SGI STL 中的绑定好友模板

bound friend templates in SGI STL

本文关键字:好友 绑定 STL SGI      更新时间:2023-10-16

我知道SGI STL和我一样古老,但我仍然想弄清楚。

在 stl_stack.h 中,有一些代码如下:

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 stl_config.h 中,__STL_NULL_TMPL_ARGS定义如下:

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

但是当我尝试用G ++ 4.9.2编译它时,编译器是这样说的:

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,这个代码现在是非法代码还是什么?

谢谢!!!

stl_stack.h有一个错误,这个错误一定是当代编译器所关注的。在operator== <>被声明为模板之前提及operator==是违法的。

要解决此问题,请在定义stack之前声明operator==。但是,如果你在这一点上声明它,你不妨定义它。声明将要求向前声明stack

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,将operator==定义添加到顶部后,您将从底部删除它。