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:slucas@dsic.upv.es] Sent: 27 January 2005 09:59 To: yeoh@cs.wisc.edu Cc: haskell-cafe@haskell.org 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.

yeoh@cs.wisc.edu 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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