Re: equal? and primitive arrays

[Jamison Hope <jrh at theptrgroup dot com>]

On Oct 8, 2011, at 2:19 AM, Per Bothner wrote:

> In arrayEquals we could perhaps simplify/optimize slightly:
>
>  String tname1 = arg1.getClass().getName();
>  String tname2 = arg2.getClass().getName();
>  // int[].class.getName() returns "[I" etc
>  if (tname1.length() == 2 && tname2.equals(tname1)) { // matching  
> primitive types
>     switch (tname1.charAt(1)) {
>          case 'Z':  return Arrays.equals((boolean[]) arg1,  
> (boolean[]) arg2);
>          ... etc ...
>  }

Attached.

> > I'm a little uncertain what the correct behavior should be for edge
> > cases like comparing a String[] to an FString[] in which each pair
> > of elements are equal?. For now, that will return #f because the
> > array component types are not equal (in the Java sense).
>
> I think the result should be #f because the types are different.

OK, that was my suspicion. It makes the implementation a little simpler,
too, because I can shortcut the test with

if (!tname1.equals(tname2)) return false;

and avoid having to further examine the two array element types to see  
if
they are "compatible" in some complicated way. (And if I deferred to the
element-wise equality checks, then any two zero-length arrays might  
end up
being equal regardless of their respective types.)

--
Jamison Hope
The PTR Group
www.theptrgroup.com

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