Calling function via pointer

Question

Here is a simple code

#include<iostream.h>
#include<stdlib.h>
#include<conio.h>
void (*func[2])(int);
void main(int n=1)
{
    int i;
    cout<<endl<<n;
    func[0]=&exit;
    func[1]=&main;
    i=++n<=10;
    (func[i])(n);
}

Here I am satisfied with the output (i.e. 1 to 10 in different lines). The only thing which confused me was that why the global pointer is of the type void (*ptr[2])(int). If possible, please explain in simple words that why this pointer was taken so specifically

Reading about the clockwise/spiral rule might help you decode it. — Some programmer dude
– Some programmer dude, Commented Jan 31, 2014 at 15:33
Also, you should know that technically your program have undefined behavior. The C++ specification says that main must be defined to return an int, and take either no arguments or two arguments (an int and an array of char pointers). — Some programmer dude
– Some programmer dude, Commented Jan 31, 2014 at 15:36
@JoachimPileborg : this is a simple code for some riddle so it does not conform to standard c++ programming rules. I got my answer in first comment. Thanks alot eveyone for your time and knowledge — Gagan93
– Gagan93, Commented Jan 31, 2014 at 15:47

Mike Seymour · Accepted Answer · 2014-01-31 15:48:21Z

3

It's not a pointer, it's an array of two pointers.

This is a function:

void func(int);

This is a pointer to a function:

void (*func)(int);

and this is an array of two pointers to functions:

void (*func[2])(int);

So func[i] points to exit if i is zero (i.e. if n is greater than 10), and points to main otherwise, where i is 1.

Note that you're not allowed to call main recursively like this, nor to give main any signature other than int main() or int main(int, char**). (At least, that's the case in modern C++; these rules presumably don't apply to the prehistoric dialect your compiler accepts).

edited Jan 31, 2014 at 15:48

answered Jan 31, 2014 at 15:32

Mike Seymour

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