Question

比如 int x=17,c; c=x++*1/6; 还有 int a=3,b=5,c=1; a*=6+(b++)-(++c); 第一个里为什么C=2啊？++的优先级

比如 int x=17,c;
c=x++*1/6;
还有
int a=3,b=5,c=1;
a*=6+(b++)-(++c);
第一个里为什么C=2啊？++的优先级不是比较高吗？为什么不是X自增1之后再运算呢？
第二个里面括号里的式子不是优先级最高么？b和c在运算过程中的值是多少啊？为什么结果是27？

Answer 1

楼主您好哦@_@
您的问题我仔细看了一下，也亲自编程运行了一下，应该可以正确解答啦，哈哈~~
第一题：
int x=17,c;
c=x++*1/6;
++的优先级比较高，这不假，但是您要看到++的位置是在x的前面还是后面，本题是在x的后面，那就是说当先完成c=x*1/6;之后再进行x++的运算，所以这句话可以分开写成：
c=x*1/6;//c=17/6=2.8333但是c是int类型，所以小数部分会被丢弃，所以c=2
x++;此时x才变成18，也就是先算出来c之后x才会完成自加运算。

第二题：
int a=3,b=5,c=1;
a*=6+(b++)-(++c);
括号里优先级最高也是没有问题的，楼主您说的对。但是您看b++的++是不是在b的后面？这就是后自加，是完成a的计算之后才会进行的运算。您看
++c的++是不是放在了c的前面？这说明++c要先运算，才会带入新的c的值来计算a
那么a*=6+(b++)-(++c);
等价于 a*=6+b-(c+1)
等价于 a=a*(6+b-(c+1))=3*(6+5-2)=27.

为了您理解的更好，我分别把第一第二题的程序附在后面供您运行哦
第一题：
typedef int zx;
#include<iostream>
using namespace std;
int main()
{
zx x=17;
zx c;
c=x++*1/6;
cout<<c<<"\n";
system("pause");
return 0;
}

第二题

typedef int zx;
#include<iostream>
using namespace std;
int main()
{
zx a=3,b=5,c=1;
a*=6+(b++)-(++c);

cout<<a<<"\n";
system("pause");
return 0;
}

Answer 2

对于第一个，这是有前置后置的区别的，后置是在参加完运算之后再自增的，前置是自增之后再参加运算的
对于第二个，括号里面的是先运算的，只不过同第一题，++ 有前置后置的区别
对于++在不同的编译器会有不同的表现，一直以来的也有争议的
比如这个式子 i+=(i++)+(++i);不同的编译器可能结果就是不一样的，尽量避免这种状况

追问

第二个算式，括号里算完b等于6才对啊，可等于6算出来结果和VC++上出来的不一样。

Answer 3

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