변수인데 왜 operation on 'variable' may be undefined 야?

view source print? 1 #include "stdio.h" 2   3 int main(int argc, char *argv[]) 4 { 5     int x = 1; 6     printf("%d %d %d\n", ++x, x, x++); 7     return 0; 8 } You're running into two issues: 1. Undefined behavior -- you are attempting to modify the value of an object more than once between sequence points, and the Standard imposes no requirement on how to handle that behavior. Basically, any expression of the forms: i = i++; j = k++ * k++; foo(i,i++,--i); invoke undefined behavior. Read up on the semantics of the "++" and "--" operators in an *authoritative* reference (the Standard would obviously be one, but also K&R2 or H&S5); they don't work the way most people think they should. 2. Order of evaluation -- AFAIK, there's no requirement that expressions in a function parameter list be evaluated in any particular order. At first blush, it *looks* like the arguments were evaluated from right to left (if x = 1, then x++ evaluates to 1, with the side effect that x == 2, and ++x evaluates to 3, with the side effect that x == 3), but given that you've invoked undefined behavior the actual reason may be something else entirely. [링크 : http://bytes.com/topic/c/answers/222558-operation-x-may-undefined]

위의 소스를 gcc -Wall 옵션을 주고 컴파일 할 경우에

$ gcc -Wall cc_test.c cc_test.c: In function ‘main’: cc_test.c:6: warning: operation on ‘x’ may be undefined cc_test.c:6: warning: operation on ‘x’ may be undefined

이런 경고가 발생한다.

분명 x는 변수인데, 그에 대한 operation이 정의되지 않았다는게 무슨 말인지 모르겠지만,
아무튼 실행을 하면

$ ./a.out 3 3 1

요런 희한한 결과가 나온다.

아무튼, calling convention과도 연관이 있어 보이는데,
c언어 특성상 right-left 로 push 하므로(가장 위에는 왼쪽 값이 오도록)
가장 먼저들어가는, 오른쪽 x++ 은 1이 들어가고
1을 더해준다음, 다음 명령어를 수행하면서(++x) 한번에 2가 증가하게 되고
그럼으로 인해 x, ++x 순으로 3이 들어가는게 아닐까 생각된다.

아니면 말구?

[링크 : http://gcc.gnu.org/onlinedocs/gcc/Warning-Options.html]


winXP SP3 32bit / VC++ 6 에서 해본 결과

2 1 1

1:    #include "stdio.h"
2:
3:
4:    int main(int argc, char *argv[])
5:    {
00401010   push        ebp
00401011   mov         ebp,esp
00401013   sub         esp,48h
00401016   push        ebx
00401017   push        esi
00401018   push        edi
00401019   lea         edi,[ebp-48h]
0040101C   mov         ecx,12h
00401021   mov         eax,0CCCCCCCCh
00401026   rep stos    dword ptr [edi]
6:        int x = 1;
00401028   mov         dword ptr [ebp-4],1
7:        printf("%d %d %d\n", ++x, x, x++);
0040102F   mov         eax,dword ptr [ebp-4]
00401032   mov         dword ptr [ebp-8],eax
00401035   mov         ecx,dword ptr [ebp-8]
00401038   push        ecx

00401039   mov         edx,dword ptr [ebp-4]
0040103C   push        edx

0040103D   mov         eax,dword ptr [ebp-4]
00401040   add         eax,1
00401043   mov         dword ptr [ebp-4],eax
00401046   mov         ecx,dword ptr [ebp-4]
00401049   push        ecx

0040104A   push        offset string "%d %d %d\n" (0042201c)

0040104F   mov         edx,dword ptr [ebp-4]
00401052   add         edx,1
00401055   mov         dword ptr [ebp-4],edx
00401058   call        printf (00401090)
0040105D   add         esp,10h
8:        return 0;
00401060   xor         eax,eax
9:    }
00401062   pop         edi
00401063   pop         esi
00401064   pop         ebx
00401065   add         esp,48h
00401068   cmp         ebp,esp
0040106A   call        __chkesp (00401110)
0040106F   mov         esp,ebp
00401071   pop         ebp
00401072   ret

linux / gcc 결과

3 3 1

08048354 <main>:
main():
/home/morpheuz/tt/cc_test.c:4
#include <stdio.h>

int main(int argc, char *argv[])
{
 8048354:       8d 4c 24 04             lea    0x4(%esp),%ecx
 8048358:       83 e4 f0                and    $0xfffffff0,%esp
 804835b:       ff 71 fc                pushl  0xfffffffc(%ecx)
 804835e:       55                      push   %ebp
 804835f:       89 e5                   mov    %esp,%ebp
 8048361:       51                      push   %ecx
 8048362:       83 ec 24                sub    $0x24,%esp
/home/morpheuz/tt/cc_test.c:5
        int x = 1;
 8048365:       c7 45 f8 01 00 00 00    movl   $0x1,0xfffffff8(%ebp)
/home/morpheuz/tt/cc_test.c:6
        printf("%d %d %d\n", ++x, x, x++);
 804836c:       8b 45 f8                mov    0xfffffff8(%ebp),%eax
 804836f:       83 45 f8 01             addl   $0x1,0xfffffff8(%ebp)
 8048373:       83 45 f8 01             addl   $0x1,0xfffffff8(%ebp)
 8048377:       89 44 24 0c             mov    %eax,0xc(%esp)
 804837b:       8b 45 f8                mov    0xfffffff8(%ebp),%eax
 804837e:       89 44 24 08             mov    %eax,0x8(%esp)
 8048382:       8b 45 f8                mov    0xfffffff8(%ebp),%eax
 8048385:       89 44 24 04             mov    %eax,0x4(%esp)
 8048389:       c7 04 24 80 84 04 08    movl   $0x8048480,(%esp)
 8048390:       e8 03 ff ff ff          call   8048298 <printf@plt>
/home/morpheuz/tt/cc_test.c:7
        return 0;
 8048395:       b8 00 00 00 00          mov    $0x0,%eax
/home/morpheuz/tt/cc_test.c:8
}
 804839a:       83 c4 24                add    $0x24,%esp
 804839d:       59                      pop    %ecx
 804839e:       5d                      pop    %ebp
 804839f:       8d 61 fc                lea    0xfffffffc(%ecx),%esp
 80483a2:       c3                      ret
 80483a3:       90                      nop
 80483a4:       90                      nop
 80483a5:       90                      nop
 80483a6:       90                      nop
 80483a7:       90                      nop
 80483a8:       90                      nop
 80483a9:       90                      nop
 80483aa:       90                      nop
 80483ab:       90                      nop
 80483ac:       90                      nop
 80483ad:       90                      nop
 80483ae:       90                      nop
 80483af:       90                      nop

objdump를 소스와 함께 디스어셈블 하기 위해서는
gcc -g 옵션으로 디버깅 정보를 주어야 한다.
[링크 : http://wikidocs.net/mybook/1473]


솔찍히 어셈을 몰라서 모르겠다 ㅋㅋ
결론 : 한줄에 ++나 -- 연산자를 중복으로 사용하지 말자.