absolutely wild guess:
the compiler builds a call graph and any function that is not part of the
call graph is eliminated
(def x {:a 1 :b 2}) does not return a function but x maybe used inside some
function, so I'm assuming the compiler just leaves it there
why would it be hard to see if x is or is not used by some function that
is in the call graph? i think that is a more involved job that building a
call graph
On Tue, Mar 10, 2015 at 5:47 PM, Mike Thompson <[email protected]>
wrote:
> On Wednesday, March 11, 2015 at 11:19:28 AM UTC+11, Mike Thompson wrote:
> > This issue from David Nolen, caught my eye:
> > https://github.com/andrewmcveigh/cljs-time/issues/21
> >
> > Feels like there is important information there, but I just don't know
> enough to interpret what's said. Can anyone help?
> >
> > If I have this:
> >
> > (def x {:a 1 :b 2})
> >
> > David is saying that x can't be dead-code-eliminated. Correct?
> >
> > If so, the solution he talks about is?
> >
> > ;; Replacing PersistentHashMaps with functions? Can't be right. I think
> I'm being too literal here
> > (def x ((fn [] {:a 1 :b2})))
> >
> > ;; more likely this?
> >
> > (defn x
> > []
> > {:a 1 :b 2})
> >
> > Ie. you must now "call" x to get the map.
> >
> > If this solution is the right one, doesn't that mean we would be
> inefficiently constructing the (potentially large) hashmap inside x on each
> call.
>
>
> Two follow-on questions:
>
> - Could this problem with PersistentHashMaps not being dead code
> eliminated be fixed with correct application of the Closure's
> "@nosideeffects" tags? Ie. if the function which creates
> PersistentHashMaps had this tag in its comment?
>
> - I assume that the following is not a problem:
>
> (defn y [] :y)
> (def x y)
>
> If both x and y are never used then both will be dead code eliminated.
>
> --
> Mike
>
>
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