How do you assign a default value to a <functional> object? - c++11

Question

I pass a function pointer to a function using functional, and use a constructor initialization to save it in a local variable for later use. How do I assign a default value to the parameter.
Example:
function<void()> BEGINFILE;
somefunct(function<void()> BEGINFILE): BEGINFILE(BEGINFILE) {}
But I can't seem to do:
void nullfunct() {}
function<void()> BEGINFILE;
void somefunct(function<void()> BEGINFILE = nullfunct): BEGINFILE(BEGINFILE) {}
or:
void nullfunct() {}
function<void()> BEGINFILE;
somefunct(function<void()> BEGINFILE) {
BEGINFILE = BEGINFILE;
}
I've also read that functional is deprecated/removed in C++17. I've tried to find what C++17 does without success.

Answer 1

I suppose that somefunct is a class (or a struct) and that with
void somefunct(function<void()> BEGINFILE): BEGINfILE(BEGINFILE) {}
do you mean a constructor of somefunct.
First (secondary) problem: constructors doesn't return values, so remove the initial void.
For the main problem I suppose that you have the problem when the default function (nullfunct()) is a non static member function.
I mean, in this case
struct somefunct
{
void nullfunct() {}
std::function<void()> bf;
somefunct (std::function<void()> bf0 = nullfunct) : bf{bf0}
{ }
};
Unfortunately, a non-static member function is something strange, very different from a regular function, and you can't assign it to a std::function.
I see three ways to solve this problem.
transform it in a static one
A static member function doesn't depend from an instance of the class so is the same type of object of a regular function and can be assigned to a std::function, so if you can transform nullfunct() in a static member, you can write
struct somefunct
{
static void nullfunct() {}
std::function<void()> bf;
somefunct (std::function<void()> bf0 = nullfunct) : bf{bf0}
{ }
};
make nullfunct() an regular function.
If you can make nullfunct() a regular (not member of a class or struct) function, it becomes compatible with std::function, so
void nullfunct() {}
struct somefunct
{
std::function<void()> bf;
somefunct (std::function<void()> bf0 = nullfunct) : bf{bf0}
{ }
};
initialize with an empty function and set with a wrapping lambda
If you can't transform somefunct() in a static member function (way 1) or in regular function (way 2), you can wrap the call of somefunct() in a lambda function that you can assign to your std::function.
Unfortunately, this lambda function has to capture the this pointer and can't do it if is defined as a default value for the argument of the constructor so the way I see is initialize the std::function with an empty std::function and, in the body of the constructor, if the member contains an empty function, assign the lambda.
I mean
struct somefunct
{
void nullfunct() {}
std::function<void()> bf;
somefunct (std::function<void()> bf0 = {}) : bf{bf0}
{ if ( not bf ) bf = [this]{ this->nullfunct(); }; }
};