I have a C code that swaps two numbers.
#include<stdio.h>
void swap(int,int);
void main( )
{
int n1,n2;
printf("Enter the two numbers to be swapped\n");
scanf("%d%d",&n1,&n2);
printf("\nThe values of n1 and n2 in the main function before calling the swap function are n1=%d n2=%d",n1,n2);
swap(n1,n2);
printf("\nThe values of n1 and n2 in the main function after calling the swap function are n1=%d n2=%d",n1,n2);}
void swap(int n1,int n2)
{
int temp;
temp=n1;
n1=n2;
n2=temp;
printf("\nThe values of n1 and n2 in the swap function after swapping are n1=%d n2=%d",n1,n2);
}
I have disassembled it using objdump and been trying to find out how the swap operation happens in machine level. I think this is the swap function.
000006b4 <swap>:
6b4: 55 push %ebp
6b5: 89 e5 mov %esp,%ebp
6b7: 53 push %ebx
6b8: 83 ec 14 sub $0x14,%esp
6bb: e8 37 00 00 00 call 6f7 <__x86.get_pc_thunk.ax>
6c0: 05 0c 19 00 00 add $0x190c,%eax
6c5: 8b 55 08 mov 0x8(%ebp),%edx
6c8: 89 55 f4 mov %edx,-0xc(%ebp)
6cb: 8b 55 0c mov 0xc(%ebp),%edx
6ce: 89 55 08 mov %edx,0x8(%ebp)
6d1: 8b 55 f4 mov -0xc(%ebp),%edx
6d4: 89 55 0c mov %edx,0xc(%ebp)
6d7: 83 ec 04 sub $0x4,%esp
6da: ff 75 0c pushl 0xc(%ebp)
6dd: ff 75 08 pushl 0x8(%ebp)
6e0: 8d 90 c0 e8 ff ff lea -0x1740(%eax),%edx
6e6: 52 push %edx
6e7: 89 c3 mov %eax,%ebx
6e9: e8 72 fd ff ff call 460 <printf@plt>
6ee: 83 c4 10 add $0x10,%esp
6f1: 90 nop
6f2: 8b 5d fc mov -0x4(%ebp),%ebx
6f5: c9 leave
6f6: c3 ret
I want to know how swap operation is happening inside registers, I know it has to be something like this.
push eax
mov eax, ebx
pop ebx
But I can't see anything similar to this. Since I'm new to these things, can someone please help me how to understand how this is happening. Full output of the objdump is here.
To get started with the assembly language you can check the following link:
http://patshaughnessy.net/2016/11/26/learning-to-read-x86-assembly-language