How to use void* as an arguement?

**Kleid-0** · 05-11-2005

I was wondering how qsort works with the combination of typeSize and void* input. I want to do arithmetic, but I'm not totally sure how to work this out...I'm thinking I could do a switch( typeSize ) and then convert to the appropriate type based on that, but I wouldn't be able to distinguish the difference between an int and an unsigned int.

Code:

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>

/*
	I want the input to be void*..not long.
*/
void hmmm( void *v, size_t typeSize )
{
	/*
		void*.c: In function `hmmm':
		void*.c:7: warning: dereferencing `void *' pointer
		void*.c:7: error: invalid use of void expression
	*/
	*v = 5;
}

int main( void )
{
	int n = 4;
	char c = 4;
	
	hmmm( &n, sizeof( n ) );
	hmmm( &c, sizeof( c ) );
	
	assert( n == 5 );
	assert( c == 5 );
	
	return 0;
}

**Thantos** · 05-11-2005

You'll have to cast the void* to another pointer type prior to dereferencing it.

As a hint:

sizeof(char) == 1

**Kleid-0** · 05-11-2005

I was so sure unsigned and signed would cause problems, but it doesn't! Thanks Thantos!

Code:

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>

void hmmm( void *v, size_t typeSize )
{	
	char *cv;
	int *nv;

	switch( typeSize )
	{
		case 1:
			cv = ( char* )v;
			*cv = 5;
			break;
		case 4:
			nv = ( int* )v;
			*nv = 5;
			break;
	}
}

int main( void )
{
	unsigned int un = 2;
	int n = 3;
	char c = 4;
	
	hmmm( &un, sizeof( un ) );
	hmmm( &n, sizeof( n ) );
	hmmm( &c, sizeof( c ) );
	
	assert( un == 5 );
	assert( n == 5 );
	assert( c == 5 );
	
	return 0;
}

**^xor** · 06-13-2005

Is it really necessary to cast the void pointer when you're not dereferencing it? GCC does not complain if you remove the cast in the code marked red. An address is an address after all, no matter what it points to.

Code:

void hmmm( void *v, size_t typeSize )
{	
	char *cv;
	int *nv;

	switch( typeSize )
	{
		case 1:
			cv = ( char* )v;
			*cv = 5;
			break;
		case 4:
			nv = ( int* )v;
			*nv = 5;
			break;
	}
}

**quzah** · 06-13-2005

You're not dereferencing it there. You're making a pointer assignment.

Quzah.

**^xor** · 06-13-2005

> You're not dereferencing it there. You're making a pointer assignment.
I know, that's why I wrote "when you're not dereferencing it".

**Kleid-0** · 06-13-2005

Hey ^xor, I think quzah was just clarifying it for me. Thank you for your comments though :], the more comments the better :].

**quzah** · 06-13-2005

Originally Posted by ^xor

> You're not dereferencing it there. You're making a pointer assignment.
I know, that's why I wrote "when you're not dereferencing it".

As I have said many many many times: You never need to typecast a void * when making a pointer assignment. You only ever need to typecast it if you're trying to dereference the void pointer itself.

Quzah.

**miroslavgojic** · 11-25-2010

Code:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>


main(void)
    {

int a;
a=5;

void* pa;
pa = &a;

void* ppa;
ppa = &pa;

void* pppa;
pppa = &ppa;

void* ppppa;
ppppa = &pppa;

void* pppppa;
pppppa = &ppppa;

system("clear");

printf("\n%d \n", a);

printf("\n%p ", pa);

printf("\n%d \n", *(int*) pa);

printf("\n%p ", ppa);

printf("\n%p ", *(void**) ppa);

printf("\n%d \n", **(int**) ppa);

printf("\n%p ", pppa);

printf("\n%p ", *(void***) pppa);

printf("\n%p ", **(void***) pppa);

printf("\n%d \n", ***(int***) pppa);

printf("\n%p ", ppppa);

printf("\n%p ", *(void****) ppppa);

printf("\n%p ", **(void****) ppppa);

printf("\n%p ", ***(void****) ppppa);

printf("\n%d \n", ****(int****) ppppa);

printf("\n%p ", pppppa);

printf("\n%p ", *(void*****) pppppa);

printf("\n%p ", **(void*****) pppppa);

printf("\n%p ", ***(void*****) pppppa);

printf("\n%p ", ****(void*****) pppppa);

printf("\n%d \n", *****(int*****) pppppa);

}

this is my solution for this matter.

**Salem** · 11-25-2010

miroslavgojic, the thread is over 5 YEARS old - please don't dig up the graveyard of old posts.

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