Hi Felipe, thanks for the centavos.
So you mean that in
let p | g = e where bs
in ...
the bindings bs should be in scope in g (and of course the variables of
p are in scope in bs). Mmh, the bindings in bs that do not uses the
pattern variables could be useful in g, but the other bindings will
still lead to non-termination. Maybe such an analysis is too
sophisticated. My lesson is to use case instead of let. Only that the
let syntax is nicer and indentation-friendlier than case, so it would be
preferable.
The greater evil is that Haskell does not have a non-recursive let.
This is source of many non-termination bugs, including this one here.
let should be non-recursive by default, and for recursion we could have
the good old "let rec".
Cheers,
Andreas
On 09.07.2013 19:23, Felipe Almeida Lessa wrote:
Well, you could use p's type for something.
let x | foo (undefined `asTypeOf` x) = 3
foo _ = True
in x
Arguably not very useful. It seems to me that the most compelling
rationale is being consistent with the cases where, instead of being a
data type, p is a function. Even so most of the time you won't be
recursing on the guard. But, since you could use something from the
where clause on the guard, and we certainly won't be restricting
recursing on the where clause, it also seems compelling to allow
recursion on the guard.
My 2 centavos, =)
On Tue, Jul 9, 2013 at 2:12 PM, Andreas Abel <andreas.a...@ifi.lmu.de> wrote:
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...
--
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/
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