I have written simple function in C,
void GetInput()
{
char buffer[8];
gets(buffer);
puts(buffer);
}
When I disassemble it in gdb's disassembler, it gives following disassembly.
0x08048464 <+0>: push %ebp
0x08048465 <+1>: mov %esp,%ebp
0x08048467 <+3>: sub $0x10,%esp
0x0804846a <+6>: mov %gs:0x14,%eax
0x08048470 <+12>: mov %eax,-0x4(%ebp)
0x08048473 <+15>: xor %eax,%eax
=> 0x08048475 <+17>: lea -0xc(%ebp),%eax
0x08048478 <+20>: mov %eax,(%esp)
0x0804847b <+23>: call 0x8048360 <gets@plt>
0x08048480 <+28>: lea -0xc(%ebp),%eax
0x08048483 <+31>: mov %eax,(%esp)
0x08048486 <+34>: call 0x8048380 <puts@plt>
0x0804848b <+39>: mov -0x4(%ebp),%eax
0x0804848e <+42>: xor %gs:0x14,%eax
0x08048495 <+49>: je 0x804849c <GetInput+56>
0x08048497 <+51>: call 0x8048370 <__stack_chk_fail@plt>
0x0804849c <+56>: leave
0x0804849d <+57>: ret
Now please look at line number three, 0x08048467 <+3>: sub $0x10,%esp
, I have only 8 bytes allocated as local variable, then why compiler is allocating 16 bytes(0x10).
Secondly, what is meaning of xor %gs:0x14,%eax
.
If it is optimization, is there any way to stop it.
Two things:
Regarding your other question that you should have asked separately, xor %gs:0x14,%eax
is clearly part of a stack protection mechanism, enabled by default. If you are using GCC, turn it off with -fno-stack-protector
.