🚀 go-pugleaf

In article <4ja3qf$p9b@solutions.solon.com>
           hirth@wand.franken.de "Claus Hirth" writes:

 >Kurt Watzka (watzka@stat.uni-muenchen.de) wrote:
 >
 >: yang@math.umass.edu (Huayong Yang) writes:
 >
 >: >void Swap(void** p1, void** p2)
 >
 >: It is not allowed to dereference a void *, not is it necessary in
 >: this example.
 >
 >I agree. But you forgot to give an answer to Huayong.
 >Please enlighten us with *your* solution to this problem.

There isn't a simple generic solution to this problem.

 >The best I could directly think of is :
 >
 >void swap(void ** x, void ** y)
 >{
 >        char * tmp;
 >
 >        tmp = *( (char**)x );
 >        *( (char**)x ) = *( (char**)y );
 >        *( (char**)y ) = tmp;

You're treating the object being swapped as if it is a char *. If it
isn't a char * then this may not work - it results in undefined behaviour.

 >}
 >
 >You would call it for 
 >        SomeType * x, * y 
 >sth like
 >        swap((void**)&x,(void**)&y);
 >to avoid "incompatible pointer type"-warnings.

The warnings are there to indicate that you're doing something very wrong.
Simply shutting up the warnings is shooting yourself in the foot. Pointer
casting like this is rarely a sensible thing to do.

-- 
-----------------------------------------
Lawrence Kirby | fred@genesis.demon.co.uk
Wilts, England | 70734.126@compuserve.com
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