Function's pointer cast

Question

Can I cast a function's pointer? This code report me an error.. Can you help me?

#include <iostream>
using namespace std;

typedef float (* MyFuncPtrType) (int, char*);
typedef void* (*p) ();

void* some_func ();

int main(int argc, char** argv)
{
    MyFuncPtrType func;

    some_func = reinterpret_cast<p>(func);

    return 0;
}

Why do you want to do that? You want some kind of hash table with functions of different signatures? — Heeryu
– Heeryu, Commented Jan 30, 2014 at 18:59
I just removed the C tag, as it doesn't make sense here. You are clearly using C++. — glglgl
– glglgl, Commented Jan 31, 2014 at 9:06

glglgl · Accepted Answer · 2014-01-31 09:07:57Z

2

First, void* some_func ();, some_func is not a function pointer. You can't assign it a new value. It should be declared as function pointer as follows:

void* (*some_func) ();

or

p   some_func;

Now just write:

some_func = (p)func;

Check working code @codepad

In C++ you do some_func = reinterpret_cast<p>(func); for c++ check this working code example @codepad

edited Jan 31, 2014 at 9:07

glglgl

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answered Jan 30, 2014 at 18:40

Grijesh Chauhan

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4 Comments

SwiftMango Over a year ago

I believe OP wanna return void * for some_func, and also C style casting is a bad choice after all.

Grijesh Chauhan Over a year ago

@texasbruce I don't know C++ So i added C style only. in C++ I think it should be: ome_func = reinterpret_cast<p>(func);

molbdnilo Over a year ago

AFAIK, calling a function through a pointer cast to a different type is undefined.

Grijesh Chauhan Over a year ago

@molbdnilo yes (not sure for c++) but in C in this answer this is calling some_func() causes undefined behaviour. I just provided casting...

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