过载运算符<<用于嵌套的类模板 [英] Overload operator<< for nested class template

问题描述

我有以下设置:

template< class T >
struct Foo {

  struct Bar {
    Bar ( const T &t ) : otherT_( t ) {}

    T otherT_;
  };

  Foo ( const T &t ) : myT_( t ) {}

  T myT_;
};

现在，我想使Foo< T >::Bar的实例可流式传输到std :: cout和朋友.我试过了:

Now, I want to make instances of Foo< T >::Bar streamable to std::cout and friends. I tried this:

template< class T >
std::ostream& operator<< ( std::ostream &os, 
                           const typename Foo< T >::Bar &bar ) {
  os << "<bar: " << bar.otherT_ << ">";
  return os;
}

但是以下代码无法编译:

But the following code does not compile:

  Foo< int > foo( 5 );
  Foo< int >::Bar bar( 7 );

  std::cout << bar << std::endl;

我猜编译器无法推断类型T或其他内容.有没有办法使operator<<这样的嵌套类实例表现良好?

I guess that the compiler is not able to deduce the type T or something. Is there a way to make such instances of the nested class behave well with operator<<?

谢谢！

推荐答案

是的，简单的方法是将operator<<放在Bar中:

Yep, the easy way is to put operator<< inside Bar:

struct Bar {
  Bar ( const T &t ) : otherT_( t ) {}

  T otherT_;

  friend std::ostream& operator<< ( std::ostream &os, const Bar &bar ) 
  {
    os << "<bar: " << bar.otherT_ << ">";
    return os;
  }
};

我在另辟...径...

I am digging the other way ...

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