Re: blackholes and exception handling

2010-05-03 Thread Simon Marlow
On 02/05/10 12:10, Sebastian Fischer wrote: Is the above output intended? Yes. Interesting. Note that catching all exceptions is rarely the right thing to do. See http://www.haskell.org/ghc/docs/6.12.2/html/libraries/base-4.2.0.1/Control-Exception.html#4 for more details. I should have

RE: Parallel Haskell: 2-year project to push real world use

2010-05-03 Thread Simon Peyton-Jones
| Does this mean DPH is ready for abuse? | | The wiki page sounds pretty tentative, but it looks like it's been awhile | since it's been updated. | | http://www.haskell.org/haskellwiki/GHC/Data_Parallel_Haskell In truth, nested data parallelism has taken longer than we'd hoped to be ready for

Re: Parallel Haskell: 2-year project to push real world use

2010-05-03 Thread Johan Tibell
On Mon, May 3, 2010 at 11:12 AM, Simon Peyton-Jones simo...@microsoft.comwrote: | Does this mean DPH is ready for abuse? | | The wiki page sounds pretty tentative, but it looks like it's been awhile | since it's been updated. | | http://www.haskell.org/haskellwiki/GHC/Data_Parallel_Haskell

Re: Parallel Haskell: 2-year project to push real world use

2010-05-03 Thread Roman Leshchinskiy
On 03/05/2010, at 22:04, Johan Tibell wrote: On Mon, May 3, 2010 at 11:12 AM, Simon Peyton-Jones simo...@microsoft.com wrote: | Does this mean DPH is ready for abuse? | | The wiki page sounds pretty tentative, but it looks like it's been awhile | since it's been updated. | |

Re: Parallel Haskell: 2-year project to push real world use

2010-05-03 Thread Christian Höner zu Siederdissen
Hi, on that topic, consider this (rather trivial) array: a = array (1,10) [ (i,f i) | i -[1..10]] where f 1 = 1 f 2 = 1 f i = a!(i-1) + a!(i-2) (aah, school ;) Right now, I am abusing vector in ST by doing this: a - new a' - freeze a forM_ [3..10] $ \i - do write a (a'!(i-1) +

Re: Parallel Haskell: 2-year project to push real world use

2010-05-03 Thread Don Stewart
choener: To summarise: I need arrays that allow in-place updates. Many of the array libraries provide both mutable and immutable interfaces, typically in ST or IO, including vector. ___ Glasgow-haskell-users mailing list

Re: Parallel Haskell: 2-year project to push real world use

2010-05-03 Thread Ben Lippmeier
You can certainly create an array with these values, but in the provided code it looks like each successive array element has a serial dependency on the previous two elements. How were you expecting it to parallelise? Repa arrays don't support visible destructive update. For many algorithms

Re: Parallel Haskell: 2-year project to push real world use

2010-05-03 Thread Ben Lippmeier
On 03/05/2010, at 10:04 PM, Johan Tibell wrote: On Mon, May 3, 2010 at 11:12 AM, Simon Peyton-Jones simo...@microsoft.com wrote: | Does this mean DPH is ready for abuse? | | The wiki page sounds pretty tentative, but it looks like it's been awhile | since it's been updated. | |

Re: Parallel Haskell: 2-year project to push real world use

2010-05-03 Thread Roman Leshchinskiy
On 04/05/2010, at 09:21, Christian Höner zu Siederdissen wrote: Hi, on that topic, consider this (rather trivial) array: a = array (1,10) [ (i,f i) | i -[1..10]] where f 1 = 1 f 2 = 1 f i = a!(i-1) + a!(i-2) (aah, school ;) Right now, I am abusing vector in ST by doing this:

Re: Parallel Haskell: 2-year project to push real world use

2010-05-03 Thread Christian Höner zu Siederdissen
* Ben Lippmeier b...@ouroborus.net [04.05.2010 02:21]: You can certainly create an array with these values, but in the provided code it looks like each successive array element has a serial dependency on the previous two elements. How were you expecting it to parallelise? actually, in

Re: Parallel Haskell: 2-year project to push real world use

2010-05-03 Thread Christian Höner zu Siederdissen
* Roman Leshchinskiy r...@cse.unsw.edu.au [04.05.2010 02:32]: On 04/05/2010, at 09:21, Christian Höner zu Siederdissen wrote: Hi, on that topic, consider this (rather trivial) array: a = array (1,10) [ (i,f i) | i -[1..10]] where f 1 = 1 f 2 = 1 f i = a!(i-1) + a!(i-2)