Correct casting to function pointer that points to a function that returns a function

Question

I was reversing a source code and I've found a function it which looks like:

consider this:

int examplefn(int x) { return x * 4; }

int (*rtx())(int)
{
    return examplefn;
}

well, Then I needed make a pointer function to rtx() to do a hook, then I've done something like this:

int (*fncptr())(int) = (int(*())(int))0xC0FFEE; 
/* 0xC0FFEE it's a sample of the memory address of the function...*/

But my compiler did not compile it, then I've tried do:

typedef int(*fnc_t())(int);

// Clearer example pointing to rtx

fnc_t* TRY_2 = (fnc_t*)&rtx;

// then has successfully compiled, ex test...

 int main()
 {
    std::cout << TRY_2()(4) << std::endl; // output: 16 ok.
 }

well, I'm getting to the point, How can I do the correct casting without use a typedef?

I searched all over the internet and I have not found anything...

Answer 1

(int(*())(int)) is a function type (the same type as the function rtx has). Your code attempts to declare a function, and cast an integer to function. However you actually want to deal with a pointer to such a function.

After: typedef int(*fnc_t())(int);, the equivalent of fnc_t *x; can be found by replacing fnc_t with (*x) in the typedef: int (*(*x)())(int). So your code could be:

int (*(*fncptr)())(int) = (int(*(*)())(int))0xC0FFEE; 

Using a series of typedefs (or equivalent usings) is certainly preferable in real code.