[Haskell-cafe] Question on "case x of g" when g is a function

[Salvador Lucas]

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 at 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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