On Tue, Apr 19, 2011 at 1:41 PM, Gregory Guthrie <span dir="ltr"><<a href="mailto:guthrie@mum.edu">guthrie@mum.edu</a>></span> wrote:<br><div class="gmail_quote"><blockquote class="gmail_quote" style="margin: 0pt 0pt 0pt 0.8ex; border-left: 1px solid rgb(204, 204, 204); padding-left: 1ex;">
Perhaps the description was unclear;<br>
<br>
F1;f1 gives result r1;r2 (not the same)<br>
F1;f2 gives r1;r2<br>
F2,f1 gives r1;r2<br></blockquote><div><br>No, you were clear, and Felipe's answer still makes sense. Since f1 and f2 have the same definition, substituting one for the other should not change anything.<br>
<br>Maybe do notation is what is confusing you. Try getting rid of the do notation and writing everything in terms of the more basic (>>) and (>>=) combinators. If (>>) could be any operator, should we expect that f1 = f1 >> f1?<br>
<br>Luke<br> <br></div><blockquote class="gmail_quote" style="margin: 0pt 0pt 0pt 0.8ex; border-left: 1px solid rgb(204, 204, 204); padding-left: 1ex;">
<br>
-------------------------------------------<br>
<div class="im">> -----Original Message-----<br>
> From: Felipe Almeida Lessa [mailto:<a href="mailto:felipe.lessa@gmail.com">felipe.lessa@gmail.com</a>]<br>
> Sent: Tuesday, April 19, 2011 2:26 PM<br>
> To: Gregory Guthrie<br>
> Cc: <a href="mailto:haskell-cafe@haskell.org">haskell-cafe@haskell.org</a><br>
> Subject: Re: [Haskell-cafe] Haskell from SML - referrential Transparency?!<br>
><br>
</div><div><div></div><div class="h5">> On Tue, Apr 19, 2011 at 4:10 PM, Gregory Guthrie <<a href="mailto:guthrie@mum.edu">guthrie@mum.edu</a>> wrote:<br>
> > and I get different results from the two executions (f1,f2), even<br>
> > though they have exactly the same definition. Reversing their order,<br>
> > gives the exact same results (i.e. the results are still different,<br>
> > and in the same original order as f2;f1). Even doing (f1;f1) gives two different results.<br>
><br>
> This shows that referential transparency is working nicely.<br>
><br>
> --<br>
> Felipe.<br>
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