Thanks, Dan and Roman, for the explanation. So I have to delete the
explanation "non-recursive let = single-branch case" from my brain.
I thought the guards in a let are assertations, but in fact it is more
like an if. Ok.
But then I do not see why the pattern variables are in scope in the
guards in
let p | g = e
The variables in p are only bound to their values (given by e) if the
guard g evaluates to True. But how can g evaluate if it has yet unbound
variables? How can ever a pattern variable of p be *needed* to compute
the value of the guard? My conjecture is that it cannot, so it does not
make sense to consider variables of g bound by p. Maybe you can cook up
some counterexample.
I think the pattern variables of p should not be in scope in g, and
shadowing free variables of g by pattern variables of p should be forbidden.
Cheers,
Andreas
On 09.07.2013 17:05, Dan Doel wrote:> The definition
>
> Just x | x > 0 = Just 1
>
> is recursive. It conditionally defines Just x as Just 1 when x > 0 (and
> as bottom otherwise). So it must know the result before it can test the
> guard, but it cannot know the result until the guard is tested. Consider
> an augmented definition:
>
> Just x | x > 0 = Just 1
> | x <= 0 = Just 0
>
> What is x?
On 09.07.2013 17:49, Roman Cheplyaka wrote:
As Dan said, this behaviour is correct.
The confusing thing here is that in case expressions guards are attached
to the patterns (i.e. to the lhs), while in let expressions they are
attached to the rhs.
So, despite the common "Just x | x > 0" part, your examples mean rather
different things.
Here's the translation of 'loops' according to the Report:
loops =
let Just x =
case () of
() | x > 0 -> Just 1
in x
Here it's obvious that 'x' is used in the rhs of its own definition.
Roman
* Andreas Abel <andreas.a...@ifi.lmu.de> [2013-07-09 16:42:00+0200]
Hi, is this a known bug or feature of GHC (7.4.1, 7.6.3)?:
I got a looping behavior in one of my programs and could not explain
why. When I rewrote an irrefutable let with guards to use a case
instead, the loop disappeared. Cut-down:
works = case Just 1 of { Just x | x > 0 -> x }
loops = let Just x | x > 0 = Just 1 in x
works returns 1, loops loops. If x is unused on the rhs, the
non-termination disappears.
works' = let Just x | x > 0 = Just 1 in 42
Is this intended by the Haskell semantics or is this a bug? I would
have assumed that non-recursive let and single-branch case are
interchangeable, but apparently, not...
Cheers,
Andreas
--
Andreas Abel <>< Du bist der geliebte Mensch.
Theoretical Computer Science, University of Munich
Oettingenstr. 67, D-80538 Munich, GERMANY
andreas.a...@ifi.lmu.de
http://www2.tcs.ifi.lmu.de/~abel/
_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
--
Andreas Abel <>< Du bist der geliebte Mensch.
Theoretical Computer Science, University of Munich
Oettingenstr. 67, D-80538 Munich, GERMANY
andreas.a...@ifi.lmu.de
http://www2.tcs.ifi.lmu.de/~abel/
_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
