C++ function that returns a pointer to a member function

Question

the following works as expected:

template<typename... Types>
auto countNumberOfTypes() { return sizeof...(Types); }

template<typename... Types>
consteval auto functionReturnsFunction() { return countNumberOfTypes<Types...> };

functionReturnsFunction<int, const double>()() == 2;

but the following does not even compile:

struct Test
{
    template<typename... Types>
    auto countNumberOfTypes() { return sizeof...(Types); }
};

template<typename... Types>
consteval auto functionReturnsFunction2() { return &Test::countNumberOfTypes<Types...>; }

// functionReturnsFunction2<int, const double>()() == 2;

error: must use ‘.*’ or ‘->*’ to call pointer-to-member function in ‘&Test::countNumberOfTypes (...)’, e.g. ‘(... ->* &Test::countNumberOfTypes) (...)’
   29 |     if (functionReturnsFunction2<int, const double>()() == 2)
      |         ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~

...any suggestions?

Please show a minimal reproducible example that produces your error. Your "does not even compile" code compiles and doesn't generate anything — Drew Dormann
– Drew Dormann, Commented Oct 31, 2023 at 20:33
You're returning a pointer to a non-static member function, which has to be called on an object. Either make the function static, or plan to have an instance of the class when calling it. — Ben Voigt
– Ben Voigt, Commented Oct 31, 2023 at 20:36
In order to call a non-static member function of class Test, you need an object of type Test. There is no such object in functionReturnsFunction2<int, const double>()(). How do you expect this to work? — n. m. could be an AI
– n. m. could be an AI, Commented Oct 31, 2023 at 20:36
Should be (Test{}.*functionReturnsFunction2<int, const double>())() == 2. — Jarod42
– Jarod42, Commented Oct 31, 2023 at 20:37
Error message seemed quite clear. ` must use ‘.*’ or ‘->*’ to call pointer-to-member function in ‘&Test::countNumberOfTypes (...)’` - ` - you are trying to call a pointer-to-member function. ` e.g. ‘(... ->* &Test::countNumberOfTypes) (...)’ it's telling you what you should have written. (something->*&Test::countNumberOfTypes)(something) - how would you recommend improving the error message? — Raymond Chen
– Raymond Chen, Commented Oct 31, 2023 at 21:16

Jarod42 · Accepted Answer · 2023-10-31 21:04:56Z

4

Pointer to member functions requires special syntax with .* or ->* to be used (and an object).

(Test{}.*functionReturnsFunction2<int, const double>())() == 2)

Alternatively, you might use std::invoke which might have more regular syntax

(std::invoke(functionReturnsFunction2<int, const double>(), Test{}) == 2)

answered Oct 31, 2023 at 21:04

Jarod42

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