Casting function pointer to functor in C++

Question

I have a function pointer type imported from another .hpp file. Something like:

typedef void (*PFN_func)(int i);

I want to create a functor of the same type:

std::function<PFN_func>

But this doesn't work. I don't want a solution like

std::function<void(int)>

Because mt function pointer definition is much more complicated

`std::function` wants a function type as its template parameter - which is not the same as pointer-to-function type. There's a difference between `void(int)` and `void(*)(int)`. `PFN_func` is a typedef for the latter. — Igor Tandetnik, Sep 09 '14 at 21:03

score 7 · Accepted Answer · edited May 23 '17 at 11:49

7

You could do this:

std::function<std::remove_pointer<PFN_func>::type>

Removing the pointer from void (*)(int) gives the function type void(int).

edited May 23 '17 at 11:49

Community

answered Sep 09 '14 at 21:06

Brian Bi

1 Answers1