Re: Printing the result of a function evaluation
Steve, I should have provided the following details ... The code I pasted was a timing function I've come across in various Haskell literature, and it "works" as is ... at least, the timing results it prints are plausible. My sole "contribution" to the code was the "print v" line, in an attempt to have it print BOTH the result of the function evaluation AND the time it took to evaluate it (and still only evaluating the function once). As noted, my "print v" prints only "()" regardless of the function -- not what I wanted. But the timing part still works. There has got to be an easy fix here, but I have yet to discover it ...
Travis, OK - I'm with ya. Well, 'speedTest (read $ head args) `seq` return ()' will evaluate your expression but it'll return (). The quick fix would be to use 'return $! speedTest (read $ head args)' - That will also evaluate the expression, but it'll return the output too. The $! is the bit that forces it the evaluation of the function (so the evaluation happens during the bit that is timed). Actions are evaluated just before they're sequenced in IO, and this is actually the source of all forcing of evaluation in Haskell (or at least, in GHC). Doing 'x `seq` return ()' is one way to tie the evaluation to x to IO's evaluation, and 'return $! x' is another way. I think the second one is equivalent to 'x `seq` return x`. Or, in your case it would be 'let x = speedTest (read $ head args) in x `seq` return x', so $! saves you a bit of typing. 1. IO says 'evaluate the IO a value!' which requires the argument to be passed to the 'return' function (but not evaluated). 2. $! says 'force evaluation of the argument, then pass it to the function. This causes the evaluation to be tied to IO, and therefore actually forces the evaluation. Clear as coffee? Steve Travis Erdman wrote:
Steve, I should have provided the following details ...
The code I pasted was a timing function I've come across in various Haskell literature, and it "works" as is ... at least, the timing results it prints are plausible.
My sole "contribution" to the code was the "print v" line, in an attempt to have it print BOTH the result of the function evaluation AND the time it took to evaluate it (and still only evaluating the function once).
As noted, my "print v" prints only "()" regardless of the function -- not what I wanted. But the timing part still works. There has got to be an easy fix here, but I have yet to discover it ...
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On Mar 8, 2010, at 19:55 , Travis Erdman wrote:
The code I pasted was a timing function I've come across in various Haskell literature, and it "works" as is ... at least, the timing results it prints are plausible.
My sole "contribution" to the code was the "print v" line, in an attempt to have it print BOTH the result of the function evaluation AND the time it took to evaluate it (and still only evaluating the function once).
The root of the problem is that seq forces its first argument but returns its second, so you're getting () back from it. I'm not sure what the approved way to fix it is. -- brandon s. allbery [solaris,freebsd,perl,pugs,haskell] allbery@kf8nh.com system administrator [openafs,heimdal,too many hats] allbery@ece.cmu.edu electrical and computer engineering, carnegie mellon university KF8NH
participants (3)
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Brandon S. Allbery KF8NH -
Stephen Blackheath [to Haskell-Beginners] -
Travis Erdman