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 [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.abel@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.abel@ifi.lmu.de http://www2.tcs.ifi.lmu.de/~abel/

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 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 [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.abel@ifi.lmu.de http://www2.tcs.ifi.lmu.de/~abel/

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 mailto:andreas.abel@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 mailto:andreas.abel@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.abel@ifi.lmu.de http://www2.tcs.ifi.lmu.de/~abel/