简单的可变参数模板函数不能实例化 [英] Simple variadic template function can't instantinate

问题描述

我知道 sizeof ...（Args ...）产生了C ++ 0x打包模板参数列表中的类型数，但我想要

I'm aware that sizeof...(Args...) yields the number of types in a C++0x packed template argument list, but I wanted to implement it in terms of other features for demonstation purposes, but it won't compile.

// This is not a solution -- overload ambiguity.
// template <typename... Args> size_t num_args ();          // Line 7
// template <>
constexpr size_t num_args ()
{
    return 0;
}

template <typename H, typename... T>
constexpr size_t num_args ()                                // Line 16
{
    return 1 + num_args <T...> (); // *HERE*
}

int main ()
{
    std :: cout << num_args <int, int, int> ();
}

此错误在 * HERE * 与

No matching function call to ...
... candidate is template<class H, class ... T> size_t num_args()

它没有看到首先定义的基本案例。转发声明模板< typename ... T> num_args（）; 在重载解析中引入歧义。

i.e. it's not seeing the base case which is defined first. Forward-declaring template<typename...T>num_args(); introduces ambiguity in overload resolution.

x.cpp:30:45: note: candidates are:
x.cpp:7:36: note: size_t num_args() [with Args = {int, float, char}, size_t = long unsigned int]
x.cpp:16:9: note: size_t num_args() [with H = int, T = {float, char}, size_t = long unsigned int]

我使用gcc 4.6。如何使这项工作？

I am using gcc 4.6. How can I make this work?

感谢。

推荐答案

没有声明基本案例。您有 num_args 函数的无模板重载，但调用函数 num_args 这将永远不会被找到，显而易见的原因：它将永远尝试实例化一个函数模板。

You didn’t declare a base case. You have a template-free overload of your num_args function but when calling a function num_args<T...>() this will never be found, for obvious reasons: it will always try to instantiate a function template.

template <>
constexpr size_t num_args<>()
{
    return 0;
}

但是，这不会工作，因为这里你专门化一个无参数函数模板和这样的模板不存在：您的其他函数模板 num_args 始终至少有一个参数 H 。

However, this won’t work either since here you’re specialising a parameterless function template and such a template doesn’t exist: your other function template num_args always has at least one argument, H.

为了真正使这项工作，你需要部分专业化，这些只存在于类模板。这是您在这里需要的。

In order to really make this work you need partial specialisations, and these only exist for class templates. So this is what you need here.

template <typename T>
struct num_args_t;

template <>
struct num_args_t {
    static size_t const value = 0;
};

template <typename H, typename T...>
struct num_args_t {
    static size_t const value = num_args_t<T...>::value + 1;
};

template <typename T...>
constexpr size_t num_args() {
    return num_args_t<T...>::value;
}

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