Another way of looking at it: case peforms pattern matching; it is _not_ an
equality test.
If you want equality tests, use if-then-else, or something like this (using
guards):
f x = case () of
_ | x == bit0 -> 0
| x == bit1 -> 1
The behaviour that initially threw me was that case works for various
literals (numbers and strings), but that's just pattern matching (see
3.17.2, item 7):
http://www.haskell.org/onlinereport/exps.html#pattern-matching
Another shorthand for f, using guards:
f x
| x == bit0 = 0
| x == bit1 = 1
> -----Original Message-----
> From: Salvador Lucas [mailto:[EMAIL PROTECTED]
> Sent: 27 January 2005 09:59
> To: [EMAIL PROTECTED]
> Cc: [email protected]
> Subject: Re: [Haskell-cafe] Question on "case x of g" when g
> is a function
>
> Because both bit0 and bit1 are free *local* variables
> within the case expression. So, they have nothing
> to do with your defined functions bit0 and bit1.
>
> Best regards,
>
> Salvador.
>
> [EMAIL PROTECTED] wrote:
>
> >Can a kind soul please enlighten me on why f bit0 and f bit1
> >both return 0?
> >
> >
> >
> >>bit0 = False
> >>bit1 = True
> >>f x = case x of
> >> bit0 -> 0
> >> bit1 -> 1
> >>
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