[Haskell-cafe] Maybe a compiler bug?

[Murray Gross]

The issue here is not whether or not the code is pretty or elegant, but 
whether or not I get correct execution of what I have, which is a correct 
statement of what I want (even if not the prettiest or most lint free), 
and I don't. There are lots of ways to work around the problem, but that 
doesn't, unfortunately, make the problem go away, and it is sure to appear 
elsewhere as the program is extended, which it will be.

It would appear that the real issue here is that someone with resources I 
don't have needs to dig into the compilers--it should not be necessary to 
use trial and error to find an alternate writing of code that is legal and 
correct (regardless of the aesthetics) but is incorrectly compiled. For 
the time being, I will use native compilation and hope that someone can 
find and fix the error so that I can use the speed advantage of 
optimization.

Best,

Murray Gross



On Tue, 6 Jan 2009, Neil Mitchell wrote:

> Hi
>
>> gTst3 right left = if (lr > ll)  then  False else True
>> 			where lr = length (right ! 2)
>> 			      ll = length (left ! 2)
>
> Running this code over HLint (http://www.cs.york.ac.uk/~ndm/hlint) says:
>
> Example.hs:8:1: Error: Redundant if
> Found:
>  if (lr > ll) then False else True
> Why not:
>  not (lr > ll)
>
> Making that change and running it again gives:
>
> Example.hs:8:1: Error: Use <=
> Found:
>  not (lr > ll)
> Why not:
>  lr <= ll
>
> Which ends up with something similar to what you came up with.
> However, if we take your final answer:
>
>> gTst3 right left = (lr <= ll)
>> 			where lr = length (right ! 2)
>> 			      ll = length (left ! 2)
>
> We get:
>
> Example.hs:8:1: Warning: Redundant brackets
> Found:
>  (lr <= ll)
> Why not:
>  lr <= ll
>
> Leaving us with the HLint 1.0 compliant (TM) :
>
> gTst3 right left = lr <= ll
> 			where lr = length (right ! 2)
> 			      ll = length (left ! 2)
>
> Thanks
>
> Neil
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