How can I make a function pointer parameter no-op by default?

Question

Given my function that takes a function pointer as parameter

void call_func(std::function<void()>func) {
    func();
}

The most straight forward way is something like

void no_op() {
    ;
}

void call_func(std::function<void()>func = no_op) {
    func();
}

Is there a cleaner way to do this so that I can avoid creating the literally useless no_op function?

Or use `nullptr` and change the body to `if (func) func();`? — Cornstalks, Dec 20 '18 at 03:17
Possible duplicate of [Does a no-op "do nothing" function object exist in C++(0x)?](https://stackoverflow.com/questions/2982592/does-a-no-op-do-nothing-function-object-exist-in-c0x) — Rufus, Dec 20 '18 at 05:17

songyuanyao · Accepted Answer · 2018-12-20T03:25:32.840

6

You can use an empty lambda, which could be also wrapped by std::function. e.g.

void call_func(std::function<void()>func = []{}) {
    func();
}

edited Dec 20 '18 at 03:25

answered Dec 20 '18 at 03:17

songyuanyao

3

Maybe force a decay first `+[]{}` - The way in the default case the `std::function` won't need to allocate any storage. – StoryTeller - Unslander Monica Dec 20 '18 at 05:56

1 Answers1