Send void type defined function pointer as argument to be called

Question

I don't know if what I am trying to do is possible but it would be grate to be. It will really prove that C++ is a strong language.

So, I have a DLL and an EXE that uses a function exported by the DLL. That function takes an argument which is, or must be a pointer to another function, and executes it. Something like this:

extern void RunFunction(void (*FunctionPonter)()) {
    (*FunctionPonter)();
}

and I call it like this:

RunFunction(&FunctionToExecute);

The function sent as an argument will not have arguments, just a void function.

So, all this works, as it is presented here. Now, I would like to go further and define (*void)() as, for example, Action.

typedef (*void)() Action;

Action FunctionToExecute() {
    // My code. This function won't return results - void
}

and I would like to send the pointer to the function in my DLL like this:

// This is how it would be declared now in the DLL
extern void RunFunction(void (ACTION) {
    (*FunctionPonter)();
}

// and this is how I would like to use it
RunFunction(FunctionToExecute);

I am using VC2010 and I don't know how could I do that, if it is possible. Thanks for your ideas, explanations and help!

This starts off well, but then becomes completely unclear. What do you want your second example to do, that your first example doesn't already do? — Oliver Charlesworth
– Oliver Charlesworth, Commented Jul 1, 2012 at 15:51
It's just to create a typedef for void (*)(), to differentiate it from other void functions that are not created to be sent as arguments. Just that! — ali
– ali, Commented Jul 1, 2012 at 15:52
C++ is indeed a strong language, if the requirements are clear and reasonable! — iammilind
– iammilind, Commented Jul 1, 2012 at 15:53
A typedef is not a new type, it's just an alias. So a void (*)() and an Action are indistinguishable to the compiler. — Oliver Charlesworth
– Oliver Charlesworth, Commented Jul 1, 2012 at 15:55
Yes, but they help me know what is a function to be sent as parameters to other functions - I will call them Action functions - and what are just simple void functions to be called by other functions — ali
– ali, Commented Jul 1, 2012 at 15:56

AnT stands with Russia · Accepted Answer · 2012-07-02 02:26:28Z

2

Firstly, the proper syntax for function type typedef is

typedef void (*Action)();

Secondly, typedef-name for function type can be used to declare functions of that type, but they cannot be used to define such functions. I.e. if FunctionToExecute is the function you wat to execute through the pointer, then it has to be defined as

void FunctionToExecute() {
  /* whatever */
}

The calling function would be defined as

void RunFunction(ACTION p) {
  p();
}

And the call to it would look as follows

RunFunction(FunctionToExecute);

That's it. Unary * and & operators are optional with function pointers.

edited Jul 2, 2012 at 2:26

answered Jul 1, 2012 at 16:22

AnT stands with Russia

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3 Comments

ali Over a year ago

Thanks. That was my mistake, defining the function to be sent as the function typedef. Thanks to everybody!

MSalters Over a year ago

"typedef-name for function types can be used to declare functions, but they cannot be used in function definitions." - that's too strict, as your own void RunFunction(Action p) { } shows. That uses the typedef-name Action in the definition. Typedefs can be used in arguments or the return type, but a typedef can't be used as the complete type of a function in a definition. The root cause is that Action FunctionToExecute() defines a function that returns an Action, not a function that is an Action.

AnT stands with Russia Over a year ago

@MSalters: Yes, that's just bad wording on my part. Meanwhile, I disagree with your "root cause". I'd say that the potential (invented) syntax for function definition through a typedef-name would look as Action FunctioToExecute { /* body */ }, i.e. no () after the name (consistent with the currently legal declaration syntax Action FunctioToExecute;). The real root cause, in my opinion, is the fact that parameter names cannot be supplied.

Oliver Charlesworth · Accepted Answer · 2012-07-01 16:05:36Z

Ok, if I understand your question correctly, you're looking for a way to only allow certain functions to be passed as arguments to your DLL function.

Using a typedef won't work, as that just creates an alias; Action and void (*)() are indistinguishable to the compiler in your code. Instead, you'll have to create a brand new type. In C++, you should strongly consider using a functor:

// Abstract functor
struct Action {
    virtual void operator() () const = 0;
};

// The DLL function
void RunFunction(const Action &action) {
    action();
}

// Define your function as a derived functor
struct FunctionToExecute : Action {
    void operator() () const { std::cout << "Hello\n"; }
};

// Run a function
RunFunction(FunctionToExecute());

Demo: http://ideone.com/5pl23.

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