Thread: Need help on pointer to VOID

>how to access and dereference a VOID pointer?
Just cast it into the correct data type and dereference it:

Code:

#include <stdio.h>

int main ( void )
{
  void *vptr;
  int a = 5;
  int b;

  vptr = &a;
  printf("vprt is %d\n", *(int *)vptr);

  b = *(int *)vptr;
  *(int *)vptr = 10;
  printf("b is %d\n", b);
  printf("vprt is %d\n", *(int *)vptr);

  return 0;
}

-Prelude

>Back to the question I posted, can you guys show
how to access and dereference a VOID pointer?
Have you read what's been written here? The second post starts with

The only difference when it comes to void* pointers is that you can't dereference a void pointer

Then quzah later said

You actually can dereference a void pointer. You just have to type cast it to whatever type you're dereferencing it as. If it points to an int, dereference it as if you were using a pointer to an integer by type casting.

>No one has illustrate this point clearly so far!
Are your trolling or what?!

In your code:
>printf("vprt is %d", *((int*)vptr));
and
>*(int *)vptr = 10;

[edit]Goddammit... beaten by miles![/edit]

Originally posted by quzah

My reply was not some personal attack on unknown. I was trying to clarify the issue.

Sorry for being so grumpy

Originally posted by quzah

What I was trying to clarify is that exact line of thought. And for the record, there is a type of 'void'. You just cannot create instances of it. If there weren't 'void', how could you use it as return values and parameters?

Quzah.

void is an incomplete type , like a structure which has been declared but whose members have not been still defined , but unlike other incomplete types a void cannot be completed ,so you can never have objects of type void . sizeof cannot be applied to any incomplete type .
I think (I might be wrong) a function taking in a void parameters nothing to do with it being a type , because a function cannot have more than one parameter if it has a void parameter and it does not have to declare a variable for the void type , which clearly marks out something like f(void) as a special syntax to indicate no parameters.

>sizeof cannot be applied to any incomplete type
I don't have my copy of the standard with me, is this a constraint violation? Because my compiler compiles and runs

printf ( "%u\n", (unsigned)sizeof ( void ) );

With nothing more than a warning that the result is 0. Lint returns no warnings.

-Prelude

I got a warning and a different size of void , I used gcc , but I cant complain , I was warned

$> cat test.c
#include <stdio.h>
int
main(void)
{
printf("sizeof(void)=%lu",(unsigned long) sizeof(void) ) ;
getchar();
return 0;
}

$> gcc -ansi -pedantic -Wall test.c
test.c: In function `main':
test.c:6: warning: sizeof applied to a void type
$>./a.out
sizeof(void)=1