Shifting pointers in a function

[herakles]

Hello!

I want to shift elements in a linked list by one element. If I do it directly in the function, where I need it, it is of no problem to me. But I come up with problems, as soon as I want to do the shifting of pointers in an own function to keep this more generic.

See this code:

// linked list of n elements present, allocated somewhere else

void shift_to_next( struct test *file, struct test *start_of_list_file ) {
	if( file->next == NULL ) file = start_of_list_file;
	else file = file->next;
}

int main(void) {
//	*start and *tmp are allocated somewhere else, the linked list can be seen as established, this would just blow up this question too much. this topic is not about crating a linked list... ;)

	tmp = start;
	shift_to_next( tmp, start );
	// at this point, tmp should point to the second element in the list
}

The pointers in the function are shifted correctly, but the pointer out of the function - in the main() program - are not shifted, they stay the same.

What would I have to do in order to get the pointers in the main-program affected?

Kind regards,
Herakles

[itsme86]

You just need to pass in a pointer to the pointer:

void shift_to_next( struct test **file, struct test *start_of_list_file )
{
  if( (*file)->next == NULL ) *file = start_of_list_file;
  else *file = (*file)->next;
}

And then call the function like this:

shift_to_next( &tmp, start );

Of course, another way would be to just make shift_to_next() return the new pointer. Then you could just have this:

struct test *shift_to_next( struct test **file, struct test *start_of_list_file )
{
  return file->next == NULL ? start_of_list_file : file->next;
}

tmp = shift_to_next( tmp, start );

[herakles]

Thank you!