Should this be correct behaviour?

[Walter Bright]

Janice Caron wrote:
> Should this be correct behaviour?

Yes.

>     float[] f = new float[1];
>     float[] g = f.dup;
>     assert(f == g); /* Passes */
>     assert(f[0] == g[0]); /* Fails */
> 
> Certainly it is correct for the second assert to fail, because f[0]
> and g[0] both contain nan, and as we all know, (nan != nan).
> Essentially, the second assert (correctly) fails because the elements
> have not been initialised, and so we can't do the compare.
> 
> My question is, shouldn't the first assert also fail?
> 
> Put another way, how can two arrays be considered equal, if their
> elements are not considered equal?

You are comparing two array *references* which point to the same array. 
The two references clearly are the same. (f == g) does not compare the 
contents of the arrays.

> I realise that everything is behaving according to spec. But is it sensible?

Yes:

float a;
float b = a;
assert(b == a); // fails

And this is how floating point works.