diff --git a/src/LRijkstra.ml b/src/LRijkstra.ml
index 58a9267cc221a0ea4f3eab5a04fbc7cd4d0054a0..30e06516342d224c2b660de567a9f0e73794576b 100644
--- a/src/LRijkstra.ml
+++ b/src/LRijkstra.ml
@@ -265,9 +265,9 @@ module Trie : sig
of the trie that has been constructed for state [s]. *)
val size: int -> int
- (* After [star] has been called a number of times, [cumulated_size()]
+ (* After [star] has been called a number of times, [total_size()]
reports the total size of the tries that have been constructed. *)
- val cumulated_size: unit -> int
+ val total_size: unit -> int
(* Every (sub-)trie has a unique identity. (One can think of it as its
address.) [compare] compares the identity of two tries. This can be
@@ -430,7 +430,7 @@ end = struct
assert (size.(s) >= 0);
size.(s)
- let cumulated_size () =
+ let total_size () =
!c
let compare t1 t2 =
@@ -449,7 +449,7 @@ end = struct
SymbolMap.find a t.transitions (* careful: may raise [Not_found] *)
let verbose () =
- Printf.eprintf "Cumulated star size: %d\n%!" (cumulated_size())
+ Printf.eprintf "Total star size: %d\n%!" (total_size())
let decode i =
let t = MenhirLib.InfiniteArray.get tries i in
@@ -499,7 +499,7 @@ let dummy : fact =
(* The lookahead symbol fits in 8 bits. In the largest grammars that we have
seen, the number of unique words is about 3.10^5, so a word should fit in
about 19 bits (2^19 = 524288). In the largest grammars that we have seen,
- the cumulated star size is about 64000, so a trie should fit in about 17
+ the total star size is about 64000, so a trie should fit in about 17
bits (2^17 = 131072). We have ample space in a 63-bit word! We allocate 8
bits for [lookahead], 30 bits for [word], and 25 bits for [position]. We
could support 32-bit machines too, but that is probably pointless. *)
@@ -1340,8 +1340,8 @@ let () =
end
(* Automaton size (i.e., number of states). *)
Lr1.n
- (* Cumulated trie size. *)
- (Trie.cumulated_size())
+ (* Total trie size. *)
+ (Trie.total_size())
(* Size of [F]. *)
(F.size())
(* Size of [E]. *)