[Haskell-cafe] Haskell symbol ~

[Ryan Ingram]

Here is another example:

> f1 n ~(x:xs) = (n, x)
> f2 n (x:xs) = (n,x)

f1 5 [] = (5, error "irrefutable pattern match failure")
f2 5 [] = error "pattern match failure"

In particular:

fst (f1 5 []) = 5
fst (f2 5 []) = error "pattern match failure"

The "~" delays the pattern match until evaluation of the variables
inside the pattern is demanded.  If the variable is never demanded,
the pattern match doesn't happen.

It's especially useful for single-constructor datatypes (like pairs)
if you want the code to be more lazy.

   -- ryan

On Wed, Aug 27, 2008 at 11:56 AM, Neil Mitchell <ndmitchell at gmail.com> wrote:
> Hi
>
>> At the same place, I found that example,
>> but wasn't wise enough to figure out
>> what it does:
>>
>> (f *** g) ~(x,y) = (f x, g y)
>>
>> Can you help me understand it?
>
> It means exactly the same as:
>
> (f *** g) xy = (f (fst xy), g (snd xy))
>
> i.e. if you call (f *** g) undefined, you will get (f undefined, g
> undefined). If the pattern was strict (i.e. no ~) you would get
> undefined.
>
> Please update the keyword wiki so it makes sense to you, after you
> have got your head round it.
>
> Thanks
>
> Neil
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