模板参数的通用引用 [英] Universal reference to template template parameter

问题描述

是否可以通过通用引用传递模板模板参数值？例如，假设一个函数（不是）在STL序列上工作的最小示例：

Is it possible to pass a template template paramater value by universal reference? Consider for example this minimal example for a function (not) working on STL sequences:

#include <iostream>
#include <vector>

template < template<typename,typename> class C, template<typename> class A, typename T >
void func(C<T, A<T>>&& c) {
    // usually I'd std::forward here, but let's just use cout...
    std::cout << c.size() << "\n";
}

int main (int argc, char const* argv[]) {
    func(std::vector<float>(2));
    std::vector<float> lv(3);
    func(lv);
}

它不会编译，因为编译器不知道如何绑定在第二次调用func中的l值（lv）。

It won't compile since the compiler doesn't know how to bind the l-value ("lv") in the second call to func. I'm a bit lost when it comes to the deduction rules for the type of C. Can anyone enlighten me?

编辑：虽然我已经知道了猜测它是不相关的问题：我使用g ++ 4.9和clang 3.5（两个repo HEADs）

Edit: Although I guess it is not relevant for the question: I used g++ 4.9 and clang 3.5 (both repo HEADs)

推荐答案

通用引用是一个口语主义，并且它总是意味着严格地，一个引用 T&& 其中 T 是

"Universal reference" is a colloquialism, and it always means, strictly, a reference T && where T is a deduced template parameter.

您可以修改您的代码，仅使用：

You can modify your code, though, to use just that:

template <typename T>
void func(T && c) { /* ... std::forward<T>(c) ... */ }

如果您关心 T 是指定表单，请添加一个特征：

If you care that T is of the specified form, add a trait:

#include <type_traits>

template <typename T>
typename std::enable_if<IsATemplate<typename std::decay<T>::type>::value>::type
func(T && c) { /* ... std::forward<T>(c) ... */ }

$ c> IsATemplate trait留作练习。

Writing the IsATemplate trait is left as an exercise.

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