[Haskell-beginners] Question re "Pattern match(es) are non-exhaustive"

[Kim-Ee Yeoh]

On Tue, Feb 10, 2015 at 11:14 PM, Frerich Raabe <raabe at froglogic.com> wrote:

> Interesting! This got me thinking - is this issue because the compiler
> doesn't (cannot?) see the implementation of e.g. (<=)?
>
>
It "sees" the implementation of (<=) for otherwise it won't be able to
generate code for it. But see below.


> I noticed that a similiar case exists with
>
>   f :: Bool -> Bool
>   f x | x = True
>       | not x = False
>
> ...which yields the same warning. I suppose this is because the compiler
> doesn't know the definition of 'not' so it doesn't understand that either
> of the two guards will always be True.


Again, it has to "know" the definition of the function "not" because it has
to generate code to call it.

But it doesn't "know" the function from any other function, say foo.
Semantically speaking, both are equally opaque.

Now you could point out that "Ah, but look at the definition of not. It
could inline, simplify, et voila, obtain

f True = True
f False = False

and hence pattern-matching is complete."

Therein lies the rub. All that inlining and simplification boils down to
evaluating the program _in_ the compiler, so if your program diverges so
would the compiler, which wouldn't be a happy thing.



-- Kim-Ee
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