c++ - 在它的构造函数中 cout 一个实例

Question

我知道这听起来很奇怪，并且有办法解决这个问题。但我只是想知道在 C++ 中是否可行。

class Item{
     public:
            string name;
     Item(string input){
            name = input
            cout << this;  // unfortunately the std::ostream& operator<<(std::ostream& outstream, Item* item) are not parsed by compiler yet. and this simply prints out its address.
                cout << *this; //I don't know how to override `cout << Item`.
     }
}

std::ostream& operator<<(std::ostream& outstream, Item* item){
    outstream << item->name;
    return outstream;
}

std::ostream& operator<<(std::ostream& outstream, Item& item){
    outstream << item.name;
    return outstream;
}

Answer 1

尝试以下操作

std::ostream & operator <<( std::ostream &outstream, const class Item &item );


class Item
{
     public:
            std::string name;
            Item( const std::string &input ) : name( input )
            {
                std::cout << *this;
            }
};

std::ostream & operator<<( std::ostream &outstream, const Item &item )
{
    outstream << item.name;
    return outstream;
}

或者如果数据成员名称将被定义为私有(private)那么你可以这样写

class Item
{
     private:
            std::string name;
     public:
            Item( const std::string &input ) : name( input )
            {
                std::cout << *this;
            }
           friend std::ostream & operator <<( std::ostream &outstream, const Item &item );

};

std::ostream & operator<<( std::ostream &outstream, const Item &item )
{
    outstream << item.name;
    return outstream;
}

重载运算符的方式相同

std::ostream & operator <<( std::ostream &outstream, const class Item *item );

如果你需要的话。然而实际上没有必要重载这个运算符。将对类对象的引用用作参数的运算符就足够了。