Does std::optional change signature of the function?

Question

I need to make an optional argument with a default value in my function. Currently the signature looks something like this:

void func(int a, std::optional<int> b = 10)

and the function behaves in the following way:

func(15, 5); // works
func(15);    // works 

The question is: If I remove the explicit initialization for the optional argument, like this:

void func(int a, std::optional<int> b)

Then It seems like the signature of the function changes

func(15, 5); // works
func(15);    // fails 

Which makes me very confused about the purpose of the std::optional in the first place. What is it good for if not for creating optional arguments?

Answer 1

What is it good for if not for creating optional arguments?

std::optional is not supposed to be used for optional argument what you expect; which requires default argument as your 1st code sample showed, std::optional won't change the language syntax.

The class template std::optional manages an optional contained value, i.e. a value that may or may not be present.

You can used it like

void func(int a, std::optional<int> b = std::nullopt) {
    if (b) {
        // if b contains a value
        ...
    } else {
        ...
    }
}

then

func(15, 5); // b will contain a value (i.e. `5`)
func(15);    // b doesn't contain a value