🚀 go-pugleaf

yang@math.umass.edu (Huayong Yang) writes:

>I know void* means an arbitrary pointer. In an interview, I was asked to
>write a routine that swaps two pointers.

>I tried with this

>void Swap(void** p1, void** p2)

This is a routine to swap two "void *". It is _not_ a general pointer-
swapping routine.

>{
>  void* p = *p1;
>  *p1 = *p2;
>  *p2 = *p;

If you swap two integers, you don't dereference your "swap" variable.
You should _not_ do so when dealing with pointers.

   *p2 = p;

would be good enough in a routine that swaps two void pointers.

>}

>but it is not correct, because void* is not allowed as operand.

It is not allowed to dereference a void *, not is it necessary in
this example.

Kurt
--
| Kurt Watzka                             Phone : +49-89-2180-6254
| watzka@stat.uni-muenchen.de
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