Syntax Analysis (Bottom-Up Parser) PPTs for Third Year Computer Science Engineering

1
1
Syntax Analysis
Part II
Chapter 4
COP5621 Compiler Construction
Copyright Robert van Engelen, Florida State University, 2007
2
Bottom-Up Parsing
• LR methods (Left-to-right, Rightmost
derivation)
– SLR, Canonical LR, LALR
• Other special cases:
– Shift-reduce parsing
– Operator-precedence parsing
3
Operator-Precedence Parsing
• Special case of shift-reduce parsing
• We will not further discuss (you can skip
textbook section 4.6)

2
4
Shift-Reduce Parsing
Grammar:
S → a A B e
A → A b c | b
B → d
Shift-reduce corresponds
to a rightmost derivation:
S ⇒rm a A B e
⇒rm a A d e
⇒rm a A b c d e
⇒rm a b b c d e
Reducing a sentence:
a b b c d e
a A b c d e
a A d e
a A B e
S
S
a b b c d e
A
A
B
a b b c d e
A
A
B
a b b c d e
A
A
a b b c d e
A
These match
production’s
right-hand sides
5
Handles
Handle
Grammar:
S → a A B e
A → A b c | b
B → d
A handle is a substring of grammar symbols in a
right-sentential form that matches a right-hand side
of a production
NOT a handle, because
further reductions will fail
(result is not a sentential form)
a b b c d e
a A b c d e
a A A e
… ?
a b b c d e
a A b c d e
a A d e
a A B e
S
6
Stack Implementation of
Shift-Reduce Parsing
Stack
$
$id
$E
$E+
$E+id
$E+E
$E+E*
$E+E*id
$E+E*E
$E+E
$E
Input
id+id*id$
+id*id$
+id*id$
id*id$
*id$
*id$
id$
$
$
$
$
Action
shift
reduce E → id
shift
shift
reduce E → id
shift (or reduce?)
shift
reduce E → id
reduce E → E * E
reduce E → E + E
accept
Grammar:
E → E + E
E → E * E
E → ( E )
E → id
Find handles
to reduce
How to
resolve
conflicts?
`

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7
Conflicts
• Shift-reduce and reduce-reduce conflicts are
caused by
– The limitations of the LR parsing method (even
when the grammar is unambiguous)
– Ambiguity of the grammar
8
Shift-Reduce Parsing:
Shift-Reduce Conflicts
Stack
$…
$…if E then S
Input
…$
else…$
Action
…
shift or reduce?
Ambiguous grammar:
S → if E then S
| if E then S else S
| other
Resolve in favor
of shift, so else
matches closest if
9
Shift-Reduce Parsing:
Reduce-Reduce Conflicts
Stack
$
$a
Input
aa$
a$
Action
shift
reduce A → a or B → a ?
Grammar:
C → A B
A → a
B → a
Resolve in favor
of reduce A → a,
otherwise we’re stuck!

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10
LR(k) Parsers: Use a DFA for
Shift/Reduce Decisions
1
2
4
5
3
0
start
a
A
C
B
a
Grammar:
S → C
C → A B
A → a
B → a
State I0:
S → •C
C → •A B
A → •a
State I1:
S → C•
State I2:
C → A•B
B → •a
State I3:
A → a•
State I4:
C → A B•
State I5:
B → a•
goto(I0,C)
goto(I0,a)
goto(I0,A)
goto(I2,a)
goto(I2,B)
Can only
reduce A → a
(not B → a)
11
DFA for Shift/Reduce Decisions
Stack
$ 0
$ 0
$ 0 a 3
$ 0 A 2
$ 0 A 2 a 5
$ 0 A 2 B 4
$ 0 C 1
Input
aa$
aa$
a$
a$
$
$
$
Action
start in state 0
shift (and goto state 3)
reduce A → a (goto 2)
shift (goto 5)
reduce B → a (goto 4)
reduce C → AB (goto 1)
accept (S → C)
Grammar:
S → C
C → A B
A → a
B → a
The states of the DFA are used to determine
if a handle is on top of the stack
State I0:
S → •C
C → •A B
A → •a
State I3:
A → a•
goto(I0,a)
12
Stack
$ 0
$ 0
$ 0 a 3
$ 0 A 2
$ 0 A 2 a 5
$ 0 A 2 B 4
$ 0 C 1
Input
aa$
aa$
a$
a$
$
$
$
Action
start in state 0
shift (goto 5)
accept (S → C)
Grammar:
S → C
C → A B
A → a
B → a
State I0:
S → •C
C → •A B
A → •a
State I2:
C → A•B
B → •a
goto(I0,A)

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13
Stack
$ 0
$ 0
$ 0 a 3
$ 0 A 2
$ 0 A 2 a 5
$ 0 A 2 B 4
$ 0 C 1
Input
aa$
aa$
a$
a$
$
$
$
Action
start in state 0
shift (goto 5)
accept (S → C)
Grammar:
S → C
C → A B
A → a
B → a
State I2:
C → A•B
B → •a
State I5:
B → a•
goto(I2,a)
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Stack
$ 0
$ 0
$ 0 a 3
$ 0 A 2
$ 0 A 2 a 5
$ 0 A 2 B 4
$ 0 C 1
Input
aa$
aa$
a$
a$
$
$
$
Action
start in state 0
shift (goto 5)
accept (S → C)
Grammar:
S → C
C → A B
A → a
B → a
State I2:
C → A•B
B → •a
State I4:
C → A B•
goto(I2,B)
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Stack
$ 0
$ 0
$ 0 a 3
$ 0 A 2
$ 0 A 2 a 5
$ 0 A 2 B 4
$ 0 C 1
Input
aa$
aa$
a$
a$
$
$
$
Action
start in state 0
shift (goto 5)
accept (S → C)
Grammar:
S → C
C → A B
A → a
B → a
State I0:
S → •C
C → •A B
A → •a
State I1:
S → C•
goto(I0,C)

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16
Stack
$ 0
$ 0
$ 0 a 3
$ 0 A 2
$ 0 A 2 a 5
$ 0 A 2 B 4
$ 0 C 1
Input
aa$
aa$
a$
a$
$
$
$
Action
start in state 0
shift (goto 5)
accept (S → C)
Grammar:
S → C
C → A B
A → a
B → a
State I0:
S → •C
C → •A B
A → •a
State I1:
S → C•
goto(I0,C)
17
Model of an LR Parser
$
an
…
ai
…
a2
a1
LR Parsing Program
(driver)
goto
action
s0
…
Xm-1
sm-1
Xm
sm output
stack
input
DFA
shift
reduce
accept
error
Constructed with
LR(0) method,
SLR method,
LR(1) method, or
LALR(1) method
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LR Parsing (Driver)
(s0 X1 s1 X2 s2 … Xm sm, ai ai+1 … an $)
stack input
Configuration ( = LR parser state):
If action[sm,ai] = shift s then push ai, push s, and advance input:
(s0 X1 s1 X2 s2 … Xm sm ai s, ai+1 … an $)
If action[sm,ai] = reduce A → β and goto[sm-r,A] = s with r=|β| then
pop 2r symbols, push A, and push s:
(s0 X1 s1 X2 s2 … Xm-r sm-r A s, ai ai+1 … an $)
If action[sm,ai] = accept then stop
If action[sm,ai] = error then attempt recovery

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Example LR Parse Table
Grammar:
1. E → E + T
2. E → T
3. T → T * F
4. T → F
5. F → ( E )
6. F → id
r5
r5
r5
r5
r3
r3
r3
r3
r1
r1
s7
r1
s11
s6
s4
s5
s4
s5
r6
r6
r6
r6
s4
s5
r4
r4
r4
r4
r2
r2
s7
r2
acc
s6
s4
s5
$
)
(
*
+
id
11
10
9
8
7
6
5
4
3
2
1
0
F
T
E
10
3
9
3
2
8
3
2
1
Shift & goto 5
Reduce by
production #1
action goto
state
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Example LR Parsing
Stack
$ 0
$ 0 id 5
$ 0 F 3
$ 0 T 2
$ 0 T 2 * 7
$ 0 T 2 * 7 id 5
$ 0 T 2 * 7 F 10
$ 0 T 2
$ 0 E 1
$ 0 E 1 + 6
$ 0 E 1 + 6 id 5
$ 0 E 1 + 6 F 3
$ 0 E 1 + 6 T 9
$ 0 E 1
Input
id*id+id$
*id+id$
*id+id$
*id+id$
id+id$
+id$
+id$
+id$
+id$
id$
$
$
$
$
Action
shift 5
reduce 6 goto 3
reduce 4 goto 2
shift 7
shift 5
reduce 6 goto 10
reduce 3 goto 2
reduce 2 goto 1
shift 6
shift 5
reduce 6 goto 3
reduce 4 goto 9
reduce 1 goto 1
accept
Grammar:
1. E → E + T
2. E → T
3. T → T * F
4. T → F
5. F → ( E )
6. F → id
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SLR Grammars
• SLR (Simple LR): a simple extension of
LR(0) shift-reduce parsing
• SLR eliminates some conflicts by
populating the parsing table with reductions
A→α on symbols in FOLLOW(A)
S → E
E → id + E
E → id
State I0:
S → •E
E → •id + E
E → •id
State I2:
E → id•+ E
E → id•
goto(I0,id) goto(I3,+)
FOLLOW(E)={$}
thus reduce on $
Shift on +

8
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SLR Parsing Table
r2
s2
r3
s3
acc
s2
$
+
id
4
3
2
1
0
E
4
1
1. S → E
2. E → id + E
3. E → id
FOLLOW(E)={$}
thus reduce on $
Shift on +
• Reductions do not fill entire rows
• Otherwise the same as LR(0)
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SLR Parsing
• An LR(0) state is a set of LR(0) items
• An LR(0) item is a production with a • (dot) in the
right-hand side
• Build the LR(0) DFA by
– Closure operation to construct LR(0) items
– Goto operation to determine transitions
• Construct the SLR parsing table from the DFA
• LR parser program uses the SLR parsing table to
determine shift/reduce operations
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Constructing SLR Parsing Tables
1. Augment the grammar with S’→S
2. Construct the set C={I0,I1,…,In} of LR(0) items
3. If [A→α•aβ] ∈ Ii and goto(Ii,a)=Ij then set
action[i,a]=shift j
4. If [A→α•] ∈ Ii then set action[i,a]=reduce A→α
for all a ∈ FOLLOW(A) (apply only if A≠S’)
5. If [S’→S•] is in Ii then set action[i,$]=accept
6. If goto(Ii,A)=Ij then set goto[i,A]=j
7. Repeat 3-6 until no more entries added
8. The initial state i is the Ii holding item [S’→•S]

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25
LR(0) Items of a Grammar
• An LR(0) item of a grammar G is a production of
G with a • at some position of the right-hand side
• Thus, a production
A → X Y Z
has four items:
[A → • X Y Z]
[A → X • Y Z]
[A → X Y • Z]
[A → X Y Z •]
• Note that production A → ε has one item [A → •]
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Constructing the set of LR(0)
Items of a Grammar
1. The grammar is augmented with a new start
symbol S’ and production S’→S
2. Initially, set C = closure({[S’→•S]})
(this is the start state of the DFA)
3. For each set of items I ∈ C and each grammar
symbol X ∈ (N∪T) such that goto(I,X) ∉ C and
goto(I,X) ≠ ∅, add the set of items goto(I,X) to C
4. Repeat 3 until no more sets can be added to C
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The Closure Operation for LR(0)
Items
1. Start with closure(I) = I
2. If [A→α•Bβ] ∈ closure(I) then for each
production B→γ in the grammar, add the
item [B→•γ] to I if not already in I
3. Repeat 2 until no new items can be added

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28
The Closure Operation
(Example)
Grammar:
E → E + T | T
T → T * F | F
F → ( E )
F → id
{ [E’ → • E] }
closure({[E’ → •E]}) =
{ [E’ → • E]
[E → • E + T]
[E → • T] }
{ [E’ → • E]
[E → • E + T]
[E → • T]
[T → • T * F]
[T → • F] }
{ [E’ → • E]
[E → • E + T]
[E → • T]
[T → • T * F]
[T → • F]
[F → • ( E )]
[F → • id] }
Add [E→•γ]
Add [T→•γ]
Add [F→•γ]
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The Goto Operation for LR(0)
Items
1. For each item [A→α•Xβ] ∈ I, add the set
of items closure({[A→αX•β]}) to goto(I,X)
if not already there
2. Repeat step 1 until no more items can be
added to goto(I,X)
3. Intuitively, goto(I,X) is the set of items
that are valid for the viable prefix γX when
I is the set of items that are valid for γ
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The Goto Operation (Example 1)
Suppose I = Then goto(I,E)
= closure({[E’ → E •, E → E • + T]})
= { [E’ → E •]
[E → E • + T] }
Grammar:
E → E + T | T
T → T * F | F
F → ( E )
F → id
{ [E’ → • E]
[E → • E + T]
[E → • T]
[T → • T * F]
[T → • F]
[F → • ( E )]
[F → • id] }

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31
The Goto Operation (Example 2)
Suppose I = { [E’ → E •], [E → E • + T] }
Then goto(I,+) = closure({[E → E + • T]}) = { [E → E + • T]
[T → • T * F]
[T → • F]
[F → • ( E )]
[F → • id] }
Grammar:
E → E + T | T
T → T * F | F
F → ( E )
F → id
32
Example SLR Grammar and
LR(0) Items
Augmented
grammar:
1. C’ → C
2. C → A B
3. A → a
4. B → a
State I0:
C’ → •C
C → •A B
A → •a
State I1:
C’ → C•
State I2:
C → A•B
B → •a
State I3:
A → a•
State I4:
C → A B•
State I5:
B → a•
goto(I0,C)
goto(I0,a)
goto(I0,A)
goto(I2,a)
goto(I2,B)
I0 = closure({[C’ → •C]})
I1 = goto(I0,C) = closure({[C’ → C•]})
…
start
final
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Example SLR Parsing Table
r4
r2
r3
s5
acc
s3
$
a
5
4
3
2
1
0
B
A
C
4
2
1
State I0:
C’ → •C
C → •A B
A → •a
State I1:
C’ → C•
State I2:
C → A•B
B → •a
State I3:
A → a•
State I4:
C → A B•
State I5:
B → a•
1
2
4
5
3
0
start
a
A
C
B
a
Grammar:
1. C’ → C
2. C → A B
3. A → a
4. B → a

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34
SLR and Ambiguity
• Every SLR grammar is unambiguous, but not
every unambiguous grammar is SLR
• Consider for example the unambiguous grammar
S → L = R | R
L → * R | id
R → L
I0:
S’ → •S
S → •L=R
S → •R
L → •*R
L → •id
R → •L
I1:
S’ → S•
I2:
S → L•=R
R → L•
I3:
S → R•
I4:
L → *•R
R → •L
L → •*R
L → •id
I5:
L → id•
I6:
S → L=•R
R → •L
L → •*R
L → •id
I7:
L → *R•
I8:
R → L•
I9:
S → L=R•
action[2,=]=s6
action[2,=]=r5
no
Has no SLR
parsing table
35
LR(1) Grammars
• SLR too simple
• LR(1) parsing uses lookahead to avoid
unnecessary conflicts in parsing table
• LR(1) item = LR(0) item + lookahead
LR(0) item:
[A→α•β]
LR(1) item:
[A→α•β, a]
36
I2:
S → L•=R
R → L•
action[2,=]=s6
Should not reduce on =, because no
right-sentential form begins with R=
split
R → L•
S → L•=R
SLR Versus LR(1)
• Split the SLR states by
adding LR(1) lookahead
• Unambiguous grammar
1. S → L = R
2. | R
3. L → * R
4. | id
5. R → L
lookahead=$
action[2,$]=r5

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37
LR(1) Items
• An LR(1) item
[A→α•β, a]
contains a lookahead terminal a, meaning α
already on top of the stack, expect to see βa
• For items of the form
[A→α•, a]
the lookahead a is used to reduce A→α only if the
next input is a
• For items of the form
[A→α•β, a]
with β≠ε the lookahead has no effect
38
The Closure Operation for LR(1)
Items
1. Start with closure(I) = I
2. If [A→α•Bβ, a] ∈ closure(I) then for each
production B→γ in the grammar and each
terminal b ∈ FIRST(βa), add the item
[B→•γ, b] to I if not already in I
3. Repeat 2 until no new items can be added
39
The Goto Operation for LR(1)
Items
1. For each item [A→α•Xβ, a] ∈ I, add the
set of items closure({[A→αX•β, a]}) to
goto(I,X) if not already there
2. Repeat step 1 until no more items can be
added to goto(I,X)

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40
Constructing the set of LR(1)
Items of a Grammar
1. Augment the grammar with a new start symbol S’
and production S’→S
2. Initially, set C = closure({[S’→•S, $]})
(this is the start state of the DFA)
3. For each set of items I ∈ C and each grammar
symbol X ∈ (N∪T) such that goto(I,X) ∉ C and
goto(I,X) ≠ ∅, add the set of items goto(I,X) to C
4. Repeat 3 until no more sets can be added to C
41
Example Grammar and LR(1)
Items
• Unambiguous LR(1) grammar:
S → L = R
| R
L → * R
| id
R → L
• Augment with S’ → S
• LR(1) items (next slide)
42
[S’ → •S, $] goto(I0,S)=I1
[S → •L=R, $] goto(I0,L)=I2
[S → •R, $] goto(I0,R)=I3
[L → •*R, =/$] goto(I0,*)=I4
[L → •id, =/$] goto(I0,id)=I5
[R → •L, $] goto(I0,L)=I2
[S’ → S•, $]
[S → L•=R, $] goto(I0,=)=I6
[R → L•, $]
[S → R•, $]
[L → *•R, =/$] goto(I4,R)=I7
[R → •L, =/$] goto(I4,L)=I8
[L → •*R, =/$] goto(I4,*)=I4
[L → •id, =/$] goto(I4,id)=I5
[L → id•, =/$]
[S → L=•R, $] goto(I6,R)=I9
[R → •L, $] goto(I6,L)=I10
[L → •*R, $] goto(I6,*)=I11
[L → •id, $] goto(I6,id)=I12
[L → *R•, =/$]
[R → L•, =/$]
[S → L=R•, $]
[R → L•, $]
[L → *•R, $] goto(I11,R)=I13
[R → •L, $] goto(I11,L)=I10
[L → •*R, $] goto(I11,*)=I11
[L → •id, $] goto(I11,id)=I12
[L → id•, $]
[L → *R•, $]
I0:
I1:
I2:
I3:
I4:
I5:
I6:
I7:
I8:
I9:
I10:
I12:
I11:
I13:

15
43
Constructing Canonical LR(1)
Parsing Tables
1. Augment the grammar with S’→S
2. Construct the set C={I0,I1,…,In} of LR(1) items
3. If [A→α•aβ, b] ∈ Ii and goto(Ii,a)=Ij then set
action[i,a]=shift j
4. If [A→α•, a] ∈ Ii then set action[i,a]=reduce
A→α (apply only if A≠S’)
5. If [S’→S•, $] is in Ii then set action[i,$]=accept
6. If goto(Ii,A)=Ij then set goto[i,A]=j
7. Repeat 3-6 until no more entries added
8. The initial state i is the Ii holding item [S’→•S,$]
44
Example LR(1) Parsing Table
r4
r5
s11
s12
r6
r2
r6
r6
r4
r4
s11
s12
r5
r5
s4
s5
r3
r6
s6
acc
s4
s5
$
=
*
id
13
12
11
10
9
8
7
6
5
4
3
2
1
0
R
L
S
13
10
4
10
7
8
3
2
1
Grammar:
1. S’ → S
2. S → L = R
3. S → R
4. L → * R
5. L → id
6. R → L
45
LALR(1) Grammars
• LR(1) parsing tables have many states
• LALR(1) parsing (Look-Ahead LR) combines
LR(1) states to reduce table size
• Less powerful than LR(1)
– Will not introduce shift-reduce conflicts, because shifts
do not use lookaheads
– May introduce reduce-reduce conflicts, but seldom do
so for grammars of programming languages

16
46
Constructing LALR(1) Parsing
Tables
1. Construct sets of LR(1) items
2. Combine LR(1) sets with sets of items that
share the same first part
[L → *•R, =]
[R → •L, =]
[L → •*R, =]
[L → •id, =]
[L → *•R, $]
[R → •L, $]
[L → •*R, $]
[L → •id, $]
I4:
I11:
[L → *•R, =/$]
[R → •L, =/$]
[L → •*R, =/$]
[L → •id, =/$]
Shorthand
for two items
in the same set
47
Example LALR(1) Grammar
• Unambiguous LR(1) grammar:
S → L = R
| R
L → * R
| id
R → L
• Augment with S’ → S
• LALR(1) items (next slide)
48
[S’ → •S, $] goto(I0,S)=I1
[S → •L=R, $] goto(I0,L)=I2
[S → •R, $] goto(I0,R)=I3
[L → •*R, =/$] goto(I0,*)=I4
[L → •id, =/$] goto(I0,id)=I5
[R → •L, $] goto(I0,L)=I2
[S’ → S•, $]
[S → L•=R, $] goto(I0,=)=I6
[R → L•, $]
[S → R•, $]
[L → *•R, =/$] goto(I4,R)=I7
[R → •L, =/$] goto(I4,L)=I9
[L → •*R, =/$] goto(I4,*)=I4
[L → •id, =/$] goto(I4,id)=I5
[L → id•, =/$]
[S → L=•R, $] goto(I6,R)=I8
[R → •L, $] goto(I6,L)=I9
[L → •*R, $] goto(I6,*)=I4
[L → •id, $] goto(I6,id)=I5
[L → *R•, =/$]
[S → L=R•, $]
[R → L•, =/$]
I0:
I1:
I2:
I3:
I4:
I5:
I6:
I7:
I8:
I9:
Shorthand
for two items
[R → L•, =]
[R → L•, $]

17
49
Example LALR(1) Parsing Table
r6
r6
r2
r4
r4
s4
s5
r5
r5
s4
s5
r3
r6
s6
acc
s4
s5
$
=
*
id
9
8
7
6
5
4
3
2
1
0
R
L
S
8
9
7
9
3
2
1
Grammar:
1. S’ → S
2. S → L = R
3. S → R
4. L → * R
5. L → id
6. R → L
50
LL, SLR, LR, LALR Summary
• LL parse tables computed using FIRST/FOLLOW
– Nonterminals × terminals → productions
– Computed using FIRST/FOLLOW
• LR parsing tables computed using closure/goto
– LR states × terminals → shift/reduce actions
– LR states × nonterminals → goto state transitions
• A grammar is
– LL(1) if its LL(1) parse table has no conflicts
– SLR if its SLR parse table has no conflicts
– LALR(1) if its LALR(1) parse table has no conflicts
– LR(1) if its LR(1) parse table has no conflicts
51
LL, SLR, LR, LALR Grammars
LL(1)
LR(1)
LR(0)
SLR
LALR(1)

18
52
Dealing with Ambiguous
Grammars
1. S’ → E
2. E → E + E
3. E → id
r2
s3/r2
s2
r3
r3
acc
s3
s2
$
+
id
4
3
2
1
0
E
4
1
Shift/reduce conflict:
action[4,+] = shift 4
action[4,+] = reduce E → E + E
When reducing on +:
yields left associativity
(id+id)+id
When shifting on +:
yields right associativity
id+(id+id)
$ 0 E 1 + 3 E 4
id+id+id$
+id$
$ 0
… …
stack input
≈ ≈ ≈
53
Using Associativity and
Precedence to Resolve Conflicts
• Left-associative operators: reduce
• Right-associative operators: shift
• Operator of higher precedence on stack: reduce
• Operator of lower precedence on stack: shift
S’ → E
E → E + E
E → E * E
E → id
$ 0 E 1 * 3 E 5
id*id+id$
+id$
$ 0
… …
stack input
reduce E → E * E
≈ ≈ ≈
54
Error Detection in LR Parsing
• Canonical LR parser uses full LR(1) parse
tables and will never make a single
reduction before recognizing the error when
a syntax error occurs on the input
• SLR and LALR may still reduce when a
syntax error occurs on the input, but will
never shift the erroneous input symbol

19
55
Error Recovery in LR Parsing
• Panic mode
– Pop until state with a goto on a nonterminal A is found,
(where A represents a major programming construct),
push A
– Discard input symbols until one is found in the
FOLLOW set of A
• Phrase-level recovery
– Implement error routines for every error entry in table
• Error productions
– Pop until state has error production, then shift on stack
– Discard input until symbol is encountered that allows
parsing to continue

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