Skip to content
GitLab
Menu
Projects
Groups
Snippets
Loading...
Help
Help
Support
Community forum
Keyboard shortcuts
?
Submit feedback
Contribute to GitLab
Sign in
Toggle navigation
Menu
Open sidebar
POTTIER Francois
menhir
Commits
68b4af50
Commit
68b4af50
authored
Jul 06, 2015
by
POTTIER Francois
Browse files
More cleanup.
parent
e7fafc9d
Changes
1
Hide whitespace changes
Inline
Sidebyside
src/Coverage.ml
View file @
68b4af50
...
...
@@ 132,11 +132,10 @@ let causes_an_error s z =
would be faster, but (according to a quick experiment) the paths thus
obtained would be really far from optimal. *)
module
P
=
CompletedNatWitness
type
property
=
Terminal
.
t
P
.
t
module
P
=
struct
include
CompletedNatWitness
type
property
=
Terminal
.
t
t
end
(*  *)
...
...
@@ 146,7 +145,7 @@ type property =
If so, [s'] is passed to the continuation [k]. Otherwise, [P.bottom] is
returned. *)
let
has_transition
s
sym
k
:
property
=
let
has_transition
s
sym
k
:
P
.
property
=
try
let
s'
=
SymbolMap
.
find
sym
(
Lr1
.
transitions
s
)
in
k
s'
...
...
@@ 155,7 +154,7 @@ let has_transition s sym k : property =
(* This computes a minimum over a set of terminal symbols. *)
let
foreach_terminal_in
toks
(
f
:
Terminal
.
t
>
property
)
:
property
=
let
foreach_terminal_in
toks
(
f
:
Terminal
.
t
>
P
.
property
)
:
P
.
property
=
TerminalSet
.
fold
(
fun
t
accu
>
(* Using [min_lazy] allows stopping if we find a path of length 0.
This is just an optimization. *)
...
...
@@ 165,7 +164,7 @@ let foreach_terminal_in toks (f : Terminal.t > property) : property =
(* This is analogous to [foreach_terminal_in], but stops as soon as a
finite value is reached, i.e., as soon as one path is found. *)
let
foreach_terminal_until_finite
(
f
:
Terminal
.
t
>
property
)
:
property
=
let
foreach_terminal_until_finite
(
f
:
Terminal
.
t
>
P
.
property
)
:
P
.
property
=
Terminal
.
fold
(
fun
t
accu
>
(* We stop as soon as we obtain a finite result. *)
P
.
until_finite
accu
(
fun
()
>
f
t
)
...
...
@@ 173,7 +172,7 @@ let foreach_terminal_until_finite (f : Terminal.t > property) : property =
(* This computes a minimum over the productions associated with [nt]. *)
let
foreach_production
nt
(
f
:
Production
.
index
>
property
)
:
property
=
let
foreach_production
nt
(
f
:
Production
.
index
>
P
.
property
)
:
P
.
property
=
Production
.
foldnt
nt
P
.
bottom
(
fun
prod
accu
>
(* Using [min_lazy] allows stopping if we find a path of length 0.
This is just an optimization. *)
...
...
@@ 256,12 +255,15 @@ module QuestionMap =
let
first
=
Analysis
.
first_prod_lookahead
(* The following function answers a question. This requires a fixed point
computation. We have a certain amount of flexibility in how much
information we memoize; if we use a recursive call to [answer], we
recompute; if we use a call to [get], we memoize. *)
(* The following function defines the analysis. *)
(* We have a certain amount of flexibility in how much information we memoize;
if we use a recursive call to [answer], we recompute; if we use a call to
[get], we memoize. As long as every direct recursive call is decreasing,
either choice is acceptable. A quick experiment suggests that memoization
everywhere is costeffective. *)
let
answer
(
q
:
question
)
(
get
:
question
>
property
)
:
property
=
let
answer
(
q
:
question
)
(
get
:
question
>
P
.
property
)
:
P
.
property
=
let
rhs
=
Production
.
rhs
q
.
prod
in
let
n
=
Array
.
length
rhs
in
...
...
@@ 297,12 +299,11 @@ let answer (q : question) (get : question > property) : property =
match
sym
with

Symbol
.
T
t
>
(* Case 2a. [sym] is a terminal symbol [t]. Our precondition
implies that [t] is equal to [a]. [w] must begin with [a]
The rest must be some word [w'] such that, by starting from
[s'] and by reading [w'], we reach our goal. The first letter
in [w'] could be any terminal symbol [c], so we try all of
them. *)
(* Case 2a. [sym] is a terminal symbol [t]. Our precondition implies
that [t] is equal to [a]. [w] must begin with [a]. The rest must
be some word [w'] such that, by starting from [s'] and by reading
[w'], we reach our goal. The first letter in [w'] could be any
terminal symbol [c], so we try all of them. *)
assert
(
Terminal
.
equal
q
.
a
t
);
(* per our precondition *)
P
.
add
(
P
.
singleton
q
.
a
)
(
...
...
@@ 314,11 +315,11 @@ let answer (q : question) (get : question > property) : property =

Symbol
.
N
nt
>
(* Case 2b. [sym] is a nonterminal symbol [nt]. For each letter [c],
for each production [prod'] associated with [nt], we
must
concatenate:
1 a word that takes us from [s], beginning with [a],
to a state
where we can reduce [prod'], looking at [c];
and 2 a
word that takes us from [s'], beginning with [c], to a state
where
we reach our original goal. *)
for each production [prod'] associated with [nt], we
concatenate:
1 a word that takes us from [s], beginning with [a],
to a state
where we can reduce [prod'], looking at [c];
2 a
word that takes us from [s'], beginning with [c], to a state
where
we reach our original goal. *)
foreach_terminal_in
(
first
q
.
prod
(
q
.
i
+
1
)
q
.
z
)
(
fun
c
>
foreach_production
nt
(
fun
prod'
>
...
...
@@ 335,7 +336,7 @@ let answer (q : question) (get : question > property) : property =
end
(* Debugging
wrapper
. TEMPORARY *)
(* Debugging. TEMPORARY *)
let
qs
=
ref
0
let
answer
q
get
=
incr
qs
;
...
...
@@ 343,21 +344,33 @@ let answer q get =
Printf
.
fprintf
stderr
"qs = %d
\n
%!"
!
qs
;
answer
q
get
(* The fixed point. *)
module
F
=
Fix
.
Make
(
Maps
.
PersistentMapsToImperativeMaps
(
QuestionMap
))
(
struct
include
P
type
property
=
Terminal
.
t
t
end
)
(
P
)
let
answer
:
question
>
property
=
let
answer
:
question
>
P
.
property
=
F
.
lfp
answer
(*  *)
(* We now wish to determine, given a state [s'] and a terminal symbol [z],
a minimal path that takes us from some entry state to state [s'] with
[z] as the next (unconsumed) symbol. *)
(* This can be formulated as a search for a shortest path in a graph. The
graph is not just the automaton, though. It is a (much) larger graph
whose vertices are pairs [s, z] and whose edges are obtained by calling
the expensive analysis above. Because we perform a backward search, from
[s', z] to any entry state, we use reverse edges, from a state to its
predecessors in the automaton. *)
(* Debugging. TEMPORARY *)
let
es
=
ref
0
exception
Success
of
property
let
backward
(
s'
,
z
)
:
property
=
let
backward
(
s'
,
z
)
:
P
.
property
=
let
module
G
=
struct
type
vertex
=
Lr1
.
node
*
Terminal
.
t
let
equal
(
s'1
,
z1
)
(
s'2
,
z2
)
=
...
...
@@ 389,6 +402,7 @@ let backward (s', z) : property =
)
(
Lr1
.
predecessors
s'
)
end
in
let
module
D
=
Dijkstra
.
Make
(
G
)
in
let
module
S
=
struct
exception
Success
of
P
.
property
end
in
try
let
_
=
D
.
search
(
fun
(
distance
,
(
v'
,
_
)
,
path
)
>
incr
es
;
...
...
@@ 396,13 +410,13 @@ let backward (s', z) : property =
Printf
.
fprintf
stderr
"es = %d
\n
%!"
!
es
;
if
Lr1
.
incoming_symbol
v'
=
None
then
let
path
=
List
.
map
snd
path
in
raise
(
Success
(
P
.
Finite
(
distance
,
Seq
.
concat
path
)))
(* TEMPORARY keep path *)
raise
(
S
.
Success
(
P
.
Finite
(
distance
,
Seq
.
concat
path
)))
(* TEMPORARY keep path *)
)
in
P
.
bottom
with
Success
p
>
with
S
.
Success
p
>
p
let
backward
s'
:
property
=
let
backward
s'
:
P
.
property
=
Printf
.
fprintf
stderr
"Attempting to reach an error in state %d:
\n
%!"
...
...
@@ 444,3 +458,4 @@ let () =
(* TEMPORARY avoid [error] token unless forced to use it *)
(* TEMPORARY implement and exploit [Lr1.ImperativeNodeMap] using an array *)
(* TEMPORARY the code in this module should run only if coverage is set *)
(* TEMPORARY gain a constant factor by memoizing [nullable_first_prod]? *)
Write
Preview
Markdown
is supported
0%
Try again
or
attach a new file
.
Attach a file
Cancel
You are about to add
0
people
to the discussion. Proceed with caution.
Finish editing this message first!
Cancel
Please
register
or
sign in
to comment