AST equality test that ignores NodeInfo

Gregory Crosswhite gcross at
Thu Feb 26 12:59:31 EST 2009

In my own code when I want to test that two ASTs are equivalent  
ignoring nodeInfos, I first run them through a function which converts  
all nodeInfos to internalNode and all idents to internalIdent.  It  
makes use of Data.Generics to accomplish this:

convertToInternal :: GenericT
convertToInternal = everywhere ((mkT convertToInternalNode) `extT`  
        convertToInternalNode :: NodeInfo -> NodeInfo
        convertToInternalNode _ = internalNode

        convertToInternalIdent :: Ident -> Ident
        convertToInternalIdent ident = internalIdent (identToString  

If you're not familiar with the "Scrap-Your-Boilerplate" style, what  
this code basically does is create a transformer which has specific  
behaviours for the two explicitly given types and acts as the identity  
on all other types, and the "everywhere" function goes through the  
data structure and applies the transformer to all of the terms/ 
subterms in the AST.


On Feb 26, 2009, at 8:40 AM, Bueno, Denis wrote:

> Hi all,
> Is there a better way than implementing my own equality operator that
> explicitly ignores NodeInfo, which is almost all boilerplate and  
> will take a
> lot of time?  Is there a way to (ab)use Data.Typeable for this?
> If I have an AST for a C file, I would like to be able to read  
> expressions
> from another file, to see if any of those expressions match any part  
> of the
> AST while I'm traversing the AST.
>                              Denis
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> Language-c mailing list
> Language-c at

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