在条件三元运算符中使用逗号时发现了什么？ [英] Something we found when using comma in condition ternary operator?

问题描述

好吧，我有一个关于三元运算符中的逗号的问题。
剪切垃圾，代码如下：

  void test_comma_in_condition（void）
 {
 int ia，ib，ic; 
 
 ia = ib = ic = 0; 
 bool condition = true; 
 
 cout<<Original：<< endl; 
 cout<<ia：<< ia<< endl; 
 cout<<ib：<< ib<< endl; 
 condition？（ia = 1，ib = 2）:( ia = 11，ib = 12）; 
 cout<<After：<< endl; 
 cout<<ia：<< ia<< endl; 
 cout<<ib：<< ib<< endl; 
 
 ia = ib = ic = 0; 
条件ia = 1，ib = 2，ic = 3：ib = 22，ia = 21，ic = 23; 
 cout<<操作必须加括号，否则你会看到...<< endl; 
 cout<<ia：<< ia<< endl; 
 cout<<ib：<< ib<< endl; 
 cout<<ic：<< ic<< endl; 
 
条件ia = 1，ib = 2，ic = 3：ia = 21，ib = 22，ic = 23; 
 cout<<操作必须加括号，否则你会看到...<< endl; 
 cout<<ia：<< ia<< endl; 
 cout<<ib：<< ib<< endl; 
 cout<<ic：<< ic<< endl; 
 
 return; 
} 
  

输出结果如下：

 原文：
 ia：0 
 ib：0 
之后：
 ia：1 
 ib：2 
操作必须在括号内，或者你会看到... 
 ia：21 
 ib：2 
 ic：23 
操作必须加括号，你会看到... 
 ia：1 
 ib：22 
 ic：23 
  

这是合法吗？

解决方案

这是操作符优先级的问题。您的表达式：

  condition？ia = 1，ib = 2，ic = 3：ib = 22，ia = 21，ic = 23; 
  

可由编译器理解为：

<$ （ia = 1，ib = 2，ic = 3）:( ib = 22）），ia = 21，ic = 23;

此时，您应该能够看到为什么得到程序输出。

Well, I had a question about comma in ternary operator. Cut the crap, the code is below:

void test_comma_in_condition(void)
{
    int ia, ib, ic;

    ia = ib = ic = 0;
    bool condition=true;

    cout<<"Original:"<<endl;
    cout<<"ia: "<<ia<<endl;
    cout<<"ib: "<<ib<<endl;
    condition?(ia=1, ib=2):(ia=11, ib=12);
    cout<<"After:"<<endl;
    cout<<"ia: "<<ia<<endl;
    cout<<"ib: "<<ib<<endl;

    ia = ib = ic = 0;
    condition?ia=1, ib=2, ic=3:ib=22,ia=21, ic=23;
    cout<<"The operation must be bracketed, or you'll see..."<<endl;
    cout<<"ia: "<<ia<<endl;
    cout<<"ib: "<<ib<<endl;
    cout<<"ic: "<<ic<<endl;

    condition?ia=1, ib=2, ic=3:ia=21, ib=22, ic=23;
    cout<<"The operation must be bracketed, or you'll see..."<<endl;
    cout<<"ia: "<<ia<<endl;
    cout<<"ib: "<<ib<<endl;
    cout<<"ic: "<<ic<<endl; 

    return;
}

The output will be like:

Original:
ia: 0
ib: 0
After:
ia: 1
ib: 2
The operation must be bracketed, or you'll see...
ia: 21
ib: 2
ic: 23
The operation must be bracketed, or you'll see...
ia: 1
ib: 22
ic: 23

Is this legal?

解决方案

This is a matter of operator precedence. Your expression:

condition?ia=1, ib=2, ic=3:ib=22,ia=21, ic=23;

is understood by the compiler as:

(condition?(ia=1, ib=2, ic=3):(ib=22)),ia=21, ic=23;

At this point you should be able to see why you get the program output.

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