casting void* to std::function

Question

I have an issue. I'm trying to convert a void* to std::function. This is just a simple example, any suggestions will be appreciated

#.h file
class Example {

public:

  Example();
  int foo(void* hi);

  int fooFunc(std::function<int(int, int)> const& arg, int x, int y) {
    foo(arg.target<void*>(), x, y);
    return 2;
  }

};


#.cpp file
Example::Example() {

  }

  int Example::foo(void * func, int x, int y)
  {
    //cast back to std::function
    func(x, y);
    std::cout << "running in foo: " << a << "\n";
    return a;
  }

Every casting i tried did not work.

I know i can send a std::function in this example, but it's for something bigger and i'm working on an example to make it work here.

The whole meaning of void*, is for sometimes to use it, in these situations, when you don't know what you will receive, and then cast it to the specific usage you need.

Thanks!

Why do you do that? What is the `void*` coming from (some `dlsym`?). Notice that closures are bigger than one pointer. — Basile Starynkevitch, Feb 19 '17 at 15:15
Some dsystem lib. If i wouldn't have to. I would not have done it — Mumfordwiz, Feb 19 '17 at 15:23
It isn't possible to store a `void*` in a `std::function`, because it isn't callable. Your call to `target` will always return a null pointer. — Joseph Thomson, Feb 19 '17 at 15:24
Please **edit your question** to improve it (don't comment it). — Basile Starynkevitch, Feb 19 '17 at 15:29
Why should i edit? It's good that people know it's not possible to do this cast. Stop giving -1 on questions, that cannot be done. It's good to know that too. — Mumfordwiz, Feb 19 '17 at 15:31

n. m. could be an AI · Answer 1 · 2019-08-02T11:16:16.377

You can't.

You can cast a data pointer to void* and then back to the same pointer type you have started with. std::function is not a pointer type, so the cast is statically invalid, and it's not the same thing you have started with. You have started with a .target of type void(*)() but it's not a data pointer, it's a function pointer, so casting it to void* and back is implementation-defined.

You can:

Ignore the issue and cast to void(*)() anyway. Will work on most (but not all) platforms.
Use void(*)() instead of void* as a universal function pointer (you can cast it to other function types).
Use whatever tools C++ offers to avoid the cast altogether.

casting void* to std::function

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