Transcript ITS 015: Compiler Construction

LR Parsing Table Costruction
Lecture 6
Syntax Analysis
1
LR parsing example
Grammar:
1. E -> E + T
2. E -> T
3. T -> T * F
4. T -> F
5. F -> ( E )
6. F -> id
2
LR parsing example
CONFIGURATIONS
STACK
INPUT
0
id * id + id $
ACTION
shift 5
3
STACK
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
0
0
0
0
0
0
0
0
0
0
0
0
0
0
id 5
F5
T2
T2*7
T 2 * 7 id 5
T 2 * 7 F 10
T2
E1
E1+6
E 1 + 6 id 5
E1+6F3
E1+6T9
E1
INPUT
id *
*
*
*
id
id
id
id
id
ACTION
+
+
+
+
+
+
+
+
+
id$
id$
id$
id$
id$
id$
id$
id$
id$
id$
$
$
$
$
shift
reduced by
reduced by
shift
shift
reduced by
reduced by
reduced by
shift
shift
reduced by
reduced by
EE+T
accept
F  id
TF
F  id
T  T*F
ET
F  id
TF
Fig. 4.32. Moves of LR parser on id * id + id.
4
LR grammars
If it is possible to construct an LR parse table for G, we
say “G is an LR grammar”.
LR parsers DO NOT need to parse the entire stack to
decide what to do (other shift-reduce parsers might).
Instead, the STATE symbol summarizes all the information
needed to make the decision of what to do next.
The GOTO function corresponds to a DFA that knows how
to find the HANDLE by reading the top of the stack
downwards.
In the example, we only looked at 1 input symbol at a
time. This means the grammar is LR(1).
5
How to construct an LR parse table?
We will look at 3 methods:
 Simple LR (SLR): simple but not very powerful
 Canonical LR: very powerful but too many states
 LALR: almost as powerful with many fewer states
yacc uses the LALR algorithm.
6
SLR (Simple LR) Parse Table Construction
7
SLR parse tables
The SLR parse table is easy to construct, but the resulting parser is
a little weak.
The table is based on LR(0) ITEMS, or just plain ITEMS.
A LR(0) item is a production G with a dot at some position on the
RHS.
The production A -> XYZ could generate the following LR(0) items:
 A -> .XYZ
 A -> X.YZ
 A -> XY.Z
 A -> XYZ.
The production A -> ε only generates 1 LR(0) item:
 A -> .
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LR(0) items
An item indicates how far we are in parsing the RHS.
A -> .XYZ means we think we’re at the beginning of an
A production, but haven’t seen an X yet.
A -> X.YZ means we think we’re in the middle of an A
production, have seen an X, and should see a Y soon.
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Augmenting the grammar G
Before we can produce an SLR parse table, we have to
AUGMENT the input grammar, G.
Given G, we produce G’, the AUGMENTED GRAMMAR
for G:
 Add a new symbol S’
 Add a new production S’ -> S (where S is the old
start symbol)
 Make S’ the new start symbol
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Item set closure
We need a new concept: the CLOSURE of a set of
LR(0) items.
If I is a set of items for grammar G’, then the CLOSURE
of I is defined recursively:
 Initially, every item in I is added to closure(I)
 If A -> α . B β is in closure(I) and B -> γ is a
production, then add the item B -> . γ to I, if not
already there.
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Itemset closure example
E’ -> E
E -> E + T | T
T -> T * F | F
F -> ( E ) | id
Closure(I) = { E’ -> . E
E -> . E + T
E -> . T
T -> . T * F
T -> . F
Initial itemset I is { E’ -> .E }
F -> . ( E )
F -> . id }
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The goto table
We also need the function goto(I,X) that takes an
itemset I and a grammar symbol X, and returns the
closure of the set of all items [ A -> α X . β ] such
that [ A -> α . X β ] is in I.
Example:
I = { [E’ -> E.], [E -> E. + T] }
goto(I,+) =
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I0: E'  · E
E·E+T
E·T
T·T*F
T·F
F  · (E)
F  · id
I1: E'  E ·
EE·+T
I2: E  T ·
TT·*F
I3: T  F ·
I4: F  (· E )
E·E+T
E·T
T·T*F
T·F
F  · (E)
F  · id
I5: F  id ·
I6: E  E + · T
T·T*F
T·F
F  · (E)
F  · id
I7: T  T * · F
F  · (E)
F  · id
I8: F  ( E · )
EE·+T
I9: E  E + T ·
TT·*F
I10: T  T * F ·
I11: F  ( E ) ·
Fig. 4.35. Canonical LR(0) collection
for grammar (4.19)
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I0
E
+
I1
I6
T
F
(
id
T
*
I2
I7
F
(
id
F
I9
*
to I7
to I3
to I4
to I5
I10
to I4
to I5
I3
(
(
I4
E
I8
)
+
T
id
id
F
I11
to I6
to I2
to I3
I5
Fig. 4.36. Transition diagram of DFA D form viable prefixes.
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Canonical LR(0) itemsets
The CANONICAL LR(0) ITEMSETS can be used to create
the states in the SLR parse table.
We begin with an initial set C = {closure({ [S’->.S] })}.
Then, foreach I in C and each grammar symbol X such
that goto(I,X) is not empty and not in C already, do
 Add goto(I,X) to C
Example: canonical LR(0) itemsets for the same grammar.
Each set in C corresponds to a state in a DFA.
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How to build the SLR parse table
1.
2.
3.
4.
Take the augmented grammar G’
Construct the canonical LR(0) itemsets C for G’
Associate a state with each itemset Ii in C
Construct the parse table as follows:
1. If A -> α . a β is in Ii and goto(Ii,a) = Ij, then set action[i,a]
to “shift j” (“a” here is a terminal)
2. If A -> α . is in Ii then set action[i,a] to “reduce A -> α” for
all a in FOLLOW(A)
3. If S’ -> S . is in Ii then set action[i,$] to “accept”
If any of the actions in the table conflict, then G is NOT SLR.
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Example SLR table construction
For the first LR(0) itemset in our favorite grammar:
I0:
E’ -> .E
E -> .E + T
E -> .T
T -> .T * F
T -> .F
F -> .(E)
F -> .id
This gives us action[0,(] = shift 4
This gives us action[0,id] = shift 5
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Using Ambiguous Grammars
19
What to do with ambiguity?
Sometimes it is convenient to leave ambiguity in G
For instance, G1:
is simpler than G2:
E -> E + E
E -> E + T | T
|E*E
E -> T * F | F
|(E)
F -> ( E ) | id
| id
But SLR(1), LR(1), and LALR(1) parsers will all have a
shift/reduce conflict for G1.
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What to do with ambiguity?
Sometimes it is convenient to leave ambiguity in G
For instance, G1:
is simpler than G2:
E -> E + E
E -> E + T | T
|E*E
E -> T * F | F
|(E)
F -> ( E ) | id
| id
But SLR(1), LR(1), and LALR(1) parsers will all have a
shift/reduce conflict for G1.
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LR(0) itemsets for G1
22
Ambiguity leads to conflicts
G1 is ambiguous, so we are guaranteed to get conflicts.
For example, in I7:
 We will add rules to “shift 4” on ‘+’ and “shift 5” on ‘*’.
 For the item E -> E+E. we will add the rule
“reduce E->E+E” to the parse table for each terminal
in FOLLOW(E).
 But! FOLLOW(E) contains + and * -- shift/reduce
conflict.
LR(1) and LALR(1) tables will have the same problems.
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Resolving the conflicts
Knowing about operator precedence and associativity,
we can resolve the conflicts.
Example: for input “id + id * id”, we will be in state 7
after processing “id + id”
STACK
INPUT
0E1+4E7
* id $
since * has higher precedence than +, we should
really shift, not reduce.
With a + next in the input, we should reduce, to enforce
left-associativity.
See Fig. 4.47 in text for a complete SLR(1) table.
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If-else ambiguity
The ambiguity of the “dangling else” creates a shiftreduce conflict in parsers for most languages.
Since the else is normally associated with the nearest if,
we resolve the conflict by shifting, instead of educing,
when we see “else” in the input.
See the LR(0) states and parse table on page 251.
This method is much simpler than writing an
unambiguous grammar.
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Non-SLR grammars
Consider the assignment grammar
1.
2.
3.
4.
5.
6.
S’ -> S
S -> L = R
S -> R
L -> * R
L -> id
R -> L
generating, e.g. S =*> id = * id
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Non-SLR grammars
Construct the initial canonical LR(0) itemset I0.
Compute I2 = goto(I0,L) and I6 = goto(I2,=).
Compute FOLLOW(L)
Compute parse table entries for I2: shift/reduce conflict!
This means in state I2, with ‘=’ in the input, we do not
know whether to shift and go to state I6 or reduce
with R -> L, since ‘=’ is in FOLLOW(L).
To correct this, we need to know more about the context
of the L we just parsed.
“Canonical LR(1)” and “LALR(1)” are powerful enough.
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Canonical LR Parse Table Construction
28
I0: S'  · S
S·L=R
S·R
L·*R
L  · id
R·L
I5: L  id ·
I6: S  L = · R
R·L
L·*R
L  · id
I1: S'  S ·
I7: L  * R ·
I2: S  L · = R
I8: R  L ·
I3: S  R ·
I9: S  L = R ·
RL·
I4: L  * · R
R·L
L·*R
L  · id
Fig. 4.37. Canonical LR(0) collection for grammar (4.20).
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More states means more memory
In SLR, we said in state i we should reduce by A -> α if
the itemset contains the item [A -> α .] and a is in
FOLLOW(A).
However, sometimes when state i is on top of the stack,
and a is next in the input, what comes BEFORE α on
the stack might invalidate the reduction A -> α.
Example from previous grammar: sentential form
“R = …” is impossible, but “* R =” is possible.
So actually, we really want to reduce by L -> * R when
we see R on stack and “=” in the input.
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LR(1) idea
Our parser needs to keep track of more state information.
How can it?
Idea: use canonical LR(0) states, but split states as
needed by adding a terminal symbol to each item.
LR(1) ITEMS take the form [A-> α.β,a], where A-> αβ is
a production in G and a is a terminal symbol or $.
The “1” refers to the length of a, the LOOKAHEAD for
each item. If length = k, we would have an LR(k) item.
In parsing, we will now only reduce αβ. to A if an item’s
lookahead symbol agrees with the next input.
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LR(1) parse table construction
We need to redefine closure(I) for a set of LR(1) items:
for each
item [A-> α.B β,a] in I
production B -> γ in G’
terminal b in FIRST(β a)
such that [B->. γ,b] is not already in I, do:
add [B->. γ,b] to I
repeat until no more items can be added to I
goto(I,X) is the same as for SLR(1).
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Example LR(1) parser construction
Begin with augmented grammar G’:
S’ -> S
S -> C C
[ what is L(G’)?? ]
C -> c C | d
The first itemset
S’ -> .S,$
S -> .CC,$
C -> .cC,c/d
C -> .d,c/d
}
I0 = closure({S’->.S,$}) = {
[ from S’->.S,$ and S->CC, B=S, α=ε, β= ε ]
[ from S’->.CC,$ and C->cC, B=C, α= ε, β=C ]
[ from S’->.CC,$ and C->d, B=C, α= ε, β=C ]
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S' → ·S , $
S → ·CC , $
C → ·cC , c /d
C → ·d , c /d
I0
S
C
S' → S·, $
I1
S → C·C , $
C → ·cC , $
C → ·d , $
I2
C
S → CC·, $
I5
c
c
C → c·C , $
C → ·cC , $
C → ·d , $
I6
C
C → cC·, $
I9
d
d
C → d·, $
I7
C
C → cC·, c /d
I8
c
c
C → c·C , c /d
C → ·cC , c /d
C → ·d , c /d
I3
d
d
C → d·, c /d
I4
Fig. 4.39. The goto graph for grammar (4.21).
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LR(1) parsers: the good news
LR(1) is quite similar to SLR(1), with one main difference:
 We only add reduce rules to the parse table when
the input matches the LOOKAHEAD for the item
 SLR(1) adds reduce rules for any terminal in the
FOLLOW set.
This means LR(1) will have fewer shift/reduce and
reduce/reduce conflicts, because it tries to reduce in
fewer situations.
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LR(1) parsers: the bad news
LR(1) parsers are powerful, able to parse almost any
unambiguous CFG used for real programming
languages.
But there is a price: the number of states is huge.
For the very simple c*dc*d language with 4 productions,
we already needed 10 LR(1) states.
For a typical PL like Pascal, the LR(1) table would
contain a few THOUSAND states!
Is there a technique as powerful with fewer states?
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action
STATE
0
c
d
s3
s4
1
goto
$
S
C
1
2
acc
2
s6
s7
5
3
s3
s4
8
4
r3
r3
5
6
r1
s6
s7
7
8
9
9
r3
r2
r2
r2
Fig. 4.40. Canonical parsing table for grammar (4.21).
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LALR Parse Table Construction
38
LALR parse tables
LALR makes smaller parse tables than canonical LR, but
still covers most common programming language
constructs.
LALR has the same number of states as the SLR parser
for the same grammar, but is more picky about when
to reduce, so fewer conflicts come up.
yacc actually constructs a LALR(1) table, not a
canonical LR(1) table.
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LALR idea
Usually, in a LR parser, there will be many states that
are identical, except for the lookahead symbol.
LALR takes these identical states and MERGES them,
forming the UNION of the lookahead symbols for the
merged items.
Algorithm: build the LR(1) itemsets, then merge
itemsets with the same CORES.
40
LALR example
I0: S’ -> .S,$
S -> .CC,$
C -> .cC,c/d
C -> .d,c/d
I1: S’ -> S.,$
I2: S -> C.C,$
C -> .cC,$
C -> .d,$
I4: C -> d.,c/d
I9: C -> cC.,$
Which LR(1) itemsets
can be merged?
I3: C -> c.C,c/d
C -> .cC,c/d
C -> .d,c/d
I5: S -> CC.,$
I6: C -> c.C,$
C -> .cC,$
C -> .d,$
I7: C -> d.,$
I8: C -> cC.,c/d
41
action
STATE
0
c
d
s36
s47
1
goto
$
S
C
1
2
acc
2
s36
s47
5
36
s36
s47
89
47
r3
r3
5
89
r3
r1
r2
r2
r2
Fig. 4.41. LALR parsing table for grammar (4.21).
42
Efficient Construction of LALR Parsing Tables
Example 4.46. Let us again consider the augmented grammar
S'  S
SL=R|R
A  * R | id
BL
The kernels of the sets of LR(0) items for this grammar are shown in Fig. 4.42.
I0: S'  · S
I1: S'  S ·
I2: S  L · = R
RL·
I3: S  R ·
I4: L  * · R
I5: L  id ·
I6: S  L = · R
I7: L  * R ·
I8: R  L ·
I9: S  L = R ·
Fig. 4.42. Kernels of the sets of LR(0) items for grammar (4.20).
43
Efficient Construction of LALR Parsing Tables
Example 4.47. Let us construct the kernels of the LALR(1) items for the grammar in
the previous example. The kernels of the LR(0) items were shown in Fig. 4.42.
When we apply Algorithm 4.12 to the kernel of set of items I0, we compute
closure ({[S'  · S, #]}), which is
S'  · S, #
S  · L = R, #
S  · R, #
L  · * R, #/=
L  · id, #/=
R  · L, #
44
I0:
FROM
S'  · S
I2: S  L · = R
I4: L  * · R
I6: S  L = · R
TO
I1:
I2:
I2:
I3:
I4:
I5:
I6:
I4:
I5:
I7:
I8:
I4:
I5:
I8:
I9:
S'  S ·
SL·=R
RL·
SR·
L*·R
L  id ·
SL=·R
L*·R
L  id ·
L*R·
RL·
L*·R
L*·R
RL·
SL=R·
Fig.4.44. Propagation of lookaheads.
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SET
ITEM
LOOKAHEADS
INIT
PASS1
PASS2
PASS3
$
$
$
$
I1: S'  S ·
$
$
$
I2: S  L · = R
$
$
$
I2: R L ·
$
$
$
I3: S  R ·
$
$
$
I0: S'  · S
I4: L  * · R
=
=/$
=/$
=/$
I5: L  id ·
=
=/$
=/$
=/$
$
$
I6: S  L = · R
I7: L  * R ·
=
=/$
=/$
I8: R  L ·
=
=/$
=/$
I9: S  L = R ·
$
Fig. 4.45. Computation of lookaheads.
46
Next time
- Yacc 사용법은 조교가 설명
- Semantic 처리 (Yacc에서 배운 것 구현 방법)
47