Well suppose data T a where T1 :: T (Int,Int,Int) T2 :: T (Int,Int)
f :: T a -> a -> Int f a b = case a of T1 -> case b of { (x,y,z) -> x+y+z } _ -> 0 g :: T a -> a -> Int f a b = case a of T2 -> case b of { (x,y) -> x+y } _ -> 0 h :: T a -> a -> Int h x y = f x y + g x y Then I think you'll get a S(SSS) demand from the f call, and S(SS) from the g call. And then you'll 'both' them. Please don’t document on Phab! In a Note in the code, or a wiki page or both. Maybe that's what you intend. Simon | -----Original Message----- | From: Ömer Sinan Ağacan [mailto:omeraga...@gmail.com] | Sent: 04 March 2016 14:26 | To: Simon Peyton Jones <simo...@microsoft.com> | Cc: ghc-devs <ghc-devs@haskell.org> | Subject: Re: 'lub' and 'both' on strictness - what does it mean for | products to have different arity? | | Simon, your GADT example clearly shows how lub sometimes needs to | handle products with different arities, but the case with bothDmd is | still confuses me. I think we need have a code like this: | | f x + g x | | Where f puts a S(SS) demand on x and g puts a S(SSS). This looks like a | bug/type error to me. Do you have any example for bothDmd too? | | (I started D1968 for documenting these) | | 2016-03-02 3:47 GMT-05:00 Simon Peyton Jones <simo...@microsoft.com>: | > | > | -- lubUse (UProd {}) Used = Used | > | lubUse (UProd ux) Used = UProd (map (`lubArgUse` | > | useTop) ux) | > | lubUse Used (UProd ux) = UProd (map (`lubArgUse` | > | useTop) ux) | > | | > | And then somewhere around that code there's this: | > | | > | Note [Used should win] | > | ~~~~~~~~~~~~~~~~~~~~~~ | > | Both in lubUse and bothUse we want (Used `both` UProd us) to | be | > | Used. | > | Why? Because Used carries the implication the whole thing is | used, | > | box and all, so we don't want to w/w it. If we use it both | > | boxed and | > | unboxed, then we are definitely using the box, and so we are | quite | > | likely to pay a reboxing cost. So we make Used win here. | > | | > | ... | > | | > | It seems like at some point the note was valid, but now the code | > | seems to be doing the opposite. Any ideas which one needs an | update? | > | > I have no idea. It seems that the entire definition of lubUse, | including the Note and the commented-out line, appeared in a single big | patch 99d4e5b4. So no clues there. | > | > However the Note has some nofib numbers. | > | > So yes, it's puzzling. The argument in the Note seems plausible. | Would you like to try reverting to "Used should win" and see what | happens to performance? Probably something will get worse and you'll | have do some digging with ticky-ticky. | > | > | > | I'm also a bit confused about why both and lub are not | commutative, | > | or if they're commutative, why do they have redundant cases. | > | > Yes: lub and both should be commutative; yell if not. (In contrast | > see Note [Asymmetry of 'both' for DmdType and DmdResult]; but that's | > well documented.) | > | > | For example, | > | lubUse has | > | this: | > | | > | lubUse UHead u = u | > | lubUse (UCall c u) UHead = UCall c u | > | | > | instead of something like: | > | | > | lubUse UHead u = u | > | lubUse u UHead = u | > | > I think this is just stylistic. I was dealing with all the cases for | UHead in the first arg, then all the cases for UCall, and so on. That | way I know I've got coverage. | > | > (And perhaps it's more efficient: we pattern match at most once on | each argument. | > | > Simon | > | > | | > | I didn't check all the cases to see if it's really commutative or | > | not, but can I assume that they need to be commutative and | simplify | > | the code? Otherwise let's add a note about why they're not | commutative? | > | | > | Thanks.. | > | | > | 2016-03-02 1:07 GMT-05:00 Ömer Sinan Ağacan | <omeraga...@gmail.com>: | > | >> Could I ask that you add this example as a Note to the relevant | > | >> functions, so that the next time someone asks this question | > | they'll >> find the answer right there? | > | > | > | > Yep, I'll do that soon. | > | > | > | > 2016-03-01 12:01 GMT-05:00 Simon Peyton Jones | > | <simo...@microsoft.com>: | > | >> Omer | > | >> | > | >> Joachim is right. The strictness analyser just looks inside | > | casts, >> so these unexpectedly ill-typed cases could show up. | For | > | example >> >> f :: T a -> a -> Int >> f x y = case x of | > | >> P1 -> case y of (a,b,c) -> a+b+c | > | >> P2 -> case y of (p,q) -> p+q | > | >> | > | >> data T a where | > | >> P1 :: T (Int,Int,Int) | > | >> P2 :: T (Int,Int) | > | >> | > | >> In the P1 branch we have that y::(Int,Int,Int), so we'll get a | > | demand S(SSS). And similarly in the P2 branch. Now we combine | them. | > | And there you have it. | > | >> | > | >> | > | >> Could I ask that you add this example as a Note to the relevant | > | functions, so that the next time someone asks this question they'll | > | find the answer right there? | > | >> | > | >> Thanks | > | >> | > | >> Simon | > | >> | > | >> | -----Original Message----- | > | >> | From: ghc-devs [mailto:ghc-devs-boun...@haskell.org] On | > | Behalf Of >> | Ömer Sinan Agacan >> | Sent: 19 February 2016 | > | 17:27 >> | To: ghc-devs <ghc-devs@haskell.org> >> | Subject: | > | 'lub' and 'both' on strictness - what does it mean for >> | | > | products to have different arity? | > | >> | | > | >> | I was looking at implementations of LUB and AND on demand | >> | > | | signatures and I found this interesting case: | > | >> | | > | >> | lubStr (SProd s1) (SProd s2) | > | >> | | length s1 == length s2 = mkSProd (zipWith | lubArgStr | > | s1 s2) | > | >> | | otherwise = HeadStr | > | >> | | > | >> | The "otherwise" case is interesting, I'd expect that to be | an | > | error. | > | >> | I'm trying to come up with an example, let's say I have a | > | case >> | expression: | > | >> | | > | >> | case x of | > | >> | P1 -> RHS1 | > | >> | P2 -> RHS2 | > | >> | | > | >> | and y is used in RHS1 with S(SS) and in RHS2 with S(SSS). | > | This >> | seems to me like a type error leaked into the demand | analysis. | > | >> | | > | >> | Same thing applies to `bothDmd` too. Funnily, it has this | > | extra >> | comment on this same code: | > | >> | | > | >> | bothStr (SProd s1) (SProd s2) | > | >> | | length s1 == length s2 = mkSProd (zipWith | bothArgStr | > | s1 s2) | > | >> | | otherwise = HyperStr -- Weird | > | >> | | > | >> | Does "Weird" here means "type error and/or bug" ? | > | >> | | > | >> | Should I try replacing these cases with panics and try to | > | validate? | > | >> | _______________________________________________ | > | >> | ghc-devs mailing list | > | >> | ghc-devs@haskell.org | > | >> | | > | >> | | > | | https://na01.safelinks.protection.outlook.com/?url=http%3a%2f%2fmai | > | >> | l.ha | > | >> | skell.org%2fcgi-bin%2fmailman%2flistinfo%2fghc- | > | >> | | > | >> | | > | | devs&data=01%7c01%7csimonpj%40064d.mgd.microsoft.com%7cdd8ae326a8e1 | > | >> | 4ef8 | > | >> | | > | | 9e8608d339520cb9%7c72f988bf86f141af91ab2d7cd011db47%7c1&sdata=sRFRK | > | >> | gj3y zQdEZT4y7KLk18cP43Rv1J%2bx8oPZyr1QzA%3d _______________________________________________ ghc-devs mailing list ghc-devs@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/ghc-devs