Re: Case expressions in STG

Hi Tom, SPJ is surely more qualified to answer than I, but I'll take a stab. In general, it is computationally infeasible to compare functions. Granted, in your example, the function isn't being compared-- and therefore the case expr is extraneous. I don't think there exists a feasible, uncontrived example, so I imagine that's why STG forbids it (though I don't know where it's specified). Cheers, Will

On Jun 18, 2014, at 9:43 AM, Tom Ellis wrote:

I am reading SPJ's seminal work "Implementing lazy functional languages on stock hardware: the Spinless Tagless G-machine" (1992). The paper is available here

http://citeseer.ist.psu.edu/viewdoc/summary?doi=10.1.1.53.3729

On p62 we see "The expression scrutinised by a case expression must eventually evaluate either to a primitive value or a constructor application.".

This doesn't make sense to me. Isn't the following a valid STG program? What am I missing? Where is it specified that the scrutinee of a case expression cannot be of a function type?

{x} \n {z} -> let f = \{x} \n {y} -> g x y in case f of f' -> f' z

Thanks,

Tom _______________________________________________ ghc-devs mailing list ghc-devs@haskell.org http://www.haskell.org/mailman/listinfo/ghc-devs