错误：case标签不会减少为整数常量 [英] error: case label does not reduce to an integer constant

问题描述

  void  flight_operator :: get_define_flight_num（ char  namecode）{
 cout<<  输入航班代码：; 
 cin>> namecode; 
  switch （namecode）{
  case    MH： case     mh：
 cout<<   Make Airlines \\\
<< endl; 
  break ; 
  case    FY ： case    fy：
 cout<<   Fire Airlines \ n<< endl; 
 默认：
 cout<<  无效的代码！<< endl; 
  break ; 
} 
 
} 

解决方案

有什么不清楚的？案例标签应为整数常量。它可以是整数类型之一，枚举类型，类似这样，但不是字符串，不是数组，或任何其他更复杂的对象或指向对象的指针。你不能使用你的MH，FY等。



你可以在成千上万的地方找到一些解释：http://bit.ly/1lLKRN4 [ ^ ]。



使用立即常量的整个想法，特别是字符串，非常糟糕。这种代码是不可支持的。这实际上是一个很大的编程反模式：

http://en.wikipedia.org/wiki/Magic_string_(programming)#Magic_strings_in_code [ ^ ]，

http ：//en.wikipedia.org/wiki/Magic_number_（编程）#Unnamed_numerical_constants [ ^ ]（是的，魔术数字也是如此）;

参见：http://en.wikipedia.org/wiki/Anti-pattern#Programming [ ^ ]。

-SA

void flight_operator::get_define_flight_num(char namecode){
    cout<<"Enter flight code:  ";
    cin>>namecode;
    switch(namecode){
        case "MH": case "mh":
            cout<<"Make Airlines\n"<<endl;
            break;
        case "FY": case "fy":
            cout<<"Fire Airlines\n"<<endl;
        default:
            cout<<"Invalid code!"<<endl;
            break;
    }

}

解决方案

What is not clear? A case label should be an integer constant. It could be one of integer types, enumeration type, something like this, but not a string, not an array, or any other more complex object or a pointer to object. You cannot use your "MH", "FY" and the like.

You could find some explanation in thousands of places: http://bit.ly/1lLKRN4[^].

The whole idea of using immediate constants, especially string, it very bad. This kind of code is not supportable. This is actually a big programming anti-pattern:
http://en.wikipedia.org/wiki/Magic_string_(programming)#Magic_strings_in_code[^],
http://en.wikipedia.org/wiki/Magic_number_(programming)#Unnamed_numerical_constants[^] (yes, magic numbers, too);
see also: http://en.wikipedia.org/wiki/Anti-pattern#Programming[^].

—SA

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