Template class function pointer type alias

Question

I am trying to make a type alias for a function (myFunction) inside myClass.

template<typename T>
using calc_t = matrix<T> (MyClass<T>::*myFunction)(const matrix<T> &X);

Where I would get a generic syntax error, missing ')' before MyClass<T>::*myFunction.

And then using it as so

calc_t<double> fptr = &MyClass<double>::myFunction;

I am not sure on the syntax to use in this specific case for when using the using type alias as opposed to a non-templated typedef.

I have looked at the following other SO questions that don't seem to cover this specific usage:

• Function pointer to member function

• Function pointer type not recognized inside template class

I have tried some other variants but to no success.

Answer 1

As an alternative, and as I already suggested in a comment, you could use std::function as it will be easier to use and more generic.

template <typename T>
class matrix { };

template <typename T>
class MyClass
{
public:
    matrix<T> myFunction(matrix<T> const&) { return {}; }
};

template<typename T>
using calc_t = std::function<matrix<T>(matrix<T> const&)>;

int main()
{
    MyClass<double> myObject;

    using namespace std::placeholders;  // For e.g. _1
    calc_t<double> myFunction = std::bind(&MyClass<double>::myFunction, myObject, _1);

    matrix<double> myFirstMatrix, mySecondMatrix;
    myFirstMatrix = myFunction(mySecondMatrix);
}

As shown above, you could use std::bind. But you could also use lambda expressions:

calc_t<double> myFunction = [&](matrix<double> const& m)
{
    return myObject.myFunction(m);
};

Or better yet (for this simple use-case anyway) use type-deduction

auto myFunction = [&](matrix<double> const& m)
{
    return myObject.myFunction(m);
};

With lambdas, type-deduction and templates you can create very generic and expressive and complex code in a simple way.