Transcript Term

Part 4 Syntax Analysis
E.g.
<sentence>  <Subject><Predicate>
<Subject>  <adjective><noun>
<Subject>  <noun>
<Predicate> <verb><Object>
<Object>  <adjective><noun>
<Object>  <noun>
“young men like pop music”
Lexical Analyzer
“(adjective, ) (noum, ) (verb, ) (adjective, ) (noum, )”
????
<sentence>  <Subject><Predicate>
Leftmost
Rightmost
 <adjective><noun><Predicate>
Derivation
Reduction
 <adjective><noun>< verb><object>
 <adjective><noun>< verb>< adjective><noun>
Leftmost <adjective><noun>< verb>< adjective><noun>
Reduction <Subject >< verb>< adjective><noun>
 <Subject >< verb>< object>
 <Subject><Predicate>
 <sentence>
How can I design and code the “derivation” or
“reduction”?
0 Approaches to implement a Syntax analyzer
1、The syntax description of programming
language constructs
– Context-free grammars
What is the definition of Context-free grammars?
Please recall it!
6
2、Why a grammar is usually used to describe the
syntax of a programming language?
– A grammar gives a precise ,yet easy-tounderstand, syntactic specification of a
programming language
– From certain classes of grammar we can
automatically construct an efficient parser that
determines if a source program is syntactically
well formed
– A properly designed grammar imparts a
structure to a programming language that
is useful for the translation of source
programs into correct object code and for
the detection of errors
– The evolved constructs can be added to a
language more easily
3、Approached to implement a syntax
analyzer
– Manual construction
– Construction by tools
4.1 The Role of the Parser
1、 Main task
– Obtain a string of tokens from the lexical
analyzer
– Verify that the string can be generated by the
grammar of related programming language
– Report any syntax errors in an intelligible
fashion
– Recover from commonly occurring errors so
that it can continue processing the remainder of
its input
2、Position of parser in compiler model
Source Lexical
program analyzer
token
Rest of
Intermediate
Parse
Parser
front end representation
tree
Get next
token
Symbol
table
3、Parsing methods
(1)Top-Down
(2)Bottom-Up
4、Syntax Error handling
1) Error levels
– Lexical, such as misspelling an identifier,
keyword, or operator
– Syntactic, such as an arithmetic expression
with unbalanced parentheses
– Semantic, such as an operator applied to an
incompatible operand
– Logical, such as an infinitely recursive call
2) Simple-to-state goals of the error handler
– It should report the presence of errors
clearly and accurately
– It should recover from each error quickly
enough to be able to detect subsequent
errors
– It should not significantly slow down the
processing of correct programs
3) Error-recovery strategies
– Panic mode
• Discard input symbols one at a time until
one of a designated set of synchronizing
tokens is found
– Phrase level
• Replace a prefix of the remaining input by
some string that allows the parser to continue
Simple Instruction of Top-Down and Bottom-Up
E.g. 1) S  xAy 2) A ** 3)A  *, and Verify “x*y”
Top-Down:
(1)Left-most derivation
 SxAy x*y
 Parser Tree
Bottom-Up:
(1)Left-most reduction
 x*y xAy
 Parser Tree
S
x
A
y
*
How codes?
PDA
Model
x*y #
x*y #
S
#
controller
rules
Sentential
form
?
controller
rules
output
#
1,3
String
1) S  xAy 2) A ** 3)A  *,
output
3,1
?
Controller
T-D PDA Controller
 IF “x” is the top symbol of the
stack and is non-terminal，then
find a production rule as
“x……” randomly，replace “x”
with the right of the rule，and
output the No of the rule——
derivation。
 IF “x” is the top symbol of the
stack and is same to that under the
reading point，then… ——
Matching。
 IF (2) fail, then make a
backtracking action to the scene
before the last derivation and
select a new rule —backtracking
 IF there no new rule, fail
 IF there is only “＃” in the stack，
and “＃” is under the reading
point ，success
B-U PDA Controller
•
IF the several top symbols in stack is
a Handling, then reduction,
•
else if “＃” is under the reading
point then fail, else Move the
symbol under reading point into stack.
•
IF there is only #S in the stack，
and “＃” is under the reading
point ，success
E.G.
x*y #
x*y #
S
#
controller
rules
output
controller
rules
#
output
E.G.
x*y #
x*y #
x
A
y
#
controller
rules
1
x
#
controller
rules
output
E.G.
x*y #
x*y #
A
y
#
controller
rules
1
*
x
#
controller
rules
output
E.G.
x*y #
x*y #
*
*
y
#
controller
rules
1,2
A
x
#
controller
rules
3
E.G.
x*y #
x*y #
*
y
#
controller
rules
1,2
y
A
x
#
controller
rules
3
E.G.
x*y #
x*y #
A
y
#
controller
rules
1
S
#
controller
rules
3,1
E.G.
x*y #
x*y #
*
y
#
controller
rules
1,3
S
#
controller
rules
3,1
E.G.
x*y #
x*y #
y
#
controller
rules
1,3
S
#
controller
rules
3,1
E.G.
x*y #
x*y #
#
controller
rules
1,3
S
#
controller
rules
3,1
Discussion
Flaw of T-D
 Left Recursion Infinite
loop
Eliminating Left Recursion
 Backtracking inefficient
1. Methods: Predictive and
Eliminating Ambiguity
2. Left common factor
Flaw of B-U
• Next
4. 2 TOP-DOWN PARSING
1、Ideas
Find a leftmost
derivation for an
input string
E (E) (E+E) (E*E+E)
( i*E+E) ( i*i+E) ( i* i+ i)
E
Construct a parse
tree for the input
starting from the
root and creating the
nodes of the parse
tree in preorder.
E
i
(
E
)
E
+
E
*
E
i
i
2、Main methods
– Predictive parsing (no backtracking)
– Recursive descent (involve backtracking)
3、Recursive descent
– A deducing procedure, which construct a
parse tree for the string top-down from S.
When there is any mismatch, the program
go back to the nearest non-terminal,
select another production to construct the
parse tree
– If you produce a parse tree at last, then
the parsing is success, otherwise, fail.
Grammar for Parsing Example
Start  Expr
Expr  Expr + Term
Expr  Expr - Term
Expr  Term
Term  Term * Int
Term  Term / Int
Term  Int
• Set of tokens is
{ +, -, *, /, Int },
where Int = [0-9][0-9]*
Parsing Example
Parse Tree
Remaining Input
Start
<int,><-,><int,  ><*,><int,  >
Sentential form
Start
Applied Production
Current Position in Parse Tree
Parsing Example
Parse
Tree
Start
Remaining Input
<int,><-,><int,  ><*,><int,  >
Expr
Sentential Form
Expr
Current Position in Parse Tree
Start  Expr
Parsing Example
Parse
Tree
Remaining Input
Start
<int,><-,><int,  ><*,><int,  >
Expr
Expr
-
Sentential Form
Term
Expr - Term
Expr  Expr + Term
Expr  Expr - Term
Expr  Term
Applied Production
Expr  Expr - Term
Parsing Example
Parse
Tree
Remaining Input
Start
<int,><-,><int,  ><*,><int,  >
Expr
Expr
Term
-
Sentential Form
Term
Term - Term
Expr  Expr + Term
Expr  Expr - Term
Expr  Term
Applied Production
Expr  Term
Parsing Example
Parse
Tree
Remaining Input
Start
<int,><-,><int,  ><*,><int,  >
Expr
Expr
Term
Int
-
Sentential Form
Term
Int - Term
Applied Production
Term  Int
Parsing Example
Parse
Tree
Remaining Input
Match
Input <int,><-,><int,  ><*,><int,  >
Token!
Sentential Form
Start
Expr
Expr
Term
Int
-
Term
Int - Term
Parsing Example
Parse
Tree
Match
Input
Token!
Start
Expr
Expr
Term
Int
-
Remaining Input
<-,><int,  ><*,><int,  >
Sentential Form
Term
- Term
Parsing Example
Parse
Tree
Match
Input
Token!
Start
Expr
Expr
Term
Int
-
Remaining Input
<int,  ><*,><int,  >
Sentential Form
Term
Term
Parsing Example
Parse
Tree
Remaining Input
Start
<int,  ><*,><int,  >
Expr
Expr
Term
Int
Term
Sentential Form
Term
*
Term*Int
Int
Applied Production
Term  Term * Int
Parsing Example
Parse
Tree
Remaining Input
Start
<int,  ><*,><int,  >
Expr
Expr
Term
Int
Term
Int
Sentential Form
Term
*
Int * Int
Int
Applied Production
Term  Int
Parsing Example
Parse
Tree
Match
Input
Token!
Start
Expr
Expr
Term
Int
Term
Int
Remaining Input
<int,  ><*,><int,  >
Sentential Form
Term
*
Int* Int
Int
Parsing Example
Parse
Tree
Match
Input
Token!
Start
Expr
Expr
Term
Int
Term
Int
Remaining Input
<*,><int,  >
Sentential Form
Term
*
* Int
Int
Parsing Example
Parse
Tree
Match
Input
Token!
Start
Expr
Expr
Term
Int
Term
Int
Remaining Input
<int,  >
Sentential Form
Term
*
Int
Int
Parsing Example
Parse
Tree
Parse
Complete!
Start
Expr
Expr
Term
Int
Term
Int
Remaining Input
<int,  >
Sentential Form
Term
*
Int
Backtracking Example
Start
<int,><-,><int,  ><*,><int,  >
Start
Backtracking Example
Start
<int,><-,><int,  ><*,><int,  >
Expr
Start  Expr
Backtracking Example
Parse
Tree
Start
<int,><-,><int,  ><*,><int,  >
Expr
Sentential Form
Expr
+
Term
Expr + Term
Expr  Expr + Term
Backtracking Example
Parse
Tree
Remaining Input
Start
<int,><-,><int,  ><*,><int,  >
Expr
Sentential Form
Expr
Term
+
Term
Term + Term
Applied Production
Expr  Term
Backtracking Example
Parse
Tree
Remaining Input
Match
Input <int,><-,><int,  ><*,><int,  >
Token!
Sentential Form
Start
Expr
Expr
Term
Int
+
Term
Int + Term
Applied Production
Term  Int
Backtracking Example
Parse
Tree
Can’t
Match
Input
Token!
Start
Expr
Expr
Term
Int 2
+
Term
Remaining Input
<-,><int,  ><*,><int,  >
Sentential Form
Int - Term
Applied Production
Term  Int
Backtracking Example
Parse
Tree
Start
Remaining Input
So
<int,><-,><int,  ><*,><int,  >
Backtrack!
Expr
Sentential Form
Expr
Applied Production
Start  Expr
Backtracking Example
Parse
Tree
Remaining Input
Start
<int,><-,><int,  ><*,><int,  >
Expr
Sentential Form
Expr
-
Term
Expr - Term
Applied Production
Expr  Expr - Term
Backtracking Example
Parse
Tree
Remaining Input
Start
<int,><-,><int,  ><*,><int,  >
Expr
Sentential Form
Expr
Term
-
Term
Term - Term
Applied Production
Expr  Term
Backtracking Example
Parse
Tree
Remaining Input
Start
<-,><int,  ><*,><int,  >
Expr
Sentential Form
Expr
Term
Int
-
Term
Int - Term
Applied Production
Term  Int
Backtracking Example
Parse
Tree
Match
Input
Token!
Start
Expr
Expr
Term
Int
-
Term
Remaining Input
<-,><int,  ><*,><int,  >
Sentential Form
Int - Term
Backtracking Example
Parse
Tree
Match
Input
Token!
Start
Expr
Expr
Term
Int
-
Term
Remaining Input
<int,  ><*,><int,  >
Sentential Form
Int - Term
How to code that? PDA models
a+b……#
S
#
Control part
Production rules
输出带
Runing
（1）if “x” is the top symbol of the stack and is nonterminal，then find a production rule as
“x……”randomly，replace “x” with the right of the
rule，and output the No of the rule——derivation。
（2） if “x” is the top symbol of the stack and is same
to that under the reading point，then… ——Matching。
（3）if (2) fail, then make a backtracking action to the
scene before the last derivation and select a new rule —
backtracking
(4) If there no new rule, fail
（5）if there is only “＃” in the stack，and “＃” is
under the reading point ，success
E.g. 1) S  xAy
2) A **
3)A  *
x*y #
S
#
controller
Production rules
output
E.g. 1) S  xAy
2) A **
3)A  *
x*y #
S
#
1) S  xAy
2) A **
3)A  *
x*y #
x
A
y
#
1
1) S  xAy
2) A **
3)A  *
x*y #
A
y
#
1
1) S  xAy
2) A **
3)A  *
x*y #
*
*
y
#
1,2
1) S  xAy
2) A **
3)A  *
x*y #
*
y
#
1,2
1) S  xAy
2) A **
3)A  *
x*y #
A
y
#
1
1) S  xAy
2) A **
3)A  *
x*y #
*
y
#
1,3
1) S  xAy
2) A **
3)A  *
x*y #
y
#
1,3
Left Recursion + Top-Down Parsing = Infinite
Loop
• Example Production: Term  Term*Num
• Potential parsing steps:
Term
Term
Term
Term *
Num
Term *
Term *
Num
Num
Backtracking parsers are not seen frequently,
because:
• Backtracking is not very efficient.
Why backtracking occurred?
A left-recursive grammar can cause a
recursive-descent parser to go into an infinite
loop.
An ambiguity grammar can cause
backtracking
Left factor can also cause a backtracking
4、Elimination of Left Recursion
1)Basic form of left recursion
Left recursion is the grammar contains the
following kind of productions.
• P P| Immediate recursion
or
• P Aa , APb Indirect recursion
2)Strategy for elimination of Left Recursion
Convert left recursion into the equivalent
right recursion
P  P|
=> P->*
=> P P’ P’ P’|
3)Algorithm
(1) Elimination of immediate left recursion
P  P|
=> P->*
=> P P’ P’ P’|
(2) Elimination of indirect left recursion
Convert it into immediate left recursion first
according to specific order, then eliminate the
related immediate left recursion
Algorithm:
– (1)Arrange the non-terminals in G in some order as
P1,P2,…,Pn, do step 2 for each of them.
– (2) for (i=1,i<=n,i++)
{for (k=1,k<=i-1,k++)
{replace each production of the form Pi Pk
by Pi 1  | 2  |……| ,n ;
where Pk 1| 2|……| ,n are all the
current Pk -productions
}
change Pi  Pi1| Pi2|…. | Pim|1| 2|….| n
into Pi  1 Pi `| 2 Pi `|……| n Pi `
Pi`1Pi`|2Pi`|……| mPi`| }
/*eliminate the immediate left recursion*/
(3)Simplify the grammar.
E.g. Eliminating all left recursion in the following
grammar:
(1) S  Qc|c (2)Q  Rb|b (3) R  Sa|a
Answer: 1)Arrange the non-terminals in the order:R,Q,S
2）for R: no actions.
for Q:Q  Rb|b
Q  Sab|ab|b
for S: S  Qc|c
S  Sabc|abc|bc|c;
then get S  (abc|bc|c)S`
S`  abcS`| 
3) Because R,Q is not reachable, so delete them
so, the grammar is :
S  (abc|bc|c)S`
S`  abcS`| 
5、Eliminating Ambiguity of a grammar
– Rewriting the grammar
stmtif expr then stmt|if expr then stmt else
stmt|other
==>
stmt matched-stmt|unmatched-stmt
matched-stmt if expr then matched-stmt
else matched-stmt|other
unmatched-stmt if expr then stmt|if expr
then matched-stmt else unmatched-stmt
6、Left factoring
– A grammar transformation that is useful
for producing a grammar suitable for
predictive parsing
– Rewrite the productions to defer the
decision until we have seen enough of the
input to make right choice
If the grammar contains the productions like
A1| 2|…. | n
Chang them into AA`
A`1|2|…. |n
7、Predictive Parsers Methods
– Transition diagram based predictive parser
– Non-recursive predictive parser
8、 Transition diagram based Predictive Parsers
1) Transition diagram
– create an initial and final(return) state
– for each production AX1X2…Xn, create
a path from initial to the final state, with
edges labeled X1,X2,..,Xn
Note: (1)There is one diagram for each nonterminal;
(2)The labels of edges are tokens or nonterminals;
(3)If the edge is labeled by a non-terminal
A, the parser instead goes to the start state
for A, without moving the input cursor
(4)When an edge labeled by a nonterminal is followed, a potentially
recursive procedure call is made
2) Transition diagram based predictive parsing
• Begins in the start state for the start symbol;
• When it is in state s with an edge labeled by terminal
a to state t, and the next input symbol is a, then the
parser moves the input cursor and goes to state t
• When it is in state s with an edge labeled by nonterminal A to state t, then the parser instead goes to
the start state for A, without moving the input cursor.
If it ever reaches the final state for A, it immediately
goes to state t, in effect having read A from the input
during the time it moved from state s to t.
9、Non-recursive Predictive Parsing
1) key problem in predictive parsing
• Determining the production to be
applied for a non-terminal
2)Basic idea of the parser
Table-driven and use stack
3) Model of a non-recursive predictive parser
Input
a+b……#
Stack
S
#
Predictive Parsing
Program
Parsing Table M
Output
4) Predictive Parsing Program
X: the symbol on top of the stack;
a: the current input symbol
If X=a=#, the parser halts and announces
successful completion of parsing;
If X=a!=#, the parser pops X off the stack
and advances the input pointer to the next
input symbol;
If X is a non-terminal, the program consults
entry M[X,a] of the parsing table M. This
entry will be either an X-production of the
grammar or an error entry.
E.g. Consider the following grammar, and
parse the string id+id*id#
1.E  TE`
2.E`  +TE`
3.E`  
4.T  FT`
5.T`  *FT`
6.T`  
7.F  id
8.F (E)
Parsing table M
id
E
(
TFT`
#
E`ε
E`ε
T`ε
T`ε
TFT`
T`ε
F i
)
ETE`
E`
+TE`
T`
F
*
ETE`
E`
T
+
T`
*FT`
F (E)
id+id*id#
E
#
Predictive Parsing
Program
Parsing Table M
Please Write down the procedure of analysis!
10、Construction of a predictive parser
1) FIRST & FOLLOW
FIRST:
• If  is any string of grammar symbols,
let FIRST() be the set of terminals
that begin the string derived from .
+ , then  is also in FIRST()
• If  
• That is :
 V*, First()＝{a|  a……,a VT }
FOLLOW:
• For non-terminal A, to be the set of
terminals a that can appear
immediately to the right of A in some
sentential form.
• That is: Follow(A)＝{a|S …Aa…,a
VT }
If S…A, then # FOLLOW(A)。
2) Computing FIRST()
(1)to compute FIRST(X) for all grammar
symbols X
• If X is terminal, then FIRST(X) is {X}.
• If X  is a production, then add  to
FIRST(X).
• If Xa is a production, then add a to
FIRST(X).
• If X is non-terminal, and X 
Y1Y2…Yk，Yj(VNVT),1j k, then
{
j=1; FIRST(X)={}; //initiate
while ( j<k and FIRST(Yj)) {
FIRST(X)=FIRST(X)(FIRST(Yj)-{})
j=j+1
}
IF (j=k and  FIRST(Yk))
FIRST(X)=FIRST(X)  {}
}
(2)to compute FIRST for any string  =X1X2…Xn，
Xi(VNVT),1i n
{i=1; FIRST()={}; //initiate
repeat
{
FIRST()=FIRST()(FIRST(Xi)-{})
i=i+1
}
until (i=n and  FIRST(Xj))
IF (i=n and  FIRST(Xn))
FIRST()=FIRST(){}
}
3) Computing FOLLOW(A)
(1) Place # in FOLLOW(S), where S is the
start symbol and # is the input right endmarker.
(2)If there is A B in G, then add
First()-{}to Follow(B).
(3)If there is A B, or AB where
FIRST() contains ，then add Follow(A)
to Follow(B).
E.g. Consider the following Grammar,
construct FIRST & FOLLOW for each nonterminals
1.E  TE`
2.E`  +TE`
3.E` 
4.T  FT`
5.T`  *FT`
6.T` 
7.F  i
8.F (E)
Answer:
First(E)=First(T)=First(F)={(, i}
First(E`)={+, }
First(T`)={*, }
Follow(E)= Follow(E`)={),#}
Follow(T)= Follow(T`)={+,),#}
Follow(F)={*,+,),#}
4) Construction of Predictive Parsing Tables
Main Idea: Suppose A  is a production
with a in FIRST(). Then the parser will
expand A by  when the current input
* , we should again
symbol is a. If  
expand A by  if the current input symbol is
in FOLLOW(A), or if the # on the input has
been reached and # is in FOLLOW(A).
– Input. Grammar G.
– Output. Parsing table M.
Method.
1. For each production A  , do steps 2 and
3.
2. For each terminal a in FIRST(), add A
 to M[A,a].
3. If  is in FIRST(), add A  to M[A,b]
for each terminal b in FOLLOW(A). If  is
in FIRST() and # is in FOLLOW(A), add
A  to M[A,#].
4.Make each undefined entry of M be error.
E.g. Consider the following Grammar,
construct predictive parsing table for it.
1.E  TE`
2.E`  +TE`
3.E` 
4.T  FT`
5.T`  *FT`
6.T` 
7.F  i
8.F (E)
Answer:
First(E)=First(T)=First(F)={(, i}
First(E`)={+, }
First(T`)={*, }
Follow(E)= Follow(E`)={),#}
Follow(T)= Follow(T`)={+,),#}
Follow(F)={*,+,),#}
i
E
(
TFT`
#
E`ε
E`ε
T`ε
T`ε
TFT`
T`ε
F i
)
ETE`
E`
+TE`
T`
F
*
ETE`
E`
T
+
T`
*FT`
F (E)
11、LL(1) Grammars
E.g. Consider the following Grammar,
construct predictive parsing table for it.
S  iEtSS` |a
S`  eS | 
E b
a
S
b
S a
i
t
#
S
iEtSS`
S`
E
e
S` eS
S` 
E b
S`ε
1)Definition
A grammar whose parsing table has no multiplydefined entries is said to be LL(1).
The first “L” stands for scanning the input from
left to right.
The second “L” stands for producing a leftmost
derivation
“1” means using one input symbol of look-ahead
s.t each step to make parsing action decisions.
Note:
(1)No ambiguous can be LL(1).
(2)Left-recursive grammar cannot be LL(1).
(3)A grammar G is LL(1) if and only if
whenever A  |  are two distinct
productions of G:
1). For no terminal a do both  and  derive
strings beginning with a.
2). At most one of  and  can derive the
empty string.
* then  does not derive any string
3). If  ,
beginning with a terminal in FOLLOW(A).
12、Transform a grammar to LL(1) Grammar
– Eliminating all left recursion
– Left factoring
13、Error recovery in predictive parsing
Panic-mode error recovery
Phrase-level recovery
4. 3 BOTTOM-UP Parsing
1、Basic idea of bottom-up parsing
Shift-reduce parsing
– Operator-precedence parsing
• An easy-to-implement form
– LR parsing
• A much more general method
• Used in a number of automatic parser
generators
2、Basic concepts in Shift-reducing Parsing
– Handles
– Handle Pruning
3、Stack implementation of Shift-Reduce parsing
Input
……#
Stack
#
Parsing Program
Parsing Table M
Output
4. 4 Operator-precedence parsing
1、The definition of an operator grammar
– The grammar has the property that no
production right side is  or has two adjacent
non-terminals.
– E.g. E E+E|E-E|E*E|E/E|(E)|i
2、Precedence relations
– Three disjoint precedence relations
, between certain pairs of terminals.
Terminals a,b, with the following forms:“…ab…”,
“…aQb…”, and Q if non-terminal. Then the
relationship of a and b is:
1) a b
a yields precedence to b
2) a b
a has the same precedence as b
3) a b
a takes precedence over b
4) for some terminals,we might have none of these
relations.
RS
LS
+
*
(
)
id
+
*
(
)
id
#
Related Grammar: EE+F|F F  F*G|G G (E)|id
#
3、Using Operator-Precedence Relations
Delimit the handle of a right sentential form,
with marking the left end, appearing in the
interior of the handle, and marking the right
end.
• Let’s analyze id+id+id*id# according to
Operator-Precedence Relations.
4、Operator-precedence parsing Algorithm
– Input. An input string w and a table of
precedence relations.
– Output. If w is well formed , a skeletal
parse tree, with a placeholder nonterminal E labeling all interior nodes;
otherwise, an error indication.
– Method. Initially, the stack contains # and
the input buffer the string w#.
Algorithm
Set ip to point to the first symbol of w#;
While (1) {
if (# is on top of the stack an ip points to #) /*success*/
return;
else {
let a be the topmost terminal symbol on the stack;
let b be the symbol pointed to by ip;
if (a b || a b) /*Shift*/
{
push b onto the stack;
advance ip to the next input symbol;
}
Algorithm
else if a b /*reduce*/
do {
pop the stack}
while the top stack terminal is not related by to
the terminal most recently popped
else error()
}
}
5、Construct the operator-precedence
relationship table
– Construct the FIRSTVT and LASTVT for
each non-terminals in the grammar.
– Find out the relations between each of the
terminals.
FIRSTVT(P)=
{ a|P a…or P Qa…，a VT; P,Q VN}
LASTVT(P)=
{ a|P  … a or P … aQ，a VT; P,Q VN}
Construct FIRSTVT(P)
(1) If the productions are like P a… or P
Qa… , then a FIRSTVT(P)
(2) If a FIRSTVT(Q), and there is a
production like P Q… in the grammar,
then a FIRSTVT(P)
– If there is such string as …aP…at the
right side of a production, for each of the
terminals belong to FIRSTVT(P), the
relation is a b;
– If there is such string as …Pb… at the
right side of a production, for each of the
terminals belong to LASTVT(P), the
relation is a b.
– If there is such string as …aPb… or …ab…
at the right side of a production, then a b.
Notes: We assume the precedence of a unary
operator is always higher than that of a binary
operator
E.g. Construct the operator-precedence
relationship table
S if Eb then E else E
E E+T|T
T T*F|F
F i
Eb b
Answer: add a production S’#S#
FIRSTVT(S)={if}
FIRSTVT(E)={+,*,i}
FIRSTVT(T)={*,i}
FIRSTVT(F)={i}
FIRSTVT(Eb )={b}
LASTVT(S)={else,+,*,i}
LASTVT(E)={+,*,i}
LASTVT(T)={*,i}
LASTVT(F)={i}
LASTVT(Eb)={b}
if then else
if
then
else
+
*
i
b
#
+
*
i
b
#
6、Advantages of Operator-precedence
parsing
– Simplicity, easy to construct by hand
7、Disadvantages of Operator-precedence parsing
– It is hard to handle tokens like the unary
operators
– Since the relationship between a grammar for
the language being parsed and the operatorprecedence parser itself is tenuous, one cannot
always be sure the parser accepts exactly the
desired language.
– Only a small class of grammars can be parsed
using operator-precedence techniques.
Exercises:
4.14, 4.27
4. 5 LR parsers
1、LR parser
– An efficient, bottom-up syntax analysis
technique that can be used to parse a large
class of context-free grammars
– LR(k)
• L: left-to-right scan
• R: construct a rightmost derivation in
reverse
• k: the number of input symbols of look
ahead
2、Advantages of LR parser
– It can recognize virtually all programming language
constructs for which context-free grammars can be
written
– It is the most general non backtracking shift-reduce
parsing method
– It can parse more grammars than predictive parsers
can
– It can detect a syntactic error as soon as it is possible
to do so on a left-to-right scan of the input
3、Disadvantages of LR parser
– It is too much work to construct an LR parser
by hand
– It needs a specialized tool,YACC, help it to
generate a LR parser
4、Three techniques for constructing an LR
parsing
– SLR: simple LR
– LR(1): canonical LR
– LALR: look ahead LR
5、The LR Parsing Model
input
a+b……#
LR Parsing Program
S0
stack
Parsing table
output
Note: 1)The driver program is the same for all LR
parsers; only the parsing table changes from one
parser to another
2)The parsing program reads characters from
an input buffer one at a time
3)Si is a state, each state symbol summarizes
the information contained in the stack below
it
4)The current input symbol are used to index
the parsing table and determine the shiftreduce parsing decision
5)In an implementation, the grammar symbols
need not appear on the stack
6、The parsing table
state
0
1
2
3
4
5
i
S5
+
S6
r2
r4
ACTION
*
(
S4
)
#
accept
S7
r4
S5
r2
r4
r2
r4
S4
r6
GOTO
E
T
F
1
2
3
r6
8
r6
r6
2
3
– Action: a parsing action function
• Action[S,a]: S represent the state currently
on top of the stack, and a represent the
current input symbol. So Action[S,a] means
the parsing action for S and a.
Action: a parsing action function
• Shift
– The next input symbol is shifted onto the top of
the stack
– Shift S, where S is a state
• Reduce
– The parser knows the right end of the handle is
at the top of the stack, locates the left end of the
handle within the stack and decides what nonterminal to replace the handle. Reduce by a
grammar production A 
• Accept
– The parser announces successful completion of
parsing.
• Error
– The parser discovers that a syntax error has
occurred and calls an error recovery routine.
Goto: a goto function that takes a state and
grammar symbol as arguments and
produces a state
E.g. the parsing action and goto functions of an
LR parsing table for the following grammar.
E  E+T
E T
T T*F
T F
F (E)
Fi
state
0
1
2
3
4
5
6
7
8
9
10
11
i
S5
+
S6
r2
r4
ACTION
*
(
S4
)
#
accept
S7
r4
S5
r2
r4
r2
r4
S4
r6
r6
S5
S5
8
r6
r3
r5
S11
r1
r3
r5
2
3
9
3
10
r6
S4
S4
S6
r1
r3
r5
GOTO
E
T
F
1
2
3
r1
r3
r5
1)Sj means shift and stack state j, and the
top of the stack change into（j,a）;
2)rj means reduce by production numbered j;
3)Accept means accept
4)blank
means error
Moves of LR parser on i*i+i
State stack
Sym bol stack
input
action
0
#
i*i+ i#
Shift
05
#i
*i+ i#
R educe by 6
03
#F
*i+ i#
R educe by 4
02
#T
*i+ i#
Shift
027
#T *
i+ i#
Shift
0275
#T *i
+ i#
R educe by 6
02710
#T *F
+ i#
R educe by 3
02
#T
+ i#
R educe by 2
01
#E
+ i#
Shift
016
#E +
i#
Shift
0165
#E + i
#
R educe by 6
0163
#E + F
#
R educe by 4
0169
#E + T
#
R educe by 1
01
#E
#
A ccept
Action conflict
• Shift/reduce conflict
– Cannot decide whether to shift or to reduce
• Reduce/reduce conflict
– Cannot decide which of several reductions to make
Notes: An ambiguous grammar can cause
conflicts and can never be LR,e.g.
If_stmt syntax (if expr then stmt [else stmt])
7、The algorithm
– The next move of the parser is determined by
reading the current input symbol a, and the
state S on top of the stack,and then consulting
the parsing action table entry action[S,a].
– If action[Sm,ai]=shift S`,the parser executes a
shift move ,enter the S` into the stack,and the
next input symbol ai+1 become the current
symbol.
– If action[Sm,ai]=reduce A , then the parser
executes a reduce move. If the length of  is ,
then delete  states from the stack, so that the
state at the top of the stack is Sm-  . Push the
state S’=GOTO[Sm- ,A] and non-terminal A
into the stack. The input symbol does not
change.
– If action[Sm,ai]=accept, parsing is completed.
– If action[Sm,ai]=error, the parser has
discovered an error and calls an error
recovery routine.
8、LR Grammars
– A grammar for which we can construct a
parsing table is said to be an LR grammar.
9、The difference between LL and LR grammars
– LR grammars can describe more languages
than LL grammars
10、Types of LR grammars
– LR(0), SLR, LR(1), LALR
– Note:the LR parsing algorithm is the same,but
parsing table is different.
Discussion
• Can we regard a parsing table as a FA.
• What is the FA doing? State? Action?
11、Canonical LR(0)
1）LR(0) item
– An LR(0) item of a grammar G is a
production of G with a dot at some position of
the right side.
• Such as: A  XYZ yields the four items:
– A•XYZ . We hope to see a string
derivable from XYZ next on the input.
– AX•YZ . We have just seen on the
input a string derivable from X and that
we hope next to see a string derivable
from YZ next on the input.
– AXY•Z
– AX YZ•
• The production A generates only one
item, A•.
• Each of this item is a viable prefixes
2) Construct the canonical LR(0) collection
(1)Define a augmented grammar
• If G is a grammar with start symbol S,the
augmented grammar G` is G with a new
start symbol S`, and production S` S
• The purpose of the augmented grammar is
to indicate to the parser when it should stop
parsing and announce acceptance of the
input.
(2)the Closure Operation
• If I is a set of items for a grammar G, then
closure(I) is the set of items constructed
from I by the two rules:
– Initially, every item in I is added to closure(I).
– If A•B is in CLOSURE(I), and B is a
production, then add the item B• to CLOSURE(I);
Apply this rule until no more new items can be added
to CLOSURE(I).
(3)the Goto Operation
• Form: goto(I, X),I is a set of items and X is
a grammar symbol
• goto(I, X)is defined to be the CLOSURE(J)，
X ( VN VT), J={all items like AX•|
A•XI}。
3)The Sets-of-Items Construction
void ITEMSETS-LR0()
{ C:={CLOSURE(S` •S)} /*initial*/
do
{ for (each set of items I in C and each
grammar symbol X )
IF (Goto(I,X) is not empty and not in C)
{add Goto(I,X) to C}
}while C is still extending
}
e.g. construct the canonical collection of sets of LR(0)
items for the following augmented grammar.
S` E E aA|bB A cA|d
B cB|d
Answer:1、the items are：
1. S` •E
2. S` E•
3. E  •aA
4. E  a•A
5. E  aA•
6. A  •cA
7. A  c•A
8. A  cA • 9. A  •d
10. A  d•
11. E  •bB 12. E  b•B
13. E  bB•
14. B  •cB 15. B  c•B
16.B  cB•
17. B  •d
18. B  d•
c
c
2:Ea•A
A •cA
A •dc
a
0: S`•E
E •aA
E •bB
4:Ac•A
A •cA
A •d
E
b
d
d
A
8:Ac A •
10:A d •
6:EaA •
1: S` E •
3: Eb•B
B •cB
B •d
B
5: Bc•B
B •cB
B •d
d
11:B d •
B
9:BcB •
c
c
A
7:EbB•
d
12、SLR Parsing Table Algorithm
– Input. An augmented grammar G`
– Output. The SLR parsing table functions
action and goto for G`
– Method.
– (1) Construct C={I0,I1,…In}, the collection of
sets of LR(0) items for G`.
– (2) State i is constructed from Ii. The parsing
actions for state i are determined as follows:
(a) If [A•a] is in Ii and goto(Ii,a)= Ij,
then set ACTION[i,a]=“Shift j”, here a must
be a terminal.
(b) If [A• ]Ik, then set
ACTION[k,a]=rj for all a in follow(A); here A
may not be S`, and j is the No. of production
A .
– (3) The goto transitions for state I are
constructed for all non terminals A using the
rule: if goto (Ii,A)= Ij, then goto[i,A]=j
– (4) All entries not defined by rules 2 and 3 are
made “error”
– (5) The initial state of the parser is the one
constructed from the set of items containing
[S`  S•].
– If any conflicting actions are generated by the
above rules, we say the grammar is not
SLR(1).
e.g. construct the SLR(1) table for the
following grammar
0. S` E
1. E  E+T
2. E T
3. T T*F
4.T F
5. F (E)
6. F  i
i
I0：S’E
T I2：E T
E E+T
T  T*F
E T
T T*F E I1：S’ E
E E+T
T F
F (E)
(
F i
I4：F’(E)
E E+T
F
i
E T
i
T T*F
I5：F i
T F
F (E)
F
I3：T F
F i
(
T
I2
I5
* I7：T T*F F I10：T T*F 
F (E)
(
I4
F i
*
I9：E E+T 
I
：
E
E+T
+ 6
T
TT  * F
T T*F
(
T F
F (E)
F i
E I8：F  (E)
E E+T
)
F
I3
i
I5
I11：F (E)
state
0
1
2
3
4
5
6
7
8
9
10
11
i
S5
+
S6
r2
r4
ACTION
*
(
S4
)
#
accept
S7
r4
S5
r2
r4
r2
r4
S4
r6
r6
S5
S5
8
r6
r3
r5
S11
r1
r3
r5
2
3
9
3
10
r6
S4
S4
S6
r1
r3
r5
GOTO
E
T
F
1
2
3
r1
r3
r5
E.G.
1. S` S
2. S L=R
3. S R
4. L *R
5. L  i
6. R L
0: S`•S
S •L=R
S •R
L •*R
L •I
R •L
S
L 1: S`S•
2: SL•=R
R L•
R
*
i
3:SR•
4:L*•R
R •L
* L •*R
L •i
7:L*R•
R
L
8:RL•
i
5:Li •
i
6: SL=•R
=
R •L
L •*R
L •i
* L
R
9:SL=R•
state
=
0
1
2
3
4
5
6
7
8
9
ACTION
i
*
S5
S4
S
1
R
3
8
7
8
9
acc
r6
r3
S6/ r6
S5
S4
r5
r5
S5
r4
r6
#
GOTO
L
2
S4
r4
r6
r2
Notes: In the above grammar , the shift/reduce
conflict arises from the fact that the SLR parser
construction method is not powerful enough to
remember enough left context to decide what
action the parser should take on input = having
seen a string reducible to L. That is
“R=“ cannot be a part of any right sentential
form. So when “L” appears on the top of stack
and “=“ is the current character of the input
buffer , we can not reduce “L” into “R”.
• 在SLR方法中，若I中有A,当读头为
aFollow(A)，但是也不一定能够采用A
归约，因为为栈顶时，栈里也可能有
viable prefix “”，而“”作为活前缀未
必允许归约为A，因为可能没有一个句型
含有Aa,
• 例如“R=”不是任何活前缀。
Method-LR(1)
• 每个LR(0)项目添加展望信息：句柄之后可
能跟的k个终结符。
• (A•, a）的含义：预期当栈顶句柄
形成后，在读头下读到a。此时，
在栈内，还未入栈，即它展望了句柄
后的一个符号。
• 若存在规范推导S`A，其中
称规范句型的活前缀(记作)，且
aFirst()，则LR(1)项目(A•,a)对于活
前缀是有效的。注：1)如果bFirst(),即使
bFollow(A),项目( A •,a)也是无效的。
13、LR(1) item
• How to rule out invalid reductions?
– By splitting states when necessary, we can
arrange to have each state of an LR parser
indicate exactly which input symbols can
follow a handle  for which there is a
possible reduction to A.
• Item (A•,a) is an LR(1) item, “1” refers to
the length of the second component, called the
look-ahead of the item.
Note：
1)The look-ahead has no effect in an item of the
form (A•,a), where  is not ,but an item of
the form (A•,a) calls for a reduction by
A only if the next input symbol is a.
2)The set of such a’s will always be a proper
subset of FOLLOW(A). Why?
14、Valid LR(1) item
Formally, we say LR(1) item (A•,a) is
valid for a viable prefix  if there is a derivation
S`A, where
– = ,and
– Either a is the first symbol of , or  is  and
a is #.
15、Construction of the sets of LR(1) items
– Input. An augmented grammar G`
– Output. The sets of LR(1) items that are the
set of items valid for one or more viable
prefixes of G`.
– Method. The procedures closure and goto and
the main routine items for constructing the
sets of items.
function closure(I);
{ do { for (each item (A•B,a) in I,
each production B   in G`,
and each terminal b in FIRST(a)
such that (B•  ,b) is not in I )
add (B•  ,b) to I;
}while there is still new items add to I;
return I
}
function goto(I, X);
{ let J be the set of items (AX•,a) such
that (A• X ,a) is in I ;
return closure(J)
}
Void items (G`);
{C={closure({ (S`•S,#)})};
do { for (each set of items I in C and each
grammar symbol X
such that
goto(I, X) is not empty and not in C )
add goto(I, X) to C
} while there is still new items add to C;
}
e.g.compute the items for the following
grammar:
1. S` S
2. S CC
3. C cC|d
Answer: the initial set of items is I0：
I0
S` •S,#
S•CC,#
C•cC, c|d
C•d,c|d
Now we compute goto(I0,X) for the various values of
X. And then get the goto graph for the grammar.
I0: S' -> •S, #
I6: C -> c•C, #
S -> •CC, #
C -> •cC, #
C -> •cC, c/d
C -> •d, #
C -> •d, c/d
I1: S' -> S•, #
I8: C -> cC•, c/d
I2: S -> C•C, #
C -> •cC, #
C -> •d, #
I3: C -> c•C, c/d
C -> •cC, c/d
C -> •d, c/d
I5: S -> CC•, #
I7: C -> d•, #
I9: C -> cC•, #
I4: C -> d•, c/d
s
C
C
c
c
C
d
c
d
c
d
C
d
16、Construction of the canonical LR parsing table
– Input. An augmented grammar G`
– Output. The canonical LR parsing table
functions action and goto for G`
– Method.
(1) Construct C={I0,I1,…In}, the collection of
sets of LR(1) items for G`.
(2) State i is constructed from Ii. The parsing
actions for state i are determined as follows:
a) If [A•a,b] is in Ii and goto(Ii,a)= Ij, then
set ACTION[i,a]=“Shift j”, here a must be a
terminal.
b) If [A• ,a]Ii, A!=S`,then set
ACTION[i,a]=rj; j is the No. of production
A .
c) If [S`•S,#]is in Ii, then set ACTION[i,#] to
“accept”
(3) The goto transitions for state i are determined as
follows: if goto (Ii,A)= Ij, then goto[i,A]=j.
(4) All entries not defined by rules 2 and 3 are made
“error”
(5) The initial state of the parser is the one constructed
from the set of items containing [S`•S,#].
– If any conflicting actions are generated by the above
rules, we say the grammar is not LR(1).
E.g.construct the canonical parsing table
for the following grammar:
1. S` S
2. S CC
3. C cC
4. C d
state
0
1
2
3
4
5
6
7
8
9
c
S3
Action
d
S4
goto
#
S
1
C
2
acc
S6
S3
r3
S7
S4
r3
5
8
r1
S6
S7
9
r3
r2
r2
r2
Notes:
1)Every SLR(1) grammar is an LR(1) grammar
2)The canonical LR parser may have more
states than the SLR parser for the same
grammar.
17、LALR(lookahead-LR)
1)Basic idea
Merge the set of LR(1) items having the same
core
(1)When merging, the GOTO sub-table can be merged
without any conflict, because GOTO function just
relies on the core
(2) When merging, the ACTION sub-table can also be
merged without any conflicts, but it may occur the case
of merging of error and shift/reduce actions. We
assume non-error actions
(3)After the set of LR(1) items are merged, an error may
be caught lately, but the error will eventually be caught,
in fact, it will be caught before any more input symbols
are shifted.
(4)After merging, the conflict of reduce/reduce
may be occurred.
2)The sets of LR(1) items having the same core
– The states which have the same items but the
look-ahead symbols are different, then the
states are having the same core.
Notes: We may merge these sets with common
cores into one set of items.
18、An easy, but space-consuming LALR table construction
• Input. An augmented grammar G`
• Output. The LALR parsing table functions action and goto
for G`
• Method.
– (1) Construct C={I0,I1,…In}, the collection of sets of
LR(1) items.
– (2) For each core present among the set of LR(1) items,
find all sets having that core, and replace these sets by
their union.
– (3) Let C`={J0,J1,…Jm}be the resulting sets of
LR(1) items. The parsing actions for state I
are constructed from Ji. If there is a parsing
action conflict, the algorithm fails to produce
a parser, and the grammar is not a LALR.
– (4) The goto table is constructed as follows.
– If J is the union of one or more sets of LR(1)
items, that is , J= I1I2  …  Ik then the
cores of goto(I1,X), goto(I2,X),…,
goto(Ik,X)are the same, since I1,I2,…In all
have the same core. Let K be the union of all
sets of items having the same core as goto
(I1,X). then goto(J,X)=k.
If there is no parsing action conflicts , the given
grammar is said to be an LALR(1) grammar
sta
te
0
1
2
3
4
5
6
7
8
9
Action
goto
c d # S C
S3 S4
1 2
acc
S6 S7
5
S3 S4
8
r3 r3
r1
S6 S7
9
r3
r2 r2
r2
Parsing string ccd
4. 6 Using ambiguous grammars
1、Using Precedence and Associativity to
Resolve Parsing Action Conflicts
Grammar: EE+E|E*E|(E)|i
E E+T|T
T T*F|F
F (E)|i
i+i+i*i+i
With LR idea，according other conditions，
analyze ambiguity Grammar。Steps：
1、Construct LR(0) parsing table；
2、if Conflicts happens, solve them with SLR ；
3、The rest conflicts are solved by other
conditions
E.g：
E` E
E E+E|E*E|(E)|I
1) LR(0) Parsing Table
2) SLR
E.G
I1: E` E•
E E•+E
E E•*i
Re－Shift conflict
3)Other conflicts
E.g：I7， E` E＋E•
E E•+E
E E•*E
Re－Shift conflict
state
0
1
2
3
4
5
6
7
8
9
i
S3
＋
S4
S3
r4
S3
S3
S4
r1/S4
r2/S4
r3
ACTION
*
（
S2
S5
S2
r4
S2
S2
S5
S5/r1
r2/S5
r3
）
#
GOTO
S
1
acc
6
r4
r4
7
8
S9
r1
r2
r3
r1
r2
r3
For ACTION[7, *]，
reduction or shift?
“Shift” because “*” is superior
For ACTION[7, +]

reduction or shift?
“Shift” because the left “*” is superior
状态
0
1
2
3
4
5
6
7
8
9
i
S3
S3
S3
S3
ACTION
*
＋
（
S2
S4
S5
S2
r4
r4
S2
S2
S4
S5
r1(S4) S5 (r1)
r2(S4) r2(S5)
r3
r3
）
#
GOTO
S
1
acc
6
r4
r4
7
8
S9
r1
r2
r3
r1
r2
r3
2、The “Dangling-else” Ambiguity
Grammar:
S’S
S if expr then stmt else stmt
|if expr then stmt
|other
S’S
S iSeS|iS|a
state
0
1
2
3
4
5
6
ACTION
i
S2
e
a
S3
#
GOTO
S
1
acc
S2
r4
r3
S2
r1
r4
S5/r3
r1
S3
r4
r3
S3
r1
4
r4
6
state
0
1
2
3
4
5
6
ACTION
i
S2
e
a
S3
#
GOTO
S
1
acc
S2
r4
r3
S2
r1
r4
S5/r3
r1
S3
r4
r3
S3
r1
4
r4
6
4. 7 Parser Generator Yacc
1、Creating an input/output translator with Yacc
Yacc
specification
translate.y
y.tab.c
Yacc
y.tab.c
Compiler
C
a.out
Compiler
input
a.out
output
2、Three parts of a Yacc source program
declaration
%%
translation rules
%%
supporting C-routines
Notes: The form of a translation rule is as
followings:
<Left side>: <alt> {semantic action}
Syntax Analysis
Context-Free
Grammar
Push-down
Automation
Specification
Tool
Top-down
DerivationMatching
Recursivedescent
Table-driven
Top-down,
Skill
Bottom-UP
Methods
Bottom-Up
Shift-Reducing
Predictive
Precedence
First,Follow
FIRSTVT
LASTVT
LR Parsing
Layered
Automation
SLR(1)
LR(1)
LALR(1)
Recursive Descent Analyses
Advantages: Easy to write programs
Disadvantages: Backtracking, poor efficiency
a
Skills : First, Follow
Disadvantages: More preprocesses(Elimination of left
recursions , Extracting
maximum common left factors)
A
……….
Predictive Analyses : predict the
production which is used when a
non-terminated occurs on top of
the analyses stack
Controller
LL(1) Parse
Table
First() A
Follow(A) A
Bottom-up ---Operator Precedence Analyses
Skills : Shift– Reduce ,
FIRSTVT, LASTVT
Disadvantages: Strict grammar limitation, poor reduce
mechanism
b
Simple LR Analyses : based on
determined FA, state stack and
symbol stack (two stacks)

E
a
Controller
Skills : LR item and Follow(A)
….
Disadvantages: cannot solve the
problems of shift-reduce conflict
and reduce-reduce conflict
OP Parse
Table
FIRSTVT() A 
LR(1) analyses
LASTVT() A 
SLR(1) Parser:
b
a

i
….
#
0
symbol state
Controller
SLR(1)
Parse Table
LR items (Shift items, Reducible items)
LR item –extension (AB)
(B)
Follow(A) A 
Canonical LR Analyses(LR(1))
Skills : LR(1) item and Look-ahead symbol
Disadvantages: more states
LALR(1)
Skills : Merge states with the same core
Disadvantages: maybe cause reduce-reduce
conflict
LR(1) Parser:
b
a

i
….
#
0
symbol state
Controller
LR(1) Parse
Table
LR items (Shift items, Reducible items)
LR item –extension (AB,a)
(B,first(a) )
Generation of Parse Tree
E.g. construct the parse tree for the string
“i+i*i” under SLR(1) of the following
grammar
0. S` E
1. E  E+T
2. E T
3. T T*F
4.T F
5. F (E)
6. F  i
state
0
1
2
3
4
5
6
7
8
9
10
11
i
S5
+
S6
r2
r4
ACTION
*
(
S4
)
#
accept
S7
r4
S5
r2
r4
r2
r4
S4
r6
r6
S5
S5
8
r6
r3
r5
S11
r1
r3
r5
2
3
9
3
10
r6
S4
S4
S6
r1
r3
r5
GOTO
E
T
F
1
2
3
r1
r3
r5
E
E
T
T
T
F
F
F
i + i * i
Exercises
• Constructing the related LL(1) parsing table.
Pb S d
SS ; A|A
AB|C
Ba
CD|D e A
DE B
Ei F t
Fb
• Please show that the following operator
grammar is whether an operator precedence
grammar by constructing the related parsing
table.
SS ; G|G
GG(T)|H
Ha|(S)
TT+S|S
• Please construct a LR(1) parsing table for the
following two ambiguous grammar with the
additional conditions:.
Sif S else S|if S|S;S|a that else dangles with the
closest previous unmatched if , ; has the property of
left associative
CC and C|C or C|b that or has higher precedence
than that of and, and has the property of right
associative, or has the property of right associative.