Using typedef function pointer in C

Question

I am trying to use function pointer, which is somewhat like this:

#include "stdio.h"

typedef void (*func)(int*, int*);


void func1(int *a, int*b)
{
    printf("Func1\n");
}

void func2(int *a, int*b)
{
    printf("Func2\n");
}

void func3(int *a, int*b)
{
    printf("Func3\n");
}

int main()
{
    int i;
    func f[] = {
      func1, func2, func3
    };
    printf("Hello\n");

    for(i=0; i< 3; i++)
    {
        func fn = f[i];
        *(fn)(&i, &i);
    }
    return 0;
}

I am always getting error: "void value not ignored as it ought to be"

Do not know, how to overcome this. Can anybody please help?

P0W · Accepted Answer · 2013-09-26 16:41:01Z

1

You need to use like following :

(*fn)(&i, &i);

answered Sep 26, 2013 at 16:41

P0W

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Comments

Kerrek SB · Accepted Answer · 2013-09-26 16:41:05Z

1

It's either (*fn)(&i, &i), or just fn(&i,&i). Otherwise you're trying to dereference the result of the function call!

(Alternatively, five-star-programmers often say (*****fn)(&i, &i).

answered Sep 26, 2013 at 16:41

Kerrek SB

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Comments

Rahul Tripathi · Accepted Answer · 2013-09-26 16:41:24Z

0

This error means you assign the return of a void function to something, which is an error.

Try this:-

(*fn)(&i, &i);

answered Sep 26, 2013 at 16:41

Rahul Tripathi

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Comments

Laszlo Valko · Accepted Answer · 2013-09-26 16:41:30Z

0

Try this:

(*fn)(&i, &i);

answered Sep 26, 2013 at 16:41

Laszlo Valko

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Comments

volferine · Accepted Answer · 2013-09-26 16:44:28Z

0

Don't write *(fn)(...) - simple fn(...) is enough, no need to dereference anything. What you have means something like "Dereference the return value of fn(...)", but you obviously can't do that.

answered Sep 26, 2013 at 16:44

volferine

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