SGI STL中的绑定朋友模板

Question

我知道SGI STL和我自己一样老，但我还是想弄清楚。

在stack.h中，有如下代码：

template <class T, class Sequence = Deque<T> >
class Stack {
    friend bool operator== __STL_NULL_TMPL_ARGS (const Stack&, const Stack&);
    friend bool operator< __STL_NULL_TMPL_ARGS (const Stack&, const Stack&);
protected:
    Sequence c;
};

template <class T, class Sequence>
bool operator== (const Stack<T, Sequence>& x, const Stack<T, Sequence>& y){
    return x.c == y.c;
}

template <class T, class Sequence>
bool operator< (const Stack<T, Sequence>& x, const Stack<T, Sequence>& y){
    return x.c < y.c;
}

在config.h中，__STL_NULL_TMPL_ARGS的定义如下：

# ifdef __STL_CLASS_PARTIAL_SPECIALIZATION
#   define __STL_TEMPLATE_NULL template<>
# else
#   define __STL_TEMPLATE_NULL
# endif

但是，当我试图用G++ 4.9.2编译它时，编译器说：

In file included from stack.cpp:1:0:
stack.h:13:22: error: declaration of ‘operator==’ as non-function
  friend bool operator== __STL_NULL_TMPL_ARGS (const Stack&, const Stack&);
                      ^
stack.h:13:22: error: expected ‘;’ at end of member declaration
In file included from iterator.h:3:0,
                 from deque.h:6,
                 from stack.h:5,
                 from stack.cpp:1:
stl_config.h:111:31: error: expected unqualified-id before ‘<’ token
 # define __STL_NULL_TMPL_ARGS <>
                               ^
stack.h:13:25: note: in expansion of macro ‘__STL_NULL_TMPL_ARGS’
  friend bool operator== __STL_NULL_TMPL_ARGS (const Stack&, const Stack&);

我不知道为什么完全相同的代码不能在我的电脑上编译，这是现在的非法代码还是什么？

非常感谢！

Answer 1

stl_stack.h有一个错误，这肯定是在当代编译器的关注之下。在将operator== <>声明为模板之前，提及operator==是非法的。

要解决这个问题，请在定义operator==之前声明stack。但是如果你在这一点上声明它，你也可以定义它。声明将需要stack的前向声明。

template <class T, class Sequence>
class stack; // Forward declare for sake of operator== declaration.

template <class T, class Sequence>
bool operator==(const stack<T, Sequence>& x, const stack<T, Sequence>& y) {
  return x.c == y.c;
}

#ifndef __STL_LIMITED_DEFAULT_TEMPLATES
template <class T, class Sequence = deque<T> >
#else
…

(当然，在将operator==定义添加到顶部之后，您将从底部删除它。)