[英]Additional template arguments for friend declaration in template class?
In order for some global function为了一些全球function
template<typename T, int count>
void func (const Obj<T>& obj) {
for (int i = 0; i < count; i++)
std::cout << obj.value << std::endl;
}
to be able to access to the private field value
of some templated class能够访问某些模板化 class 的私有字段value
template<typename T>
class Obj {
public:
Obj (T value);
private:
T value;
};
template<typename T>
Obj<T>::Obj (T value) : value(value) {}
we need to declare func<T, count>
a friend of Obj<T>
.我们需要将func<T, count>
声明为Obj<T>
的朋友。 But func<T, count>
has to be declared before we can make it a friend of Obj<T>
, and for this we need to forward-declare Obj<T>
.但是func<T, count>
必须先声明,然后才能使其成为Obj<T>
的朋友,为此我们需要前向声明Obj<T>
。 The resulting code looks like this生成的代码如下所示
// Forward declarations
template<typename T>
class Obj;
template<typename T, int count>
void func (const Obj<T>& obj);
// Obj<T>
template<typename T>
class Obj {
public:
Obj (T value);
template<int count>
friend void func<T, count> (const Obj<T>& obj);
private:
T value;
};
template<typename T>
Obj<T>::Obj (T value) : value(value) {} // <-- ERROR
// func<T>
template<typename T, int count>
void func (const Obj<T>& obj) {
for (int i = 0; i < count; i++)
std::cout << obj.value << std::endl;
}
But this makes gcc complain about the "invalid use of template-id 'func' in declaration of primary template", so how do I actually declare func<T, count>
a friend of Obj<T>
for every count
?但这使得 gcc 抱怨“在主模板声明中无效使用模板 ID 'func'”,那么我如何实际声明func<T, count>
为Obj<T>
的每个count
的朋友? According to this answer I just need to replace the friend declaration with this根据这个答案,我只需要用这个替换朋友声明
template<typename T1, int count>
friend void func (const Obj<T1>& obj);
As far as I know this would make func<T1, count>
a friend of Obj<T>
regardless of whether T1
and T
match, which is absurd.据我所知,这将使func<T1, count>
成为Obj<T>
的朋友,无论T1
和T
是否匹配,这是荒谬的。 Is it possible to declare func<T, count>
a friend of Obj<T>
and no other Obj<T1>
?是否可以将func<T, count>
声明为Obj<T>
的朋友而没有其他Obj<T1>
? (preferably in a way that keeps the definition of func<T, count>
outside the definition of Obj<T>
) (最好将func<T, count>
的定义保持在Obj<T>
的定义之外)
(I know I could just make count
a real parameter, but the example above is just a simplification of my real code. In reality I'm trying to overload std::basic_ostream<CharT, Traits>& operator<< (std::basic_ostream<CharT, Traits>& stream, const Obj<T>& obj)
for some class Obj<T>
in a way that allows operator<<
to access private fields of Obj<T>
.) (我知道我可以让count
成为一个真实的参数,但上面的例子只是我真实代码的简化。实际上我试图重载std::basic_ostream<CharT, Traits>& operator<< (std::basic_ostream<CharT, Traits>& stream, const Obj<T>& obj)
用于某些 class Obj<T>
以允许operator<<
访问Obj<T>
的私有字段。)
The friend
declaration must match any possible forward declaration, and of course the definition, including template arguments. friend
元声明必须匹配任何可能的前向声明,当然还有定义,包括模板 arguments。
That means you need eg这意味着你需要例如
template<typename U, int count>
friend void func(const Obj<U>& obj);
It doesn't matter if the class template argument T
and the function template argument U
are different, as the calls will be made to the correct function anyway. class 模板参数T
和 function 模板参数U
是否不同并不重要,因为无论如何都会对正确的 function 进行调用。
Example:例子:
Obj<int> int_obj;
Obj<float> float_obj;
func<X>(int_obj); // Will call void func<int, X>(int_obj)
func<X>(float_obj); // Will call void func<float, X>(float_obj)
As an alternative, you can define the function inline in the class definition, and then you don't need to provide the T
or U
template arguments:作为替代方案,您可以在 class 定义中定义function 内联,然后您不需要提供T
或U
模板 arguments:
template<int count>
friend void func(const Obj<T>& obj)
{
// Implementation...
}
And in neither case you should really have a forward declaration of func
(as mentioned in my comment).在这两种情况下,您都应该真正拥有func
的前向声明(如我的评论中所述)。
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