[Haskell-cafe] How efficient is read?

[Chris Eidhof]

There is the ChristmasTree package (http://hackage.haskell.org/package/ChristmasTree) which provides a very fast read alternative by deriving grammars for each datatype. If you want to know the speed differences, see http://www.cs.uu.nl/wiki/bin/view/Center/TTTAS for more information (it's in the Haskell Do You Read Me paper, see section 5 for a comparison of efficiency).

-chris

On 9 mei 2010, at 05:32, Tom Hawkins wrote:

> I have a lot of structured data in a program written in a different
> language, which I would like to read in and analyze with Haskell.  And
> I'm free to format this data in any shape or form from the other
> language.
> 
> Could I define a Haskell type for this data that derives the default
> Read, then simply print out Haskell code from the program and 'read'
> it in?  Would this be horribly inefficient?  It would save me some
> time of writing a parser.
> 
> -Tom
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