（X++）+（X++）+（++x）怎么算这个表达式的值啊？？ - 编程论坛

#2

zhcosin2011-09-09 17:55

有必要把一个程序写得如此矫揉造作吗？编程的时候完全可以避免这种容易混淆的写法。

#3

laznrbfe2011-09-09 19:46

程序代码：

#include<stdio.h>
void main()
{
    int x1=1,x2=1;
    int s1,s2;
    s1=(x1++)+(x1++)+(x1++);
    printf("x1=%d\n",x1);
    s2=(++x2)+(++x2)+(++x2);
    printf("x2=%d\n",x2);
    printf("s1=%d,s2=%d\n",s1,s2);
}

x1=4
x2=4
s1=3,s2=10

执行s1=(x1++)+(x1++)+(x1++);时，x1++是先运算，后加1，即：s1=x1+x1+x1;x1=x1++;x1=x1++;x1=x1++;
执行s2=(++x2)+(++x2)+(++x2);时，x2++是先加1，再运算。即：x2=++x2;x2=++x2;s2=x2+(++x2);
我是这样理解的。

#4

waterstar2011-09-10 10:03

把这个问题移到c语言编程模块去

#5

yuccn2011-09-10 14:14

void main()
{
00412050  push        ebp
00412051  mov         ebp,esp
00412053  sub         esp,0F0h
00412059  push        ebx
0041205A  push        esi
0041205B  push        edi
0041205C  lea         edi,[ebp-0F0h]
00412062  mov         ecx,3Ch
00412067  mov         eax,0CCCCCCCCh
0041206C  rep stos    dword ptr es:[edi]
    int x1=1,x2=1;
0041206E  mov         dword ptr [x1],1
00412075  mov         dword ptr [x2],1
    int s1,s2;
    s1=(x1++)+(x1++)+(x1++);
0041207C  mov         eax,dword ptr [x1]
0041207F  add         eax,dword ptr [x1]
00412082  add         eax,dword ptr [x1]
00412085  mov         dword ptr [s1],eax
00412088  mov         ecx,dword ptr [x1]
0041208B  add         ecx,1
0041208E  mov         dword ptr [x1],ecx
00412091  mov         edx,dword ptr [x1]
00412094  add         edx,1
00412097  mov         dword ptr [x1],edx
0041209A  mov         eax,dword ptr [x1]
0041209D  add         eax,1
004120A0  mov         dword ptr [x1],eax
    printf("x1=%d\n",x1);
004120A3  mov         esi,esp
004120A5  mov         eax,dword ptr [x1]
004120A8  push        eax
004120A9  push        offset string "x1=%d\n" (416660h)
004120AE  call        dword ptr [__imp__printf (4192F0h)]
004120B4  add         esp,8
004120B7  cmp         esi,esp
004120B9  call        @ILT+355(__RTC_CheckEsp) (411168h)
    s2=(++x2)+(++x2)+(++x2);
004120BE  mov         eax,dword ptr [x2]
004120C1  add         eax,1
004120C4  mov         dword ptr [x2],eax
004120C7  mov         ecx,dword ptr [x2]
004120CA  add         ecx,1
004120CD  mov         dword ptr [x2],ecx
004120D0  mov         edx,dword ptr [x2]
004120D3  add         edx,1
004120D6  mov         dword ptr [x2],edx
004120D9  mov         eax,dword ptr [x2]
004120DC  add         eax,dword ptr [x2]
004120DF  add         eax,dword ptr [x2]
004120E2  mov         dword ptr [s2],eax
    printf("x2=%d\n",x2);
004120E5  mov         esi,esp
004120E7  mov         eax,dword ptr [x2]
004120EA  push        eax
004120EB  push        offset string "x2=%d\n" (416658h)
004120F0  call        dword ptr [__imp__printf (4192F0h)]
004120F6  add         esp,8
004120F9  cmp         esi,esp
004120FB  call        @ILT+355(__RTC_CheckEsp) (411168h)
    printf("s1=%d,s2=%d\n",s1,s2);

如果你能看懂汇编就立即了。。这个是反汇编出来的。

#6

yuccn2011-09-10 14:19

    s1=(x1++)+(x1++)+(x1++);
0041207C  mov         eax,dword ptr [x1]
0041207F  add         eax,dword ptr [x1]
00412082  add         eax,dword ptr [x1]
00412085  mov         dword ptr [s1],eax
00412088  mov         ecx,dword ptr [x1]
0041208B  add         ecx,1
0041208E  mov         dword ptr [x1],ecx
00412091  mov         edx,dword ptr [x1]
00412094  add         edx,1
00412097  mov         dword ptr [x1],edx
0041209A  mov         eax,dword ptr [x1]
0041209D  add         eax,1
004120A0  mov         dword ptr [x1],eax
也就是 s = x1+x1+x1 = 3；
x1++;x1++;x1++;//

s1=(x1++)+(x1++)+(x1++); // s1=3，x1=4；

    s2=(++x2)+(++x2)+(++x2);
004120BE  mov         eax,dword ptr [x2]
004120C1  add         eax,1
004120C4  mov         dword ptr [x2],eax
004120C7  mov         ecx,dword ptr [x2]
004120CA  add         ecx,1
004120CD  mov         dword ptr [x2],ecx
004120D0  mov         edx,dword ptr [x2]
004120D3  add         edx,1
004120D6  mov         dword ptr [x2],edx
004120D9  mov         eax,dword ptr [x2]
004120DC  add         eax,dword ptr [x2]
004120DF  add         eax,dword ptr [x2]
004120E2  mov         dword ptr [s2],eax

也就是x2=x2+1;x2=x2+1;x2=x2+1; s2=x2+x2+x2。这时候，x2=4.s2=12.。

#7

爱德华2011-09-16 22:24

回复 2楼 zhcosin

不是我想搞得更清楚点吗.

#8

omijialang2011-09-17 16:33

发生未定义行为吧（c语言不确定这样做的结果）应该是。（好像听老师说过）

#9

艹蛋灬爱情2011-10-25 16:30

放到编程软件上运行一下