[Haskell-cafe] a really juvenile question .. hehehehe ;^)
ryani.spam at gmail.com
Mon Oct 6 12:48:20 EDT 2008
On Mon, Oct 6, 2008 at 3:30 PM, Arnar Birgisson <arnarbi at gmail.com> wrote:
>> "undefined" is evaluated to see if it matches the
>> constructor "A". But we don't even get to check, because undefined
>> throws an exception during its evaluation.
>> In the "tb" case, (B x) always matches because B is a newtype. x gets
>> bound to undefined, but never evaluated.
> And this happens because data values are basically pattern matched at
> run-time but newtype values are matched at compile-time, effectively
> turning tb into an Int -> Bool function?
Yep, that's exactly it.
> That explains pretty well why newtype can have only one constructor.
I never thought of it that way, but yes, it really does!
Also, you can get the same behavior out of "ta" if you write it like this:
ta ~(A x) = True
The translation looks something like this:
f ~(A x) = e
f a = e
where x = case a of (A v) -> v -- a,v fresh variables not mentioned in e
f a = let (A x) = a in e -- "a" some fresh variable not mentioned in e
This delays the pattern-matching (and thus, the evaluation of "a")
lazily until "x" is demanded, at which point "a" might throw an
exception or infinite loop, or, if the type has more than one
constructor, fail to pattern match (which also throws an exception).
If "x" is never demanded then neither is "a".
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