Basically, I would like to have the following semantic :
#include <functional>
#include <iostream>
class test
{
public:
void add(std::function<void()> f)
{
f();
}
void operator()()
{
++x;
}
int x = 33;
};
int main()
{
test t;
t.add(t);
// wanted: x == 34 instead: x == 33 since add(t) copies it
}
I understand std::function wraps a copy of the callable object but is there any way to get a reference to a callable object using std::function?
You want to use the std::ref
template function to create a reference wrapper for your instance:
std::reference_wrapper
is a class template that wraps a reference in a copyable, assignable object.Function templates
ref
andcref
are helper functions that generate an object of typestd::reference_wrapper
, using template argument deduction to determine the template argument of the result.
You would use it like this:
t.add(std::ref(t));