When declaring a function that has an argument pointer to function, is there such a way to restrict the argument func such that it belongs to a class

Question

Let's say that I have the function

doSomething(char* (*getterOne)())
{
   //do something here
}

although I've named the parameter of doSomething "getterOne", I have no way of verifying that the function is going to be a getter ( I think? ). So I wonder, is there a way to explicitly specify the kind of function that can be passed as an argument? For example, if I have the class "Cat", is there a way to say that the function doSomething accepts as parameter function, that belongs to the class Cat, and if so, how can I use it like this:

doSomething(char* (*getterOne)())
{
     Cat cat;
     cat.getterOne(); // where getterOne will be the getter that I pass as a parameter and do something with it

}

Also if anyone asks why I use char* instead of string, it is for a school project and we are not allowed to use string.

Does this answer your question? https://stackoverflow.com/questions/1485983/calling-c-member-functions-via-a-function-pointer — 463035818_is_not_an_ai, May 01 '22 at 18:07

score 1 · Accepted Answer · answered May 01 '22 at 18:33

A function pointer and a pointer-to-member-function are two different things. If you make a function that accepts a pointer-to-member-function you have no choice but to specify which class it belongs to.

Here is an example.

#include <iostream>

struct Cat {
    char* myGetter() {
        std::cout << "myGetter\n";
        return nullptr; 
    }
};

void doSomething(char* (Cat::*getterOne)())
{
     Cat cat;
     (cat.*getterOne)();
}

int main() {
    doSomething(&Cat::myGetter);
}

The syntax for using a pointer-to-member is .* or ->* and the extra brackets are needed.

When declaring a function that has an argument pointer to function, is there such a way to restrict the argument func such that it belongs to a class

1 Answers1