我想执行以下操作:
template <typename T>
struct foo
{
template <typename S>
friend struct foo<S>;
private:
// ...
};
但我的编译器 (VC8) 卡住了:
but my compiler (VC8) chokes on it:
error C3857: 'foo<T>': multiple template parameter lists are not allowed
我想要所有 T
的 template struct foo
朋友的所有可能的实例化.
I'd like to have all possible instantiations of template struct foo
friends of foo<T>
for all T
.
我该如何完成这项工作?
How do I make this work ?
这个
template <typename T>
struct foo
{
template <typename>
friend struct foo;
private:
// ...
};
似乎可以编译,但正确吗?朋友和模板的语法非常不自然.
seems to compile, but is it correct ? Friends and templates have very unnatural syntax.
template<typename> friend class foo
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?
this will however make all templates friends to each other. But I think this is what you want?
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