Seq question

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Lee Short

17 Apr 2012 17 Apr '12

4:56 p.m.

I've got the following code, and I'm not sure why the seq seems to be doing nothing. The code should never terminate if the seq were forcing the full evaluation of y. Instead, it runs just fine (though stack overflows if I ask for too many elements of primes). I didn't see anything in the tutorial section of haskell.org that seems to explain this -- pointers to a useful source would be welcome. sieve :: [Int] -> [Int] sieve xs = let ys = filter (\z -> 0 /= z `mod` head xs) xs in seq ys $ head xs:sieve ys primes = sieve [2..] thanks Lee

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Tim Perry

17 Apr 17 Apr

5:41 p.m.

seq evaluates to Weak Head Normal Form (WHNF). WHNF is the first contructor. So your use of seq only evaluates the first number and the cons. I.E., it evaluates to: s:(Thunk) You are expecting seq to work like deepseq and evaluate the entire list. Also, mod is slower than rem and the meaning doesn't matter in a sieve so change mod to rem for faster code. Tim On Tue, Apr 17, 2012 at 1:56 PM, Lee Short wrote:

...

I've got the following code, and I'm not sure why the seq seems to be doing nothing. The code should never terminate if the seq were forcing the full evaluation of y. Instead, it runs just fine (though stack overflows if I ask for too many elements of primes). I didn't see anything in the tutorial section of haskell.org that seems to explain this -- pointers to a useful source would be welcome.

sieve :: [Int] -> [Int] sieve xs = let ys = filter (\z -> 0 /= z `mod` head xs) xs in seq ys $ head xs:sieve ys

primes = sieve [2..]

thanks Lee

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Brent Yorgey

18 Apr 18 Apr

10:45 a.m.

On Tue, Apr 17, 2012 at 02:41:15PM -0700, Tim Perry wrote:

...

seq evaluates to Weak Head Normal Form (WHNF). WHNF is the first contructor. So your use of seq only evaluates the first number and the cons. I.E., it evaluates to: s:(Thunk)

Actually, it doesn't even force the first number. You just get Thunk : Thunk -Brent

Henk-Jan van Tuyl

19 Apr 19 Apr

5:31 a.m.

On Wed, 18 Apr 2012 16:45:05 +0200, Brent Yorgey wrote:

...

On Tue, Apr 17, 2012 at 02:41:15PM -0700, Tim Perry wrote:

...
seq evaluates to Weak Head Normal Form (WHNF). WHNF is the first contructor. So your use of seq only evaluates the first number and the cons. I.E., it evaluates to: s:(Thunk)

Actually, it doesn't even force the first number. You just get

Thunk : Thunk

A nice way to demonstrate this, is the following GHCi session: Prelude> undefined `seq` print "OK" *** Exception: Prelude.undefined Prelude> [undefined] `seq` print "OK" "OK" Regards, Henk-Jan van Tuyl -- http://Van.Tuyl.eu/ http://members.chello.nl/hjgtuyl/tourdemonad.html Haskell programming --

Tim Perry

5:13 p.m.

Henk-Jan van Tuly is correct. Oops! Sorry I was mis-informed. By the way, there is an excellent post about this topic on StackOverflow.com. It covers, seq, weak-head-normal-form, normal form, and thunks. See here: http://stackoverflow.com/questions/6872898/haskell-what-is-weak-head-normal-... Hope that makes up for my earlier comment. Tim On Thu, Apr 19, 2012 at 2:31 AM, Henk-Jan van Tuyl wrote:

...

On Wed, 18 Apr 2012 16:45:05 +0200, Brent Yorgey wrote:

On Tue, Apr 17, 2012 at 02:41:15PM -0700, Tim Perry wrote:

...
...
seq evaluates to Weak Head Normal Form (WHNF). WHNF is the first contructor. So your use of seq only evaluates the first number and the cons. I.E., it evaluates to: s:(Thunk)

Actually, it doesn't even force the first number. You just get

Thunk : Thunk

A nice way to demonstrate this, is the following GHCi session: Prelude> undefined `seq` print "OK" *** Exception: Prelude.undefined Prelude> [undefined] `seq` print "OK" "OK"

Regards, Henk-Jan van Tuyl

-- http://Van.Tuyl.eu/ http://members.chello.nl/**hjgtuyl/tourdemonad.html http://members.chello.nl/hjgtuyl/tourdemonad.html Haskell programming --

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Daniel Fischer

6:23 p.m.

On Thursday 19 April 2012, 23:13:48, Tim Perry wrote:

...

Henk-Jan van Tuly is correct. Oops! Sorry I was mis-informed.

Note, however, that in the case here, the data-dependencies force the evaluation of the first number in order to determine the outermost constructor. So actually, your

...

So your use of seq only evaluates the first number and the cons. I.E., it evaluates to: s:(Thunk)

was factually correct (in this special case).

...

By the way, there is an excellent post about this topic on StackOverflow.com. It covers, seq, weak-head-normal-form, normal form, and thunks. See here: http://stackoverflow.com/questions/6872898/haskell-what-is-weak-head-no rmal-form/6889335#6889335

Hope that makes up for my earlier comment.

Tim

On Thu, Apr 19, 2012 at 2:31 AM, Henk-Jan van Tuyl wrote:

...
On Wed, 18 Apr 2012 16:45:05 +0200, Brent Yorgey

wrote: On Tue, Apr 17, 2012 at 02:41:15PM -0700, Tim Perry wrote:

...
...
seq evaluates to Weak Head Normal Form (WHNF). WHNF is the first contructor. So your use of seq only evaluates the first number and the cons. I.E., it evaluates to: s:(Thunk)

Actually, it doesn't even force the first number. You just get

Thunk : Thunk

A nice way to demonstrate this, is the following GHCi session: Prelude> undefined `seq` print "OK" *** Exception: Prelude.undefined Prelude> [undefined] `seq` print "OK" "OK"

Regards, Henk-Jan van Tuyl

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Brent Yorgey
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Seq question

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Tim Perry

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