C++ 如何将模板化结构/类声明为朋友？
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How to declare a templated struct/class as a friend?
提问by Alexandre C.
I'd like to do the following:
我想做以下事情：
template <typename T>
struct foo
{
template <typename S>
friend struct foo<S>;
private:
// ...
};
but my compiler (VC8) chokes on it:
但我的编译器 (VC8) 窒息了：
error C3857: 'foo<T>': multiple template parameter lists are not allowed
I'd like to have all possible instantiations of template struct foofriends of foo<T>for all T.
我想拥有for all的template struct foo朋友的所有可能的实例。foo<T>T
How do I make this work ?
我如何使这项工作？
EDIT: This
编辑：这个
template <typename T>
struct foo
{
template <typename>
friend struct foo;
private:
// ...
};
seems to compile, but is it correct ? Friends and templates have very unnatural syntax.
似乎可以编译，但它正确吗？朋友和模板的语法非常不自然。
回答by Anycorn
template<typename> friend class foo
this will however make all templates friends to each other. But I think this is what you want?
然而，这将使所有模板彼此成为朋友。但我想这就是你想要的吗？
