[Haskell-cafe] GHC bug? Let with guards loops
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 AbelDu 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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Re: [Haskell-cafe] GHC bug? Let with guards loops
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 Tue, Jul 9, 2013 at 10:42 AM, Andreas Abel andreas.a...@ifi.lmu.dewrote:
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 AbelDu 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/ http://www2.tcs.ifi.lmu.de/~abel/
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Re: [Haskell-cafe] GHC bug? Let with guards loops
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 AbelDu 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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Re: [Haskell-cafe] GHC bug? Let with guards loops
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 AbelDu 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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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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Re: [Haskell-cafe] GHC bug? Let with guards loops
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...
Cheers,
Andreas
--
Andreas AbelDu 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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Andreas AbelDu 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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Felipe.
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Re: [Haskell-cafe] GHC bug? Let with guards loops
With pattern guards, it's difficult to say whether it is never 'useful' to
have things like the following work:
C x | C' y z - f x = ...
But I'd also shy away from changing the behavior because it causes a lot of
consistency issues. In
let
f vs1 | gs1 = es1
h vs2 | gs2 = es2
...
we have that f and h are in scope in both gs1 and gs2. Does it make sense
to call f in gs1? It's easy to loop if you do. So should f not be in scope
in gs1, but h is, and vice versa for gs2? But they're both in scope for es1
and es2?
And if we leave the above alone, then what about the case where there are
no vs? Is that different? Or is it only left-hand patterns that get this
treatment?
Also, it might have some weird consequences for moving code around. Like:
let Just x | x 0 = Just 1
let Just x | y 0 = Just 1
y = x
let Just x | b = Just 1
where b = x 0
let Just x | b = Just 1
b = x 0
These all behave the same way now. Which ones should change?
If Haskell had a non-recursive let, that'd probably be a different story.
But it doesn't.
On Tue, Jul 9, 2013 at 1:12 PM, Andreas Abel andreas.a...@ifi.lmu.dewrote:
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 AbelDu 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/ http://www2.tcs.ifi.lmu.de/~abel/
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Theoretical Computer Science, University of Munich
Oettingenstr. 67, D-80538 Munich, GERMANY
andreas.a...@ifi.lmu.de
http://www2.tcs.ifi.lmu.de/~**abel/ http://www2.tcs.ifi.lmu.de/~abel/
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Re: [Haskell-cafe] GHC bug? Let with guards loops
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 AbelDu 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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Re: [Haskell-cafe] GHC bug? Let with guards loops
On 09.07.2013 19:56, Dan Doel wrote:
With pattern guards, it's difficult to say whether it is never 'useful'
to have things like the following work:
C x | C' y z - f x = ...
But I'd also shy away from changing the behavior because it causes a lot
of consistency issues. In
let
f vs1 | gs1 = es1
h vs2 | gs2 = es2
...
we have that f and h are in scope in both gs1 and gs2. Does it make
sense to call f in gs1? It's easy to loop if you do. So should f not be
in scope in gs1, but h is, and vice versa for gs2? But they're both in
scope for es1 and es2?
If f and h are really mutually recursive, then they should not be in
scope in gs1 and gs2.
If the first thing you do in the body of f is calling f (which happens
if f appears in gs1), then you are bound to loop. But of course, if vs
are not just variables but patterns, then the first thing you do is
matching, so using f in gs1 could be fine.
I am getting on muddy grounds here, better retreat. I was thinking only
of non-recursive let.
In the report
http://www.haskell.org/onlinereport/haskell2010/haskellch3.html#x8-440003.12
it says that
let p = e1 in e0= case e1 of ~p - e0
where no variable in p appears free in e1
but this applies only for patterns p without guards, and I would have
expected to be true also for patterns with guards.
And if we leave the above alone, then what about the case where there
are no vs? Is that different? Or is it only left-hand patterns that
get this treatment?
Yes, it is only about the things defined by the let binding (in this
case, f and g). The variables in vs1 are bound by calling f, but by f's
body.
Also, it might have some weird consequences for moving code around. Like:
let Just x | x 0 = Just 1
Non-recursive.
let Just x | y 0 = Just 1
y = x
Recursive.
let Just x | b = Just 1
where b = x 0
Recursive?
let Just x | b = Just 1
b = x 0
Recursive. (Like 2.)
These all behave the same way now. Which ones should change?
Only the first one? Hard to tell.
If Haskell had a non-recursive let, that'd probably be a different
story. But it doesn't.
Definitely agree.
On Tue, Jul 9, 2013 at 1:12 PM, Andreas Abel andreas.a...@ifi.lmu.de
mailto: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
mailto: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 =
