Is it safe to change a function pointer (std::function) inside a called function?

Question

I have a std::function pointing to a function. Inside this function I change the pointer to another function.

std::function fun;

         


        

Answer 1

It may come back to bite you if you do it without due consideration and in code documentation, but there is no logical reason why it won't work.

In c++, the address of a function is not needed, either within the function in the return coding.

If it didn't work in some language the complier probably wouldn't accept it - if it's a half decent compiler.

Answer 2

This is legal with function pointers.

When you assign or construct a std::function with a target, it creates a copy of the target. In the case of assigning a function to the std::function, this in effect stores a function pointer as the target object.

When you invoke operator(), it is required to return what would happen if you invoked that the target with the arguments.

Within that "body" of the copy of the function object stored as a copy in the std::function, if you reassign to the std::function, this destroys the old target function object.

Destroying a function pointer has no impact on the validity of code executed within the function pointed to.

However, if you had stored function objects (lambdas, manual ones, other std::functions, std::bind, etc), at the point of assignment you'd run into the usual rules of running a method in a class when this is destroyed. In short, you would no longer be able to do anything that relied on "local state" of your instance.

std::function<void()> fun;
struct bob {
  std::string name;
  bob* next = 0;
  void operator()() const {
    std::cout << name << "\n";
    if (next) fun = *next;
    // undefined behavior:
    // std::cout << name << "\n";
  }
};
bob foo = {"foo"};
bob bar = {"bar", &foo};

int main() {
  fun = bar;
  fun();
  fun();
}

live example.

So as you can see, this can be fragile.