Business
Jobs
  • About Us
  • Solutions
    • Job Postings
      Post your job and receive qualified candidates in 48h.
    • Candidate Assessments
      500+ technical and psychological tests, plus anti-fraud.
    • Headhunting
      Tailor-made executive search from start to finish.
    • Payroll + EOR
      Payroll dispersal and EOR across 15+ LATAM countries.
  • Pricing
  • Jobs

0

230
Views
What do the instructions mov %edi and mov %rsi do?

I've written a basic C program that defines an integer variable x, sets it to zero and returns the value of that variable:

#include <stdio.h>

int main(int argc, char **argv) {
    int x;
    x = 0;
    return x;
}

When I dump the object code using objdump (compiled on Linux X86-64 with 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 the function prologue, but before we set x to 0 at address 0x000000000040047f there are two instructions that move %edi and %rsi onto the stack. What are these for?

In addition, unlike where we set x to 0, the mov instruction as shown in GAS syntax does not have a suffix.

If the suffix is not specified, and there are no memory operands for the instruction, GAS infers the operand size from the size of the destination register operand.

In this case, are -0x14(%rsbp) and -0x20(%rbp) both memory operands and what are their sizes? Since %edi is a 32 bit register, are 32 bits moved to -0x14(%rsbp) whereas since %rsi is a 64 bit register, 64 bits are moved to %rsi,-0x20(%rbp)?

over 4 years ago · Santiago Trujillo
2 answers
Answer question

0

In this simple case, why don't you ask your compiler directly? For GCC, clang and ICC there's the -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, they save argv and argv onto the stack by using the "old" frame pointer method since new architectures allow subtracting/adding from/to the stack pointer directly, thus omitting the frame pointer (-fomit-frame-pointer).

over 4 years ago · Santiago Trujillo Report

0

Purpose of ESI & EDI registers?

Based on this and the context, I'm not an expert, but my guess is these are capturing the main() input parameters. EDI takes a standard width, which would match the int argc, whereas RSI takes a long, which would match the char **argv pointer.

over 4 years ago · Santiago Trujillo Report
Answer question
Find remote jobs

Discover the new way to find a job!

Top jobs
Top job categories
Business
Post vacancy Pricing Sales
Legal
Terms and conditions Privacy policy
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Show me some job opportunities
There's an error!