使运算符<<虚拟？ [英] Making operator<< virtual?

问题描述

我需要使用虚拟<<运算符。然而，当我尝试写：

  virtual friend ostream& operator<<（ostream& os，const Advertising& add）; 
  

我得到编译器错误

错误1错误C2575：'operator <<：：
只有成员函数和基数可以是
virtual

$ b $

解决方案

这个设置的问题是，运算符< ;&你定义的是一个自由函数，它不能是虚拟的（它没有接收者对象）。为了使函数成为虚拟的，它必须被定义为某个类的成员，这在这里是有问题的，因为如果定义operator<<作为类的成员，那么操作数的顺序将是错误的：

  class MyClass {
 public：
 virtual ostream&运算符<< （ostream& out）const; 
}; 
  

表示

  MyClass myObject; 
 cout<< myObject; 
  

无法编译，但

  MyClass myObject; 
 myObject<< cout; 
  

将是合法的。

这可以应用软件工程的基本定理 - 任何问题都可以通过添加另一层间接来解决。而不是使得操作员< virtual，考虑向类添加一个新的虚拟函数，如下所示：

  class MyClass {
 public：
 virtual void print（ostream& where; const; 
};然后，定义运算符<<< p>< / b> as 
  ostream&运算符<< （ostream& out，const MyClass& mc）{
 mc.print（out）; 
 return out; 
} 
  
这样操作符<自由函数具有正确的参数顺序，但是操作符<<可以在子类中自定义。
 
 
 
希望这有助于！
 
I need to use a virtual << operator. However, when I try to write:
virtual friend ostream & operator<<(ostream& os,const Advertising& add);
I get the compiler error

  
Error 1 error C2575: 'operator <<' :
  only member functions and bases can be
  virtual
How can I turn this operator virtual?
解决方案
The problem with this setup is that the operator<< you defined above is a free function, which can't be virtual (it has no receiver object).  In order to make the function virtual, it must be defined as a member of some class, which is problematic here because if you define operator<< as a member of a class then the operands will be in the wrong order:
class MyClass {
public:
    virtual ostream& operator<< (ostream& out) const;
};
means that
MyClass myObject;
cout << myObject;
will not compile, but
MyClass myObject;
myObject << cout;
will be legal.

To fix this, you can apply the Fundamental Theorem of Software Engineering - any problem can be solved by adding another layer of indirection.  Rather than making operator<< virtual, consider adding a new virtual function to the class that looks like this:
class MyClass {
public:
    virtual void print(ostream& where) const;
};
Then, define operator<< as
ostream& operator<< (ostream& out, const MyClass& mc) {
    mc.print(out);
    return out;
}
This way, the operator<< free function has the right parameter order, but the behavior of operator<< can be customized in subclasses.

Hope this helps!

                        
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