在CRTP中推断模板成员函数的返回类型 [英] Inferring return type of templated member functions in CRTP

问题描述

是否可以推断CRTP基类中模板成员函数的返回类型？

，它返回类型失败。考虑下面的例子。

While inferring argument types works well, it fails with the return type. Consider the example below.

#include <iostream>

template <typename Derived>
struct base
{
  template <typename R, typename T>
  R f(T x)
  {
    return static_cast<Derived&>(*this).f_impl(x);
  }
};

struct derived : base<derived>
{
  bool f_impl(int x)
  {
    std::cout << "f(" << x << ")" << std::endl;
    return true;
  }
};

int main()
{
  bool b = derived{}.f(42);
  return b ? 0 : 1;
}

这会产生以下错误：

  bool b = derived{}.f(42);
           ~~~~~~~~~~^
crtp.cc:7:5: note: candidate template ignored: couldn't infer template argument 'R'
  R f(T x)
    ^
1 error generated.

我直观的假设是，如果编译器能够推断类型 int 为 f 的参数，它也应该用于返回 bool ，因为两种类型在模板实例化时间是已知的。

My intuitive assumption is that if the compiler is capable of inferring type int for the argument to f, it should also work for the return bool, because both types are known at template instantiation time.

我尝试使用尾随返回类型函数语法，但后来找不到一个工作表达式放在 decltype

I tried using the trailing return type function syntax, but then failed to find a working expression to put in decltype.

对于函数有一个或多个模板参数，DietmarKühl提供了一个基于延迟使用间接层模板实例化的解决方案。不幸的是，当基类函数没有任何参数时，这不起作用，像这样：

For the case where the function has one or more templated arguments, Dietmar Kühl provided a solution based on delaying template instantiation using on layer of indirection. Unfotunately, this does not work when a base class function does not have any argument, like this:

template <typename R>
R g()
{
  return static_cast<Derived&>(*this).g_impl();
}

尝试使用相同的技术失败，因为不存在依赖类型。如何处理这种情况？

Attempts to use the same technique fail because there exist no dependent types. How does one handle this case?

正如Johannes Schaub指出的，C ++ 11功能默认模板参数，所以总是可以使 g 依赖于任意类型，然后应用Dietmar的解决方案：

As pointed out by Johannes Schaub, C++11 features default template arguments, so it is always possible to make g dependent on an arbitrary type and then apply Dietmar's solution:

template <typename T = void>
auto g() -> typename g_impl_result<Derived, T>::type
{
  return static_cast<Derived&>(*this).g_impl();
}

EDIT 3

这个问题在C ++ 14中不存在，因为我们对正常函数有返回类型的扣除，允许我们简单写：

EDIT 3

This problem does not exist in C++14 anymore, since we have return type deduction for normal functions, allowing us to write simply:

template <typename Derived>
struct base
{
  template <typename T>
  auto f(T x)
  {
    return static_cast<Derived&>(*this).f_impl(x);
  }

  auto g()
  {
    return static_cast<Derived&>(*this).g_impl();
  }
};

struct derived : base<derived>
{
  bool f_impl(int x)
  {
    return true;
  }

  double g_impl()
  {
    return 4.2;
  }
};

推荐答案

额外的间接是你的朋友：

An extra indirection is your friend:

template <typename D, typename T>
struct f_impl_result
{
    typedef decltype(static_cast<D*>(0)->f_impl(std::declval<T>())) type;
};

template <typename Derived>
struct base
{
    template <typename T>
    auto f(T x) -> typename f_impl_result<Derived, T>::type
    {
        return static_cast<Derived&>(*this).f_impl(x);
    }
};

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