根据模板启用类的成员 [英] enable class's member depending on template

问题描述

我已经知道您可以使用std::enable_if

I already know that you can enable (or not) a class's method using std::enable_if

例如:

template<size_t D, size_t E>
class Field
{
  ...

  size_t offset(const std::array<float,D>& p) const
  {
    ...
  }

  template<typename TT = size_t>
  typename std::enable_if<D!=E, TT>::type
  offset(const std::array<float,E>& p) const
  {
    return offset(_projection(p));
  }

  ...
};

这有助于在特定情况下无法调用无效的函数，并消除重载错误……对我来说，这非常好！

This helps not being able to call function that are invalid in a specific case as well as removing overloading errors ... which, to me, is very nice !

我想走得更远，只有在需要时，让班上的一些成员才在场.这样，如果我尝试使用否则将不会启动的对象，就会出现错误

I'd like to go further and make some of my class's members being present only if the are needed. That way I would get an error if I try to use an objected which would have otherwise not been initiated

我试图做

template<size_t D, size_t E>
class Field
{
  ...

  template<typename TT = projectionFunc>
  typename std::enable_if<D!=E, TT>::type _projection;
}

但是编译器告诉我:

erreur: data member ‘_projection’ cannot be a member template

有什么办法可以实现我想要的吗?

Is there any way to achieve what I want ?

推荐答案

将数据成员保留在一个单独的类中，然后可以根据需要对其进行专门化.

Hold the data members in a separate class that you can then specialize as needed.

template<size_t D, size_t E>
class Field {
    template<size_t, size_t> struct Field_Members {
        int _projection;
    };
    template<size_t V> struct Field_Members<V, V> { };
    Field_Members<D, E> m;
};

，然后使用m._projection等.

Field_Members不必是嵌套的类模板；您可以根据需要将其移到外面.也可以从中继承Field，但是它将成为一个从属基础，并且您必须编写this->_projection，因此它不会节省太多键入内容.

Field_Members doesn't have to be a nested class template; you can move it outside if desired. It is also possible to have Field inherit from it, but then it'd be a dependent base, and you'd have to write this->_projection, so it doesn't save much typing.

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