Simon Peyton-Jones wrote:
> You might think that unnecessary bangs shouldn't lead to unnecessary work --
> if GHC knows it's strict *and* you bang the argument, it should still only be
> evaluated once. But it can happen. Consider
>
> f !xs = length xs
>
> Even though 'length' will evaluate its argument, f nevertheless evaluates it
> too.
I'm replying to a guru here, so I should keep my voice low; but I'd like to
point out what might result in a misunderstanding for other readers of
haskell-cafe. Contrasting both the bang pattern and the usage of length causing
f to be strict, one might suppose that a bang pattern, when used on a list, will
cause it to be evaluated in the same way as length does. However,
> the *first* thing length does is evaluate its argument,
and it will furthermore evaluate the argument list recursively, as much as is
necessary to determine its length. On the other hand, given
g !xs = ()
evaluating g [0..] will terminate, because g is only strict in the constructor
of its argument, which is (:). The list data type itself, however, is
non-strict.
Kalman
----------------------------------------------------------------------
Free pop3 email with a spam filter.
http://www.bluebottle.com/tag/5
_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
