为什么strcmp的回报不同？

Question

这是C代码,我用gcc编译

char *a="a";
char *d="d";
printf("%d\n", strcmp("a", "d"));
printf("%d\n", strcmp(a, "d"));
printf("%d\n", strcmp(a, d));

当我用-O输出编译时

-1
-3
-1

当我编译没有-O那么输出是

-1
-3
-3

为什么输出不同,代码是strcmp什么？

Answer 1

为什么输出不同

因为重要的是返回值的符号(正,负或零).strcmp()不需要返回+1或-1,也不必返回一致的值.我怀疑在第一种和第三种情况下,编译器优化掉调用strcmp()并将-1置于返回值的位置.在第二种情况下,我认为该函数实际上是被调用的.

strcmp的代码是什么？

从它似乎返回第一个不同角色的字符代码之间的差异的事实推断出来,我会说这是glibc strcmp():

int
 strcmp (p1, p2)
      const char *p1;
      const char *p2;
 {
   register const unsigned char *s1 = (const unsigned char *) p1;
   register const unsigned char *s2 = (const unsigned char *) p2;
   unsigned char c1, c2;

   do
     {
       c1 = (unsigned char) *s1++;
       c2 = (unsigned char) *s2++;
       if (c1 == '\0')
     return c1 - c2;
     }
   while (c1 == c2);

   return c1 - c2;
 }

编辑: @AndreyT不相信我,所以这里是为我生成的汇编GCC 4.2(OS X 10.7.5 64位Intel,默认优化级别 - 无标志):

    .section    __TEXT,__text,regular,pure_instructions
    .globl  _main
    .align  4, 0x90
_main:
Leh_func_begin1:
    pushq   %rbp
Ltmp0:
    movq    %rsp, %rbp
Ltmp1:
    subq    $32, %rsp
Ltmp2:
    leaq    L_.str(%rip), %rax
    movq    %rax, -16(%rbp)
    leaq    L_.str1(%rip), %rax
    movq    %rax, -24(%rbp)
    movl    $-1, %ecx             ; <- THIS!
    xorb    %dl, %dl
    leaq    L_.str2(%rip), %rsi
    movq    %rsi, %rdi
    movl    %ecx, %esi
    movq    %rax, -32(%rbp)
    movb    %dl, %al
    callq   _printf               ; <- no call to `strcmp()` so far!
    movq    -16(%rbp), %rax
    movq    %rax, %rdi
    movq    -32(%rbp), %rsi
    callq   _strcmp               ; <- strcmp()
    movl    %eax, %ecx
    xorb    %dl, %dl
    leaq    L_.str2(%rip), %rdi
    movl    %ecx, %esi
    movb    %dl, %al
    callq   _printf               ; <- printf()
    movq    -16(%rbp), %rax
    movq    -24(%rbp), %rcx
    movq    %rax, %rdi
    movq    %rcx, %rsi
    callq   _strcmp               ; <- strcmp()
    movl    %eax, %ecx
    xorb    %dl, %dl
    leaq    L_.str2(%rip), %rdi
    movl    %ecx, %esi
    movb    %dl, %al
    callq   _printf               ; <- printf()
    movl    $0, -8(%rbp)
    movl    -8(%rbp), %eax
    movl    %eax, -4(%rbp)
    movl    -4(%rbp), %eax
    addq    $32, %rsp
    popq    %rbp
    ret
Leh_func_end1:

    .section    __TEXT,__cstring,cstring_literals
L_.str:
    .asciz   "a"

L_.str1:
    .asciz   "d"

L_.str2:
    .asciz   "%d\n"

    .section    __TEXT,__eh_frame,coalesced,no_toc+strip_static_syms+live_support
EH_frame0:
Lsection_eh_frame:
Leh_frame_common:
Lset0 = Leh_frame_common_end-Leh_frame_common_begin
    .long   Lset0
Leh_frame_common_begin:
    .long   0
    .byte   1
    .asciz   "zR"
    .byte   1
    .byte   120
    .byte   16
    .byte   1
    .byte   16
    .byte   12
    .byte   7
    .byte   8
    .byte   144
    .byte   1
    .align  3
Leh_frame_common_end:
    .globl  _main.eh
_main.eh:
Lset1 = Leh_frame_end1-Leh_frame_begin1
    .long   Lset1
Leh_frame_begin1:
Lset2 = Leh_frame_begin1-Leh_frame_common
    .long   Lset2
Ltmp3:
    .quad   Leh_func_begin1-Ltmp3
Lset3 = Leh_func_end1-Leh_func_begin1
    .quad   Lset3
    .byte   0
    .byte   4
Lset4 = Ltmp0-Leh_func_begin1
    .long   Lset4
    .byte   14
    .byte   16
    .byte   134
    .byte   2
    .byte   4
Lset5 = Ltmp1-Ltmp0
    .long   Lset5
    .byte   13
    .byte   6
    .align  3
Leh_frame_end1:


.subsections_via_symbols

和原始源代码:

#include <stdio.h>
#include <string.h>

int main()
{
    const char *a = "a";
    const char *d = "d";
    printf("%d\n", strcmp("a", "d"));
    printf("%d\n", strcmp(a, "d"));
    printf("%d\n", strcmp(a, d));

    return 0;
}

它产生的输出(截图有更好的证明):