FUndamentals of c -_-

Question

#include<stdio.h>

int main()
{
    int a=2;
    int j=++a  + ++a + ++a + ++a ; 

    printf("%d",j);
}

//please tell me the execution of this fragment ... //i am using GCC ubuntu (12.04) compiler. -_-

so can u explain !! i mean exact parsing step for this ... out put is coming 19 O__O — Souradeep Das, Mar 09 '13 at 20:09
it's undefined behavior, your compiler can do whatever it wants to. Don't try to understand why you get 19 — alrikai, Mar 09 '13 at 21:12
The key is to understand that C code is parsed and assembly code is generated and then compiled. The result depends strictly on what assembly code the compiler generates — Salvatore, Mar 09 '13 at 21:26

Salvatore · Answer 1 · 2013-03-09T21:09:25.547

I generated the assembly code (gcc for windows on WIndows 7) [gcc -S test.c] and I get the following:

    .file   "increment.c"
    .def    ___main;    .scl    2;  .type   32; .endef
    .section .rdata,"dr"
LC0:
    .ascii "%d, %d\12\0"
    .text
    .globl  _main
    .def    _main;  .scl    2;  .type   32; .endef
_main:
LFB6:
    .cfi_startproc
    pushl   %ebp
    .cfi_def_cfa_offset 8
    .cfi_offset 5, -8
    movl    %esp, %ebp
    .cfi_def_cfa_register 5
    andl    $-16, %esp
    subl    $32, %esp
    call    ___main
    movl    $2, 28(%esp)   ;28(%esp) = 2 (a=2)
    incl    28(%esp)       ;28(%esp) = 3 (a++)
    incl    28(%esp)       ;28(%esp) = 4 (a++)
    movl    28(%esp), %eax ;%eax = 4 (eax = a = 4)
    leal    (%eax,%eax), %edx = %eax + %eax = 8 (edx = a + a) 
    incl    28(%esp)       ;28(%esp) = 5 (a++)
    movl    28(%esp), %eax ;%eax = 5 (eax = a = 5)
    addl    %eax, %edx  ;%edx = %eax + %edx = 5 + 8 = 13 (edx = a + a + a)
    incl    28(%esp) ;28(%esp) = 6 (a++)
    movl    28(%esp), %eax ; %eax = 6 (eax = a = 6)
    addl    %edx, %eax    ;%eax = %edx + %eax = 13 + 6 = 19 (eax = a + a + a + a)
    movl    %eax, 24(%esp) 
    movl    28(%esp), %eax 
    movl    %eax, 8(%esp) 
    movl    24(%esp), %eax 
    movl    %eax, 4(%esp)
    movl    $LC0, (%esp)
    call    _printf
    leave
    .cfi_restore 5
    .cfi_def_cfa 4, 4
    ret
    .cfi_endproc
LFE6:
    .def    _printf;    .scl    2;  .type   32; .endef

After computing the result (see comments) the printf function is called with 19 as result.

Anyway different compilers can give different results. The final result depends on the way the compiler arranges instructions.

FUndamentals of c -_-

1 Answers1