与其他模板类中的所有模板类成为朋友
看下面的代码：
template <typename T, int d>
class Grid {
//Following line is what I need to change
template<int d2> friend class Iterator<T,d,d2>;
}
template <typename T, int d, int d2>
class Iterator{
//some code that use private fields of Grid<T,d>
}
template <typename T, int d>
class Iterator<T,d,0>{
//This specialized class also need to use private parts of Grid<T,d>
}
专用和非专用迭代器都应该有权访问私有部分。 行：
template<int d2> friend class Iterator<T,d,d2>;
编译时出现错误：部分专业化“迭代器”声明的朋友
有人知道如何替换它吗？
编辑： 感谢@Xeo 评论，我能够做出解决方法：
template<typename TT, int dd, int d2> friend class Iterator;
但是，这使朋友可以访问所有迭代器模板，而不仅仅是那些具有匹配的第一个和第二个模板参数的模板。
Look at following code:
template <typename T, int d>
class Grid {
//Following line is what I need to change
template<int d2> friend class Iterator<T,d,d2>;
}
template <typename T, int d, int d2>
class Iterator{
//some code that use private fields of Grid<T,d>
}
template <typename T, int d>
class Iterator<T,d,0>{
//This specialized class also need to use private parts of Grid<T,d>
}
Both specialized and not specialized Iterator should have access to private parts.
Line:
template<int d2> friend class Iterator<T,d,d2>;
does not compile with error: partial specialization `Iterator' declared friend
Does anybody know how to replace it?
EDIT:
Thanks to @Xeo comment i was able to make a workaround:
template<typename TT, int dd, int d2> friend class Iterator;
However this gives friend access to all Iterator templates not only to those one that have matching first and second template parameter.
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这只是一个解决方法，但它正在起作用：
This is just a workaround but it is working: