是否有一种方法来声明模板化函数的类型名称？ [英] Is There a Way to Declare a typename for a Templatized Function?

问题描述

所以，我有这种模板化的功能（我知道这很难看。）

So, I have this templatized function (which I know is ugly to look at.)

我的意图是不默认模板参数，我的目的是创建一个 typename 派生自 T

My intention was not to default the template parameter though, my intention was to create a typename derived from T that could be used in caster that the user could not assign to.

我可以在 caster 问题是如何为用户不能作为参数传递的模板化函数创建 typename ？

My question is how do I create a typename for a templatized function which the user cannot pass as an argument?

例如：

template <typename T>
typename R = std::conditional<sizeof(T) == 4, char, short>;
R foo(T bar){return R(bar);}

不编译，但这是我想完成的行为。

Clearly this code doesn't compile, but that's the behavior I'd like to accomplish. Is a functor the only way to do this?

推荐答案

在C ++ 14中，这是一个使用返回类型推导优化解决的方法。

In C++14, this is elegantly solved using return type deduction.

// C++14
#include <type_traits>

template <typename T>
decltype(auto)
foo(T bar)
{
  using R = std::conditional_t<sizeof(T) == 4, char, short>;
  return static_cast<R>(bar);
}

在C ++ 11中，您必须重复类型计算。

In C++11, you'd have to repeat the type computation.

// C++11
#include <type_traits>

template <typename T>
typename std::conditional<sizeof(T) == 4, char, short>::type
foo(T bar)
{
  using R = typename std::conditional<sizeof(T) == 4, char, short>::type;
  return static_cast<R>(bar);
}

使用 decltype 找出类型。

// C++11
#include <type_traits>

template <typename T>
typename std::conditional<sizeof(T) == 4, char, short>::type
foo(T bar)
{
  using R = decltype(foo(bar));
  return static_cast<R>(bar);
}

但是坦率地说，使用简单的默认类型参数有什么问题？ p>

But frankly, what is wrong with using a simple default type parameter?

// C++11
#include <type_traits>

template <typename T,
          typename R = typename std::conditional<sizeof(T) == 4, char, short>::type>
R
foo(T bar)
{
  return static_cast<R>(bar);
}

注意，我已经替换了 R在 return 语句中使用 static_cast 来隐藏编译器关于缩小转换的警告，你确定这是你想要的吗？

Note that I have replaced the value initialization of R in your return statement with a static_cast to silence the compiler warning about the narrowing conversion. Are you sure that this is what you want, though?

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