模板别名如何影响模板参数的扣除？ [英] How do template aliases affect template parameter deduction?

问题描述

在C ++ 03中，模板参数扣除在一些上下文中不会发生。例如：

In C++03, template parameter deduction does not occur in some contexts. For example:

template <typename T> struct B {};

template <typename T>
struct A
{
    typedef B<T> type;
};

template <typename T>
void f(typename A<T>::type);

int main()
{
    B<int> b;
    f(b);  // ERROR: no match
}

这里， int 不会推导出 T ，因为嵌套类型如 A< T> :: type 是一个非推导的上下文。

Here, int is not deduced for T, because a nested type such as A<T>::type is a non-deduced context.

如果我写了这样的函数：

Had I written the function like this:

template <typename T> struct B {};

template <typename T>
void f(B<T>);

int main()
{
    B<int> b;
    f(b);
}

一切都很好，因为 B< T& code> 是一个推导的上下文。

everything is fine because B<T> is a deduced context.

然而，在C ++ 11中，模板别名可用于伪装嵌套类型在语法类似于第二个例子。例如：

In C++11, however, template aliases can be used to disguise a nested type in syntax similar to the second example. For example:

template <typename T> struct B {};

template <typename T>
struct A
{
    typedef B<T> type;
};

template <typename T>
using C = typename A<T>::type;

template <typename T>
void f(C<T>);

int main()
{
    B<int> b;
    f(b);
}

在这种情况下模板参数扣除是否有效？换句话说，模板别名是一个推导的上下文还是一个非推导的上下文？

Would template argument deduction work in this case? In other words, are template aliases a deduced context or a non-deduced context? Or do they inherit the deduced/non-deduced status of whatever they alias?

推荐答案

在其他方面，您可以继承他们所有别名的推论/未推论状态词，是模板别名推导的上下文还是非推导的上下文？

In other words, are template aliases a deduced context or a non-deduced context?

它们可以推演为等效代码，而不使用模板别名。例如

They are as deducible as the equivalent code without using template aliases. For example

template<typename T>
using ref = T&;

template<typename T>
void f(ref<T> r);

现在您可以调用 f（x）和 T 将被完全推断。在 f 的定义时间， ref< T> 被类型 。 T& 是一个推导的上下文。

Now you can call f(x) and T will be deduced perfectly fine. At the definition time of f already, ref<T> is replaced by type T&. And T& is a deduced context.

在您的情况下， C< T> 被 typename A< T& ：type ，这是 T 的非推演上下文，因此 T

In your case C<T> is replaced by typename A<T>::type, and that is a non-deduced context for T, so T cannot be deduced.

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