Can I declare pointer to function with default parameters? If can, the how?

Question

https://code.sololearn.com/c113l1QsG3ZI/?ref=app

Answer 1

https://en.cppreference.com/w/cpp/language/default_arguments
"Default arguments are only allowed in the parameter lists of function declarations and lambda-expressions, (since C++14) and are not allowed in the declarations of pointers to functions, references to functions, or in typedef declarations."
I don't know what you need this for but you could make them overloads and have 2 function pointers/1 function pointer then cast to the overloaded type, or use a lambda/functor/another function that takes a default parameter that invokes your function through the function pointer (I don't see the point of doing this).

Answer 2

Chirag Kumar see line with comment (//)

Answer 3

UraL ohh ok didn't notice that.

I think you cant. I don't know the perfect explanation so can't help :)

But others will guide you !!

Answer 4

jtrh I don't need it. It was in SoloLearn challenge

Answer 5

jtrh thanx