At 9:22 AM -0700 7/14/06, Dave Whipp wrote:
Darren Duncan wrote:
Assuming that all elements of $a and $b are themselves immutable to all levels of recursion, === then does a full deep copy like eqv. If at any level we get a mutable object, then at that point it turns into =:= (a trivial case) and stops.


  ( 1, "2.0", 3 ) === ( 1,2,3 )

True or false?

That would be false, because a Str does not === an Int.

I should point out, though, that in at least some situations where you are ===, you would have more control of the creation of the structures in the first place and they will probably have been created in a more strict fashion that required the data to be of corresponding types in the first place. Eg, each Seq would have only been allowed to store Ints in the first place. The best place to normalize input is as early as possible, after all, like when the Ints were input and before they were put in the Seq to be compared.

More imprtantly, how do I tell perl what I mean? The best I can think of is:

  [&&] (@a »==« @b)
Vs
  [&&] (@a »eq« @b)

But this only works for nice flat structures. For arbitrary tree structures, we probably need adverbs on a comparison op (I think Larry mentioned this a few posts back) ... but if we're going with adverbs do we really need 5 different base operators? Are all of the 5 so common that it would be clumbersome to require adverbs for their behavior?

Also, when sorting things, maybe deep inequalities would be useful, too.

I will punt on that one.

-- Darren Duncan

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