void* pointers

**Drogin** · 09-04-2009

I have this mysterious code...

Code:

void swap(void *v[], int i, int j) {
 void *temp;
 temp = v[i];
 v[i] = v[j];
 v[j] = temp;     
}

What the heck does this do?
I was suprised it even compiled.

The function suggests something is swapped.
If I use my logical sense, I would say this swaps what is in v[i] and v[j].
But what I don't get is:

how can something like this be legal:

Code:

 temp = v[i];

when v is a void* pointer.
I mean..isnt it impossible to find exactly where in memory to find element nr i, when we don't know the size of the thing we're storing in our array?

and secondly, exactly what does this do in practice?
The book im reading suggests it "exchanges two pointers"...but this just doesent make sense to me.
Would anyone be so kind as to try to explain it to me?

**Salem** · 09-04-2009

Well if you replaced 'void*' with 'int', would it make sense?

It's just swapping [i] and [j] elements of an array.
That it's an array of void* doesn't make any difference.

**bithub** · 09-04-2009

>> What the heck does this do?
You have an array of void pointers. That code just swaps the pointer at one index with a pointer at a different index.

>> when v is a void* pointer.
v is not a void pointer. v is an array of void pointers.

**Drogin** · 09-04-2009

Ah, so this code are only swapping elements of an array of *void pointers..?

If so, that sure helps.
The book had just written about how void* can be used to point to all kinds of types, so I assumed for some reason that the book were showing me a universial method of swapping 2 elements in an array of whatever type.
Which I couldnt get to make sense :P

**MK27** · 09-04-2009

Originally Posted by Drogin

I assumed for some reason that the book were showing me a universial method of swapping 2 elements in an array of whatever type.

That is exactly what it is.

**Drogin** · 09-04-2009

Code:

That is exactly what it is.

No way.

Code:

  char array[3]; // Array of 3 chars
  
  // Set all elements to 0
  int i;
  for(i=0; i > 3; i++) {
    array[i] = 0;        
  }

 // Fill in some values
  array[0] = 'a';
  array[1] = 'b';

  swap(array,0,1);

Are you saying this would swap array[0] and array[1], in the way I expect?
I think not, because the function expects an array of pointers...not normal types as I first thought

**bithub** · 09-04-2009

It only works with an array of pointers -- not arrays of any type.

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