c - What do the instructions mov %edi and mov %rsi do? -

i've written basic c program defines integer variable x, sets 0 , returns value of variable:

#include <stdio.h>  int main(int argc, char **argv) {     int x;     x = 0;     return x; } 

when dump object code using objdump (compiled on linux x86-64 gcc):

0x0000000000400474 <main+0>:    push   %rbp 0x0000000000400475 <main+1>:    mov    %rsp,%rbp 0x0000000000400478 <main+4>:    mov    %edi,-0x14(%rbp) 0x000000000040047b <main+7>:    mov    %rsi,-0x20(%rbp) 0x000000000040047f <main+11>:   movl   $0x0,-0x4(%rbp) 0x0000000000400486 <main+18>:   mov    -0x4(%rbp),%eax 0x0000000000400489 <main+21>:   leaveq  0x000000000040048a <main+22>:   retq 

i can see function prologue, before set x 0 @ address 0x000000000040047f there 2 instructions move %edi , %rsi onto stack. these for?

in addition, unlike set x 0, mov instruction shown in gas syntax not have suffix.

if suffix not specified, , there no memory operands instruction, gas infers operand size size of destination register operand.

in case, -0x14(%rsbp) , -0x20(%rbp) both memory operands , sizes? since %edi 32 bit register, 32 bits moved -0x14(%rsbp) whereas since %rsi 64 bit register, 64 bits moved %rsi,-0x20(%rbp)?

in simple case, why don't ask compiler directly? gcc, clang , icc there's -fverbose-asm option.

main:     pushq   %rbp    #     movq    %rsp, %rbp  #,     movl    %edi, -20(%rbp) # argc, argc     movq    %rsi, -32(%rbp) # argv, argv     movl    $0, -4(%rbp)    #, x     movl    -4(%rbp), %eax  # x, d.2607     popq    %rbp    #     ret 

so, yes, save argv , argv onto stack using "old" frame pointer method since new architectures allow subtracting/adding from/to stack pointer directly, omitting frame pointer (-fomit-frame-pointer).