C++:重载operator =

Question

好的,所以我有一个类'弱键入'IE它可以存储许多不同的类型定义为:

#include <string>

class myObject{
   public:
      bool isString;
      std::string strVal;

      bool isNumber;
      double numVal;

      bool isBoolean;
      bool boolVal;

      double operator= (const myObject &);
};

我想像这样重载赋值运算符:

double myObject::operator= (const myObject &right){
   if(right.isNumber){
      return right.numVal;
   }else{
      // Arbitrary Throw.
      throw 5;
   }
}

所以我可以这样做:

int main(){
   myObject obj;
   obj.isNumber = true;
   obj.numVal = 17.5;
   //This is what I would like to do
   double number = obj;
}

但当我这样做时,我得到:

error: cannot convert ‘myObject’ to ‘double’ in initialization 

在任务.

我也尝试过:

int main(){
   myObject obj;
   obj.isNumber = true;
   obj.numVal = 17.5;
   //This is what I would like to do
   double number;
   number = obj;
}

我得到了:

error: cannot convert ‘myObject’ to ‘double’ in assignment

有什么我想念的吗？或者根本不可能通过重载进行转换operator=.

Answer 1

重载operator=分配时,改变行为,以你的类的对象.

如果要为其他类型提供隐式转换,则需要提供转换运算符,例如

operator double() const
{
    if (!isNumber)
        throw something();
    return numVal;
}

Answer 2

您真正想要的是转换运算符.

operator double() const { return numVal; }
operator int() const { ...

那就是说,你可能会喜欢boost :: variant.