On Thu, Dec 29, 2011 at 11:14 AM, Gregg Reynolds <d...@mobileink.com> wrote: > > On Dec 29, 2011, at 11:01 AM, Iustin Pop wrote: > >> And to clarify better my original email: yes, (bar x) always gives you >> back the same IO action; > > More precisely: the same *type*. >
I'm confused - what do you mean by "type"? I don't think that Iustin's statement needs any sort of qualifier - (bar x) always returns the same IO action when called with the same value for x, no matter how many times you call it. Antoine _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
