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(* BZ#932 *)
(* Expected time < 1.00s *)
(* Let n be the number of let-in. The complexity comes from the fact
that each implicit arguments of f was in a larger and larger
context. To compute the type of "let _ := f ?Tn 0 in f ?T 0",
"f ?Tn 0" is substituted in the type of "f ?T 0" which is ?T. This
type is an evar instantiated on the n variables denoting the "f ?Ti 0".
One obtain "?T[1;...;n-1;f ?Tn[1;...;n-1] 0]". To compute the
type of "let _ := f ?Tn-1 0 in let _ := f ?Tn 0 in f ?T 0", another
substitution is done leading to
"?T[1;...;n-2;f ?Tn[1;...;n-2] 0;f ?Tn[1;...;n-2;f ?Tn[1;...;n-2] 0] 0]"
and so on. At the end, we get a term of exponential size *)
(* A way to cut the complexity could have been to remove the dependency in
anonymous variables in evars but this breaks intuitive behaviour
(see Case15.v); another approach could be to substitute lazily
and/or to simultaneously substitute let binders and evars *)
Variable P : Set -> Set.
Variable f : forall A : Set, A -> P A.
Time Check
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
let _ := f _ 0 in
f _ 0.
¤ Dauer der Verarbeitung: 0.19 Sekunden
(vorverarbeitet)
¤
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