If he really wanted to use a case-expression, he could write it this way:
f x = case x of
False -> 0
True -> 1
--Ham
At 1:02 PM +0100 2005/1/27, Henning Thielemann wrote:
On Thu, 27 Jan 2005 [EMAIL PROTECTED] wrote:
Can a kind soul please enlighten me on why f bit0 and f bit1
bo
m: Salvador Lucas [mailto:[EMAIL PROTECTED]
> Sent: 27 January 2005 09:59
> To: [EMAIL PROTECTED]
> 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
>
On Thu, 27 Jan 2005 [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
If you compile with 'ghc -Wall' GHC should report that the ident
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
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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