Q:

How to call any class function using its pointer if we know arguments and return type?

How to create a function pointer to any class function knowing only arguments and return type? How to call this function later?
I read about std::function, however I do not have an idea how to implement it without using specific class name like "std::function<void(const ClassName&, int)> f_add_display = &ClassName::func;"
The example below is NOT for compiling, it is only to show the idea I mean:
class collection {
    p* ...; //pointer to any(!) class function with known arguments and return type
} _collection;

class One {
public:
    ...
    bool Foo_1(int, int) {};
    void saveFuncAddr() {_collection.p = this::Foo_1};
};

class Two {
public: 
    bool Foo_2(int, int) {};
    void saveFuncAddr() {_collection.p = this::Foo_2};
};

int main() {
    one* = new One();
    one->saveFuncAddr();
    bool res1 = (*_collection.p)(1, 2);

    two* = new Two();
    two->saveFuncAddr();
    bool res2 = (*_collection.p)(1, 2);
}

A:

First, identifiers beginning with an underscore is reserved in the global namespace, so you shouldn't declare a name like _collection.
You can use lambdas to wrap your member functions:
#include <functional>

struct Collection {
    std::function<bool(int, int)> p; 
} collection;

class One {
public:
    bool Foo_1(int, int) {return true;}
    void saveFuncAddr() 
    {
        collection.p = [this](int a, int b){return this->Foo_1(a, b);};
    }
};

class Two {
public: 
    bool Foo_2(int, int) {return true;}
    void saveFuncAddr() 
    {
        collection.p = [this](int a, int b){return this->Foo_2(a, b);};
    }
};

int main() {
    auto one = new One();
    one->saveFuncAddr();
    bool res1 = (collection.p)(1, 2);
    delete one;

    auto two = new Two();
    two->saveFuncAddr();
    bool res2 = (collection.p)(1, 2);
    delete two;
}

