Function prototype, without definition

Question

I'am developing a library, that can't use Classes. I have a function I call when a get a KeyDown event and i want to call a function that i set the prototype for before in the library. The thing is, I don't want to be mandatory for the user to define this function if he doesn't want to handle any input. But the linker won't let me define a prototype without a definition or set a default empty definition. What can i do? I'am using C++ by the way...

This are the prototypes /

void keyDown(int virtual_keyCode);
void keyUp(int virtual_keyCode);

**Why** can’t you use classes? That’s a … *bizarre* … requirement. — Konrad Rudolph, Nov 05 '19 at 20:02
fwiw, the problem is not the declaration without a definition, but rather that you call the function in your code somewhere without providing a definition for it — 463035818_is_not_an_ai, Nov 05 '19 at 20:02
got curious about weak functions and found [this](https://stackoverflow.com/questions/35507446/what-are-weak-functions-and-what-are-their-uses-i-am-using-a-stm32f429-micro-co), seems to be a possible solution for your problem — 463035818_is_not_an_ai, Nov 05 '19 at 20:04
https://stackoverflow.com/questions/35507446/what-are-weak-functions-and-what-are-their-uses-i-am-using-a-stm32f429-micro-co — 463035818_is_not_an_ai, Nov 05 '19 at 20:31
I've used a weak reference but it still gives me an error when i redefine it. — Francisco Giancarelli, Nov 05 '19 at 20:35
Defining a prototype function, which is called in the code, but it's not necessary for the user to define it. Sounds like a bug report in the making. You should rethink your design. And you say you are using C++, but cannot use classes... what's going on??? You're making your life unnecessarily difficult — JHBonarius, Nov 05 '19 at 20:43
@FranciscoGiancarelli Can you add your attempt at making a weak function to the question? — HolyBlackCat, Nov 05 '19 at 21:26

JHBonarius · Accepted Answer · 2019-11-05T21:23:57.107

Redefining a function is not possible in C++. Read this about the One Definition Rule.

Only one definition of any variable, function, class type, enumeration type, concept (since C++20) or template is allowed in any one translation unit (some of these may have multiple declarations, but only one definition is allowed).

Overloading of class functions is possible with inheritance. Overloading is not the same as redefinition.

You can use function pointers, which you can reassign, e.g.

#include<functional>
#include<iostream>

std::function<void(int)> keyUp = [](int) {};

void unused() {
    keyUp(2);
}

void KeyUp(int i) {
    std::cout << i << i << '\n';
}

int main() {
    unused(); // no print
    keyUp = [](int i) { std::cout << i << '\n'; };
    unused(); // prints 2
    keyUp = KeyUp;
    unused(); // prints 22
}

It doesn't look the same, but effectively its the same...

Function prototype, without definition

1 Answers1