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fn x => let val rec f = fn () => x 1 and g = fn () => x = x in 1 end;
fn x => let fun f () = x 1 and g () = x = x in 1 end;
fn x => let fun f () = (x 1; g()) and g () = (x = x; f x) in 1 end;
fn x => let val rec f = fn () => (x 1; g()) and g = fn () => (x = x; f x) in 1 end;
fn x => let val rec f = fn () => x = x and g = fn () => x 1 in 1 end;
fn x => let fun f () = x = x and g () = x 1 in 1 end;
fn x => let fun f () = (x = x; g()) and g () = (x x; f x) in 1 end;
fn x => let val rec f = fn () => (x = x; g()) and g = fn () => (x 1; f x) in 1 end;
(* 2012-8-4 ohori compiler/util/main/SCCFun.sml:
SCC returns
* the non-connected forests in reversed order
* elements in a connected componets in reversed order
The strightforwardly coding the original SCC algorithm
results in this; i.e. the original scc algorithm visits
the G^T in the decreasing order of the finished time in G.
So when we use scc in fundecls, we do the followng.
* We make the reference ege (instead of presedence edge)
f -> g iff fun f x = ... g
so that f-componet comes earlier than g-componet in the result of
ssc.
This is consistent of the original order of occurrences.
* We then take the reversal of the reversal of ssc.
The inner reversal is necessary since the scc make a list
of elements in each scc-componet in reverse order ot their
"discovery time".
*)
(* 2012-8-4 ohori
Fixed by 4369:a93a16990a4b
*)
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