C ++模板部分专业化：为什么我不能匹配可变参数模板中的最后一个类型？ [英] C++ template partial specialization: Why cant I match the last type in variadic-template?

问题描述

我尝试编写一个 IsLast 类型特征，以检查给定类型是否是 std :: tuple ，但是下面的代码无法编译。我知道如何解决这个问题，但我很好奇编译器为什么不喜欢它。我想必须对我不知道的可变参数模板的专业化有一些规定。

I try to write a IsLast type traits to check if a given type is the last one in a std::tuple, but the code below does not compile. I know how to get around it but I am curious why the compiler does not like it.I guess there must be some rule on specialization of variadic-template that I am not aware of.

代码位于： https://godbolt.org/g/nXdodx

错误消息：

error: implicit instantiation of undefined template
 'IsLast<std::tuple<std::__cxx11::basic_string<char>, int>, int>'

关于特殊化声明的警告：

There is also a warning on specialization declaration:

警告：类模板部分特殊化包含无法推导的模板参数；

warning: class template partial specialization contains template parameters that cannot be deduced; this partial specialization will never be used

#include <tuple>
#include <string>

/////////This works
template<typename TP, typename T>
struct IsFirst;

template<typename U, typename ...V, typename T>
struct IsFirst <std::tuple<U, V...>, T>
{
  enum {value = false};
};

template<typename U, typename ...V>
struct IsFirst <std::tuple<U, V...>, U>
{
  enum {value = true};
};

////////This doesn't compile
template<typename TP, typename T>
struct IsLast;

template<typename ...U, typename V, typename T>
struct IsLast <std::tuple<U..., V>, T>
{
  enum {value = false};
};

template<typename ...U, typename V>
struct IsLast <std::tuple<U..., V>, V>
{
  enum {value = true};
};


int main()
{
  using T = std::tuple<std::string, int>;
  bool v1 = IsFirst<T, std::string>::value;
  bool v2 = IsLast <T, int>::value;
}

推荐答案

在类模板中，参数包必须位于所有其他模板参数之后，因此不允许这样。

In a class template, the parameter pack must come after all other template parameters, so something like this is not allowed.

template<typename U, typename ...V, typename T>
struct IsFirst <std::tuple<U, V...>, T>
{
  enum {value = false};
};

在功能模板中，仅在可以推导的情况下，包之后才能有其他模板参数。类模板不允许这样做，因为它们不允许推导。

In a function template, there can be other template parameters after the pack only if they can be deduced. This is not allowed for class templates because they do not allow deduction.

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