Unit-3 Parsing Theory (Syntax Analyzer) PREPARED BY: PROF. HARISH I RATHOD COMPUTER ENGINEERING...

Unit-3 Parsing Theory (Syntax Analyzer)PREPARED BY:

PROF. HARISH I RATHOD

COMPUTER ENGINEERING DEPARTMENT

GUJARAT POWER ENGINEERING & RESEARCH INSTITUTE

COMPILER DESIGN (170701)

Introduction

• Syntax analysis is the second phase after lexical analysis in compiler design.

• It basically checks the syntax of the language.

• It takes the token from lexical analyzer and groups them in such a way that some programming structure can be recognized.

GPERI – CD - UNIT-3 2

Introduction

• After grouping the tokens if any syntax cannot be recognized then syntactic error will be generated.

• It is a major component of the front end of a compiler.• For the syntactic specification of a programming language, use a

notation called context free grammar.


Role of the parser

• It obtains a string of tokens from the lexical analyzer.

• Group the tokens to identify large structure in the program.

• It should be report any syntax error in the program.

• It should be recover from the error so that it can continue to process the rest of the input.


Role of the parser

• .


Lexical analyzer Parser

Symbol Table

Source Program

Token

getNextToken

Parse Tree

Syntax Error

Context-Free Grammar

• Grammar involves four quantities:• Terminals,• Non-terminals,• A start symbol and• Production.

• One non-terminal is selected as a start symbol.• Each production consist of a non-terminal, followed by an

arrow () or (:=) followed by a string of non-terminals and terminals.



• A context free grammar (CFG) is defined:

• As 4-tuples (VN, ∑, P, S).• Where:• VN = Set of non-terminals• ∑ = Set of terminals.• S = A start symbol.• P = Set of production rules.

• One non-terminal finite string of terminals and/or non-terminals.



• Example.

stmt if ( expr ) stmt else stmt• Where:• Non-terminals: stmt, expr• Terminals: if, ( , ), else• Start symbol: stmt



• Example.expression -> expression + termexpression -> expression – termexpression -> termterm -> term * factorterm -> term / factorterm -> factor



• Example:factor -> ( expression )factor -> id



• Notational Conventions:• Terminal symbols:• Lower case letters such as a,b,c.• Operator symbols such as +, *, -, / etc.• Punctuation symbols such as parentheses, comma and so on.• The digits 0,1, ….. , 9.• Bold face string such as id or if, each of which represents a single

terminal symbol.



• Notational Conventions:• Non-terminal symbols:• Uppercase letters, such as A, B, C. • The letter S, when it appears, it usually the start symbol.• Lowercase, italic such as expr or stmt.


Derivation

• The construction of parse tree can be precise by taking a derivational view,

• In which each productions are treated as rewriting rules.• Beginning with start symbol, • Each rewriting step replace a non-terminal by the body of one

of its production.• E E + E | E * E | - E | ( E ) | id


Derivation

• list list + digit • list list – digit• list digit• digit 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9


Derivation

list=> list + digit=> list – digit + digit=> digit – digit + digit=> 9 – digit + digit=> 9 – 5 + digit=> 9 – 5 + 2


Derivation

• This is an example leftmost derivation, • because we replaced the leftmost nonterminal (underlined) in

each step. • Likewise, a rightmost derivation replaces the rightmost

nonterminal in each step.


Derivation

• Construct a CFG, for the language L = {w c w : w ϵ (a,b)*}.• Sol,

• G = (VN,∑,P,S)• Here,

• VN = {S}, ∑ = {a,b,c}• Production rule P is defined as


S -> a S aS -> b S bS -> c

Parse Tree

• The string generated by a context free grammar can be represented by a hierarchical structure called tree.

• Such tree representing derivations are called derivation trees or parse tree or syntax tree.


Parse Tree

• Characteristics of parse tree:• The root of the tree is labeled by the start symbol.• Each leaf of the tree is labeled by a terminal (token or ϵ).• Each interior node is labeled by a nonterminal.• If A → X1 X2 … Xn is a production, then node A has immediate

children X1, X2, …, Xn where Xi is a (non)terminal or ε (ε denotes the empty string)


Parse Tree - Example


list

digit

list digit

list

digit

9 - 5 + 2

Exercise

• Write a CGF, which generates strings having equal number of a’s and b’s:

• Sol:

• CGF, G = (VN,∑,P,S) where VN = {S}, ∑ = {a,b}• P is defined as:• S -> aSb • S -> bSa • S -> ^


Exercise

• Construct a CGF for the language L = {anbn : n >= 1}• Sol:

• CGF, G = (VN,∑,P,S) where VN = {S}, ∑ = {a,b}• P is defined as:• S -> aSb • S -> ab


Exercise

• Write a CGF, which generates string of balanced parenthesis.• Sol: Grammar will accept the balanced right and left

parenthesis. e.g. (), ((( ))),

• CGF, G = (VN,∑,P,S) where VN = {S}, ∑ = { ( , )}• P is given by:• S -> SS• S -> (S)• S -> ^


Exercise

• A CGF given by the productions is:• S -> a | a A S• A -> bS• Obtain the derivation tree of the word : a b a a b a a.


Exercise

• Given the grammar G = (VN,∑,P,S) where VN = {E}, S = E ,• ∑ = {id,+,*,c} and P consist of• E -> E + E | E * E | (E) | id• Obtain the derivation tree for id*id + id and (id+id)*id


Ambiguity

• A grammar is said to be ambiguous,• If there exist more than one parse tree for the same sentence.• Example:• S -> aSbS | bSaS | ϵ• For the string “abab” have two different parse tree.


Ambiguity

• A classical example of ambiguous grammar is that of:• if-then-else construct of many programming language.• Most of the language have both if-then and if-then-else versions

of the statement.• The grammar rules for it as follows:• stmt -> if condition then stmt else stmt

| if condition then stmt


Ambiguity

• Consider the following code segment:• If a>b then

if c>d then x=yelse x=z


Ambiguity

• Leftmost derivation


stmt

if condition then stmt else stmt

if condition then stmt

a>b x=z

c>d x=y

Ambiguity

• Rightmost derivation


stmt



a>b

x=zc>d x=y

else stmt

Eliminating Ambiguity

• Ambiguities may be eliminated by rewriting the grammar:• If-then-else grammar may be rewritten as:

stmt -> m_stmt | un_stmtm_stmt -> if condition then m_stmt else m_stmt

| other_stmtunm_stmt -> if condition then stmt

| if condition then m_stmt else unm_stmt



• Another technique is to modify the language a bit.• Many language require that an if should have a matching endif.• Thus the grammar is modified as• stmt -> if condition then stmt else stmt endif

| if condition then stmt endif



• Example: Grammar


E -> IE -> E + EE -> E * EE -> (E)I -> a | b | c

• Ambiguity is due to the precedence of operator, if we correct the precedence then ambiguity may be removed.

• Here two causes of ambiguity:1. The precedence of operator is not

respected.2. The sequence of identical operators can

group either from left or from right.• .


• The unambiguous grammar.


E -> TT -> FF -> IE -> E + TT -> T * FF -> (E)I -> a | b | c


• The solve parse tree for a + b * c


E




E

+ TE




E

+ TE

T




E

+ TE

T

F

I

a




E

+ TE

T

F

I

a

T * F




E

+ TE

T

F

I

a

T * F

F

I

b




E

+ TE

T

F

I

a

T * F

F

I

b

I

c

Left Recursion

• A grammar is left recursive if it has a nonterminal, say A, that has a derivation of Aα from it.

• Presence of left recursion creates difficulties while designing parsers.

• Types of left recursion:• Immediate left recursion• General left recursion


Left Recursion

• Immediate left recursion:• It happen with a nonterminal A having production rule of the

form : A -> Aα • OR• The production is recursive if the leftmost symbol on right side

is the same as non-terminal of the left side, for example:• A -> Aα


Left Recursion

• Immediate left recursion: (Continue..)• It can be eliminated by introducing a new nonterminal symbol,

say A’.• Modify the grammar:

A -> βA’A’ -> αA’ | ϵ


Left Recursion

• Immediate left recursion: (Continue..)• Thus the rule.

A -> Aα1| Aα2|…….| Aαm|β1| β1|…..…| βn

A -> β1A’| β2A’|……| βnA’

A’ -> α1A’| α2A’|……. |αmA’|ϵ


Left Recursion

• Immediate left recursion: (Continue..)• Example.

E -> E + T | TT -> T * F | FF -> (E) | id


E -> TE’E’ -> +TE’ | ϵT -> FT’T’ -> *FT’ | ϵF -> (E) | id

Left Recursion

• General left recursion: (Continue..)• If there may be no immediate left recursion, a number of

production rules may act together to give a general left recursion.

• For example:S -> AaA -> Sb | c


Here, S is left recursive, because:S -> Aa -> Sba

Left Recursion

• Algorithm eliminate left recursion:

1. Arrange non-terminals in some order say A1,A2,….,Am

2. For i = 1 to m do

for j = 1 to i-1 do

for each set of production Ai -> Ajγ and

Aj -> ᵟ1 | ᵟ2 | …….|ᵟk

replace Ai -> Ajγ by Ai -> ᵟ1γ | ᵟ2γ |…..|ᵟkγ

3. Eliminate immediate felt recursion from all production.GPERI – CD - UNIT-3 48

Left Recursion

• Example:S -> AaA -> Sb | c


The order of non-terminals S,A.For i = 1, the rule S -> Aa, no immediate left recursionFor i = 2, A -> Sb | c is modified as, A -> Aab | c, which has immediate left recursion, eliminated by modifying the rule as:A -> cA’A’ -> abA’ | ϵ

Unit-3 Parsing Theory (Syntax Analyzer) PREPARED BY: PROF. HARISH I RATHOD COMPUTER ENGINEERING...

Documents

Transcript of Unit-3 Parsing Theory (Syntax Analyzer) PREPARED BY: PROF. HARISH I RATHOD COMPUTER ENGINEERING...