what does -0x4(%rbp) means in gdb disassembly?

Question

i am currently working on gdb disassembly to help me understand more detail about the c program so i write a c program:

#include <stdio.h>

void swap(int a, int b){
        int temp = a;
        a = b;
        b = temp;
}
void main(){
        int a = 1,b = 2;
        swap(a, b);
}

I use gdb and run disass /m main to get those:

(gdb) disass /m main
Dump of assembler code for function main:
8   void main(){
   0x0000000000400492 <+0>: push   %rbp
   0x0000000000400493 <+1>: mov    %rsp,%rbp
   0x0000000000400496 <+4>: sub    $0x10,%rsp

9       int a = 1,b = 2;
   0x000000000040049a <+8>: movl   $0x1,-0x8(%rbp)
   0x00000000004004a1 <+15>:    movl   $0x2,-0x4(%rbp)

10      swap(a, b);
   0x00000000004004a8 <+22>:    mov    -0x4(%rbp),%edx
   0x00000000004004ab <+25>:    mov    -0x8(%rbp),%eax
   0x00000000004004ae <+28>:    mov    %edx,%esi
   0x00000000004004b0 <+30>:    mov    %eax,%edi
   0x00000000004004b2 <+32>:    callq  0x400474 <swap>

11  }
   0x00000000004004b7 <+37>:    leaveq 
   0x00000000004004b8 <+38>:    retq   

End of assembler dump.

My question is those -0x8(%rbp) means what?

A memory or a register?

I do know that 1 is store in -0x8(%rbp) and 2 is in -0x4(%rbp), How can i show the value in thoes kind of 'place' ? I try to use (gdb) p -0x8(%rbp) but get this:

A syntax error in expression, near `%rbp)'.

Answer 1

Registers in gdb can be referred with the prefix '$'

p *(int *)($rbp - 8)

RBP and RSP most likely refer to memory locations, specifically to stack. Other registers are more or less generic purpose registers and can point to memory too.

Answer 2

It means "the data stored when you subtract eight from the address stored in rbp". Try looking at the stack commands available in gdb: http://www.delorie.com/gnu/docs/gdb/gdb_41.html