| For two join points to be duplicates they need to not only be alpha
| equivalent but to also have the same continuation.
Yes exactly. And it would not be hard to adapt the existing CSE pass to support this. Just needs doing.
A ticket and a repo case would be really helpful.
Simon
PS: I am leaving Microsoft at the end of November 2021, at which point simonpj@microsoft.com will cease to work. Use simon.peytonjones@gmail.com instead. (For now, it just forwards to simonpj@microsoft.com.)
| -----Original Message-----
| From: ghc-devs On Behalf Of Viktor
| Dukhovni
| Sent: 24 November 2021 21:27
| To: ghc-devs@haskell.org
| Subject: Re: [EXTERNAL] Unexpected duplicate join points in "Core"
| output?
|
| On Sun, Nov 21, 2021 at 06:53:53AM -0500, Carter Schonwald wrote:
|
| > On Sat, Nov 20, 2021 at 4:17 PM Simon Peyton Jones via ghc-devs <
| > ghc-devs@haskell.org> wrote:
| >
| > > There is absolutely no reason not to common-up those to join
| points.
| > > But we can't common up some join points when we could if they were
| let's.
| > > Consider
| > >
| > > join j1 x = x+1
| > > in case v of
| > > A -> f (join j2 x = x+1 in ...j2...)
| > > B -> ....j1...
| > > C -> ....j1...
| > >
| > > Even though j2 is identical to j1's, we can't eliminate j2 in
| favour
| > > of j1 because then j1 wouldn't be a join point any more.
| >
| > In this example: why would it stop being a join point ?
| >
| > Admittedly, my intuition might be skewed by my own ideas about how
| > join points are sortah a semantic special case of other constructs.
|
| I think the point is that join points are tail calls that don't return
| to the caller. But here even though `j1` and `j2` have the same body
| j1's continuation is not the same as j2's continuation.
|
| Rather the result of `j2` is the input to `f`, but the result of j1 is
| a possible output of the whole `case` block in the B and C branches.
| For two join points to be duplicates they need to not only be alpha
| equivalent but to also have the same continuation. Something like
|
| join j1 x = x + 1 in
| join j2 y = y + 1 in
| ... j1 ...
| ... j2 ...
|
| where eliminating j2 in favour of j1 should be correct.
|
| --
| VIktor.
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