C ++抽象类操作符重载和接口执行问题 [英] C++ Abstract class operator overloading and interface enforcement question

问题描述

（从原始帖子编辑更改BaseMessage到const BaseMessage&）

Hello All，
我是C ++的新手，所以我希望你们的人可以帮助我看到我的方式的错误。

我有一个层次结构的消息，我试图使用抽象基类强制
一个接口。特别地，我想强制每个导出的消息提供重载的
<

当我尝试这样做时：

  class BaseMessage 
 {
 public：
 
 //一些非纯虚函数声明
 //一些纯虚函数声明
 
虚拟ostream& operator<<（ostream& stream，const BaseMessage& objectArg）= 0; 
 
} 
  

编译器会提示

错误：无法将参数objectArg声明为抽象类型BaseMessage

我相信还有朋友 ，但是当我试图将其声明为：

virtual friend ostream& operator<<<（ostream& stream，const BaseMessage objectArg）= 0 ;

编译器添加了一个附加错误

错误：虚函数不能是否有一种方法可以确保我的所有派生（消息）类提供一个<ostream操作符？

感谢很多，

Steve

解决方案

这个常见的惯例是在基本级别有一个 friend 输出运算符，并且调用private virtual function：

  class Base 
 {
 public：
 
 ///不要忘记这个
 virtual〜Base （）; 
 
 /// std stream interface 
 friend std :: ostream&operator<<（std :: ostream&out，const Base&b）
 {
 b.Print （out）; 
 return out; 
} 
 
 private：
 
 ///派生接口
 virtual void Print（std :: ostream&）const = 0; 
}; 
  

(edited from original post to change "BaseMessage" to "const BaseMessage&")

Hello All, I'm very new to C++, so I hope you folks can help me "see the errors of my ways".

I have a hierarchy of messages, and I'm trying to use an abstract base class to enforce an interface. In particular, I want to force each derived message to provide an overloaded << operator.

When I try doing this with something like this:

class BaseMessage
{
public:

// some non-pure virtual function declarations
// some pure virtual function declarations

virtual ostream& operator<<(ostream& stream, const BaseMessage& objectArg) = 0;

}

the compiler complains that

"error: cannot declare parameter ‘objectArg’ to be of abstract type ‘BaseMessage’

I believe there are also "friend" issues involved here, but when I tried to declare it as:

virtual friend ostream& operator<<(ostream& stream, const BaseMessage objectArg) = 0;

the compiler added an addition error

"error: virtual functions cannot be friends"

Is there a way to ensure that all of my derived (message) classes provide an "<<" ostream operator?

Thanks Much,

Steve

解决方案

The common convention for this is to have a friend output operator at the base level and have it call private virtual function:

class Base
{
public:

    /// don't forget this
    virtual ~Base();

    /// std stream interface
    friend std::ostream& operator<<( std::ostream& out, const Base& b )
    {
        b.Print( out );
        return out;
    }

private:

    /// derivation interface
    virtual void Print( std::ostream& ) const =0;
};

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