[Haskell-cafe] Why isn't pattern matching lazy by default?

[Peter Verswyvelen]

Mmm, yes of course... blush...

But shouldn't

f ~(x:xs) = rhs

give a compile-time error since neither x nor xs is used in the right 
hand side, and hence nothing will ever get pattern matched when this 
function is called, which clearly indicates a mistake? That is, if I 
understand lazy pattern matching correctly... And then in these cases 
the user would have to annotate the pattern match as being strict, so he 
is aware of the eager evaluation taking place

Oh well, the way it is now is also easy to get used to, one just has to 
know how it works (just like M-theory ;-) )

Cheers,
Peter


Neil Mitchell wrote:
> Hi
>
>   
>>>  Now why isn't pattern matching lazy by default?  This seems odd for a
>>> newbie since everything else is lazy by default.
>>>       
>
> f ~(x:xs) = rhs
> f ~[]     = rhs'
>
> Now guess what f [] does...
>
> If you use a where binding then pattern matching is lazy.
>
> Thanks
>
> Neil
>
>
>   

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