking everything directly to your C code and how x86-64 assembly works on most Unix-like systems (using the System V AMD64 calling convention).

First, let's recap the key lines from your C code that correspond to this assembly:

char const* y = argv[1];
printf("%s\n", y);

Quick reminders to set the stage:

• argv is a char** — that's a pointer to an array of string pointers (each pointer is 8 bytes in x86-64).
• main's arguments are stored on the stack after setting up the frame, so argv lives at -0x20(%rbp).
• For function calls, the first argument goes in rdi, the second in rsi, etc.

Let's unpack each ?? line: #

1. 0x00000000000006ba <+26>: mov 0x8(%rax),%rax

   • Right before this, mov -0x20(%rbp),%rax pulled argv (the second argument to main) from the stack into rax.
   • Since argv points to an array of 8-byte pointers, argv[1] is located 8 bytes past the start of the array. 0x8(%rax) computes that address, and this instruction loads the pointer to your first command-line argument into rax.

2. 0x00000000000006be <+30>: mov %rax,-0x8(%rbp)

   • This stores the pointer we just fetched (argv[1]) into the stack slot reserved for your variable y.
   • This is exactly the assembly equivalent of char const* y = argv[1]; — we're saving y's value onto the stack frame.

3. 0x00000000000006c2 <+34>: mov -0x8(%rbp),%rax

   • This loads y's value back from the stack into rax.
   • It might look unnecessary, but compilers often generate this kind of "round-trip" code when preparing values for function calls. We need this pointer in a register to pass it to the print function next.

Your big question: mov %rax,%rdi # Return value copied to 1st argument register - why?? #

• Here's the twist you missed: your compiler optimized printf("%s\n", y) to puts(y)!
• puts() only needs one argument (the string to print, and it automatically adds a newline), so passing the string pointer in rdi (the first argument register for System V AMD64) is exactly correct.
• The callq 0x560 is calling puts, not printf — your manual annotation was off here, which is why the argument count seemed wrong.

内容的提问来源于stack exchange，提问作者Average_guy