[Haskell-cafe] GHC bug? Let with guards loops

[Andreas 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 <andreas.abel at 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.abel at 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
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