How do I initialize an object using a static member function of that object's class?

Question

I have a class with some static functions. The class constructor takes a function as one of its arguments.

template <typename T>
class MyClass {
    public:
        static std::function<T(T, T)> func1 = [](T a, T b) {
            ...
        }

        static std::function<T(T, T)> func2 = [](T a, T b) {
            ...
        }
     
        MyClass(std::function<T(T, T)> func) {
            ...
        }
}

When I initialize an object from this class:

MyClass<int> myObj(MyClass.func1);

The compiler complains that:

> argument list for class template "MyClass" is missing.

What is the correct way to do this while keeping the functions in the class itself?

You have to specify the T template parameter every time that you want to access it's members: MyClass<int> myObj(MyClass<int>::func1) — Ghasem Ramezani
– Ghasem Ramezani, Commented Jul 28, 2022 at 10:29
Assign a lambda function to static member variable (c++) is not a duplicate of this question. There is no issue with creating static class members here. — Yksisarvinen
– Yksisarvinen, Commented Jul 28, 2022 at 10:39
Those are not static member functions, they are static member variables. — molbdnilo
– molbdnilo, Commented Jul 28, 2022 at 10:45

463035818_is_not_an_ai · Accepted Answer · 2022-07-28 11:49:39Z

2

MyClass is a template not a class. The class that has a static member you want to pass to the constructor is MyClass<int>.

Your code has some typos (missing ;), I had to add inline for in-class initialization of the static members, I introduced an alias, and for brevity I used struct rather than class (the class has only public members):

#include <iostream>
#include <functional>


template <typename T>
struct MyClass {
    using fun_t = std::function<T(T,T)>;
    static inline fun_t func1 = [](T a, T b) { return a+b; };
    static inline fun_t func2 = [](T a, T b) { return a-b; };
    fun_t fun;
    MyClass(std::function<T(T, T)> func) : fun(func) {}
};

int main() {
    auto x = MyClass<int>{MyClass<int>::func1};
    std::cout << x.fun(1,2);
}

As pointed out in comments by AyxanHaqverdili, this would work as well:

auto x = MyClass{ MyClass<int>::func1 };

due to CTAD (class template argument deduction) where the class templates argument is deduced from the constructors parameters. Live Demo

Also this is fine, although a bit hacky, and I don't recommend to actually use it:

MyClass<int> x { x.func1 };

In a nutshell: You can refer to a variable in its own initializer. We aren't doing something that would require x to be fully initialized, so there is no problem with accessing the (already initialized) static member via an instance of MyClass<int>. Live Demo

A less suspicious variant of this suggested by @HTNW is

MyClass<int> x{decltype(x)::func1};

Now it is more obvious that we only use the type of x and access a static member of its class.

edited Jul 28, 2022 at 11:49

answered Jul 28, 2022 at 10:52

463035818_is_not_an_ai

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7 Comments

Aykhan Hagverdili Over a year ago

MyClass{ MyClass<int>::func1 }; also works.

Aykhan Hagverdili Over a year ago

So does MyClass<int> x { x.func1 };

463035818_is_not_an_ai Over a year ago

@AyxanHaqverdili good point about CTAD, though its kind of orthogonal, and would require more explanation. And the latter is also interesting, but nasty ;). I'll give it a try...

HTNW Over a year ago

I would honestly consider MyClass<int> x{x.func1}; or MyClass<int> x{decltype(x)::func1}; as preferred. It ensures func1 comes from the right class even if the type of x changes and it clearly states the type of x.

463035818_is_not_an_ai Over a year ago

@HTNW I dont like MyClass<int> x{x.func1}; because one has to inspect the class to see that it is not a problem. With MyClass<int> x{decltype(x)::func1}; its both obvious, we only care about xs type and access a static member of the class.

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