How to define out of class functions for a class specialized using std::enable_if

Question

I have a specialization of a class called graph which is enabled only if the input is a particular type. I am not able to define out of class definitions for the functions inside that class. This question os different from some other questions on stack overflow where sfinae happens on member function. Here I want enable if on the class and just define a normal member function for this class outside the class.

Note- There are multiple graph classes with different container types. This is an example of just one.

I want to be able to define graph_func outside this class

template<typename ContainerType,
    std::enable_if<std::is_same<ContainerType, Eigen::MatrixXd>::value, int>::type = 0>
class graph
{
    .
    .
    .
    void graph_func() const;
}

I tried this and I get the error that it doesn't refer into any class

template <typename ContainerType>
void graph<ContainerType,  std::enable_if<std::is_same<Graph, Eigen::MatrixXd>::value, int>::type>::graph_func() const
{
  // definition
}

You could simplify your specialization definition to `template <> class graph`, there's no reason to use SFINAE. If you want to restrict `ContainerType` to be `Eigen::MatrixXd`, then use `static_assert`, or don't use a template at all — Piotr Skotnicki, Aug 31 '19 at 06:29
I have to use SFINAE because I have multiple graph class specializations for different container types. I don't think I can use static assert since that would occur after class selection and would eliminate the other specializations from selection. — yasht, Aug 31 '19 at 17:29
That should be included in the question itself, as together with the code you wrote it all makes it unclear. — Piotr Skotnicki, Aug 31 '19 at 17:34

康桓瑋 · Accepted Answer · 2019-08-31T10:27:45.427

Notice that std::enable_if<..., int>::type in your parameter list is an non-type template argument:

template<typename ContainerType,
    typename std::enable_if<std::is_same<ContainerType, Eigen::MatrixXd>::value, int>::type = 0>
class graph
{
    void graph_func() const;
};

You need to pass the value of that type (In here we just named it _) to the parameter list:

template <typename ContainerType,
    typename std::enable_if<std::is_same<ContainerType, Eigen::MatrixXd>::value, int>::type _>
void graph<ContainerType, _>::graph_func() const
{
  // definition
}

See live demo.

How to define out of class functions for a class specialized using std::enable_if

1 Answers1