mjhjmtu

I am trying to understand how readelf utility calculates function size. I wrote a simple program

#include <stdio.h>

int main() 
 printf("Test!n");

Now to check function size I used this (is this OK ? ):

readelf -sw a.out|sort -n -k 3,3|grep FUNC

which yielded:

 1: 0000000000000000 0 FUNC GLOBAL DEFAULT UND puts@GLIBC_2.2.5 (2)
 2: 0000000000000000 0 FUNC GLOBAL DEFAULT UND __libc_start_main@GLIBC_2.2.5 (2)
29: 0000000000400470 0 FUNC LOCAL DEFAULT 13 deregister_tm_clones
30: 00000000004004a0 0 FUNC LOCAL DEFAULT 13 register_tm_clones
31: 00000000004004e0 0 FUNC LOCAL DEFAULT 13 __do_global_dtors_aux
34: 0000000000400500 0 FUNC LOCAL DEFAULT 13 frame_dummy
48: 0000000000000000 0 FUNC GLOBAL DEFAULT UND puts@@GLIBC_2.2.5
50: 00000000004005b4 0 FUNC GLOBAL DEFAULT 14 _fini
51: 0000000000000000 0 FUNC GLOBAL DEFAULT UND __libc_start_main@@GLIBC_
58: 0000000000400440 0 FUNC GLOBAL DEFAULT 13 _start
64: 00000000004003e0 0 FUNC GLOBAL DEFAULT 11 _init
45: 00000000004005b0 2 FUNC GLOBAL DEFAULT 13 __libc_csu_fini
60: 000000000040052d 16 FUNC GLOBAL DEFAULT 13 main
56: 0000000000400540 101 FUNC GLOBAL DEFAULT 13 __libc_csu_init

Now if I check the main function's size, it shows 16. How did it arrive at that? Is that the stack size ?

Compiler used gcc version 4.8.5 (Ubuntu 4.8.5-2ubuntu1~14.04.1)

GNU readelf (GNU Binutils for Ubuntu) 2.24

The size given by readelf is the size of the binary object; for main, that’s the sequence of machine instructions which implement your function. On my system, I see

57: 00000000004004d7 21 FUNC GLOBAL DEFAULT 13 main

from readelf, which matches up nicely with the compiled code as shown by gcc -S or objdump -d:

0000000000000000 <main>:
 0: 55 push %rbp
 1: 48 89 e5 mov %rsp,%rbp
 4: bf 00 00 00 00 mov $0x0,%edi
 9: e8 00 00 00 00 callq e <main+0xe>
 e: b8 00 00 00 00 mov $0x0,%eax
 13: 5d pop %rbp
 14: c3 retq 

The 21 bytes are the bytes 55, 48, 89, e5 etc.