void variables in c functions

rubadub · 10-27-2007, 05:27 AM

Hi, i'm wanting to be able to pass any type of variable to a function (penultimately it's a wrapper for fwrite). I'm trying the void type, but am having trouble finding reference to this usage.

I keep changing and trying different ways, but here's an example:

Code:

void test(const void *i)
{
	int j = (int)i;
	printf("i contents: %d \n", j);
}

I'm about to try some variations with memcpy etc, but any help would be greatly appreciated!

tronayne · 10-27-2007, 07:45 AM

Maybe this will help -- a call to fwrite() would look like this:

Code:

(void) fwrite ((void *) buf, sizeof (char), n, stdout);

Similarly, a call to memcpy() would look something like this:

Code:

(void) memcpy ((void *) buf2, (void *) buf1, sizeof (buf1));

I think what you're trying to do is const (void *) i?

matthewg42 · 10-27-2007, 08:24 AM

In your test function, you take a pointer, but then you cast the value of the pointer (i.e. the memory address of the pointer) to an int and print that. You need to de-reference your pointer before you print it:

Code:

void test(const void *i)
{
        int *j = (int*)i;
        printf("i contents: %d \n", *j);
}

/* OR try it all in one go:
void test(const void *i)
{
        printf("i contents: %d \n", *((int*)i));
}
*/

rubadub · 10-27-2007, 08:57 AM

Cheers...
How are you calling it because I keep getting errors, e.g.

Quote:

error: void value not ignored as it ought to be

matthewg42 · 10-27-2007, 10:00 AM

Code:

int somevalue = 3142;
test(&somevalue);

rubadub · 10-27-2007, 02:07 PM

I'm still getting same error...
I'll now look into compiler because i've had some other weird results today! (gcc version 4.1.2 20070502 (Red Hat 4.1.2-12))

matthewg42 · 10-27-2007, 02:29 PM

Paste your code, preferably a minimal version which is just enough to duplicate your error. Also paste the command you use to compile/link the program.

rubadub · 10-27-2007, 02:43 PM

I'm only testing at moment, so it's brief...(and basically my code was same as your first func):

Code:

#include<stdio.h>

void test1(const void *i)
{
	int *j = (int*)i;
	printf("i contents: %d \n", *j);
}

void test2(const void *i)
{
	printf("i contents: %d \n", *((int*)i));
}

int main()
{
	int k = 6284;
	int *j = &k;
	int iret;
	
	iret = test1(&k);
	iret = test2(j);
	
	return 0;
}

I'm compiling with gcc:

Code:

gcc -ansi -c main.c -o main

...and here's the error from this setup:

Quote:

main.c: In function ‘main’:
main.c:46: error: void value not ignored as it ought to be
main.c:47: error: void value not ignored as it ought to be

rubadub · 10-27-2007, 02:47 PM

Sorry, revise the compile line:

Quote:

gcc -ansi main.c -o main

matthewg42 · 10-27-2007, 02:55 PM

Aha I see. The functions test1 and test2 are defined as returning nothing - void. In the context of a return value to a function this means "no value is returned". Yet you try to assign the return value to the variable iret, which is an int. This is the source of the error.

Either you need to change the return type of the functions to int (and return something), or get rid of these assignments.

rubadub · 10-27-2007, 03:03 PM

Yep, I likes that, originally and ultimately there is a return so... Thanks very much for that I felt like I was going mad!

matthewg42 · 10-27-2007, 03:09 PM

C does that to a man.