LL(k) Parsing Compiler Baojian Hua bjhua@ustc.edu.cn.

transcript

LL(k) Parsing

CompilerBaojian Hua

bjhua@ustc.edu.cn

Front End

source code

abstract syntax

lexical analyzer

parser

tokens

IRsemantic analyzer

Parsing

The parser translates the source program into abstract syntax trees Token sequence:

from the lexer abstract syntax trees:

compiler internal data structures for programs check (syntactic) validity of programs

Must take account the program syntax

Conceptually

token sequence

abstract

syntax treeparser

language syntax

Syntax: Context-free Grammar

Context-free grammars are (often) given by BNF expressions (Backus-Naur Form) read dragon-book sec 2.2

More powerful than RE in theory Good for defining language syntax

Context-free Grammar (CFG)

A CFG consists of 4 components: a set of terminals (tokens): T a set of nonterminals: N a set of production rules: P

s -> t1 t2 … tn with sN, and t1, …, tn (T ∪N)

a unique start nonterminal: S

Example// SLP as in Tiger book chap. 1 (simplified):

N = {S, E}T = {SEMICOLON, ID, IF, ASSIGN, …}S = S

Derivation

A derivation: Starts with the unique start nonterminal S repeatedly replacing a right-hand nonter

minal s by the body of a production rule of the nonterminal s

stop when right-hand are all terminals The final string consists of terminals o

nly and is called a sentence (program)

ExampleS -> S ; S | id := E | print (E)E -> …

x := 5;print (x)

derive me

S -> … (a choice)

Example

x := 5;print (x)

derive meS -> S ; S -> x := E ; S -> x := 5 ; S -> x := 5 ; print (E) -> x := 5 ; print (x)

S -> S ; S | id := E | print (E)E -> id | num | E + E | E * E

Another Try to Derive the same Program

x := 5;print (x)

derive me

S -> x := E -> x := 5 -> // stuck! :-(

S -> S ; S | id := E | print (E)E -> …

Derivation For same string, there may exist

many different derivations left-most derivation right-most derivation

Parsing is the problem of taking a string of terminals and figure out whether it could be derived from a CFG error-detection

Parse Trees Derivation can also be represented as trees

useful to understand AST (discussed later) Idea:

each internal node is labeled with a nonterminal

each leaf node is labeled with a terminal each use of a rule in a derivation explains how t

o generate children in the parse tree from the parent

ExampleS -> S ; S

x := 5;print (x)

derive me

x := E

print E

Parse Tree has Meanings:post-order traversalS -> S ; S

x := 5;print (x)

derive me

x := E

print E

Ambiguous Grammars

A grammar is ambiguous if the same sequence of tokens can give rise to two or more different parse trees

ExampleE -> num | id | E + E | E * E

derive me

E -> E + E -> 3 + E -> 3 + E * E -> 3 + 4 * E -> 3 + 4 * 5E -> E * E -> E + E * E -> 3 + E * E -> 3 + 4 * E -> 3 + 4 * 5

3 E * E

5E + E

Ambiguous Grammars Problem: compilers make use of parse trees

to interpret the meaning of parsed programs different parse trees have different meanings eg: 4 + 5 * 6 is not (4 + 5) * 6 languages with ambiguous grammars are DISAST

ROUS; the meaning of programs isn’t well-defined! You can’t tell what your program might do!

Solution: rewrite grammar to equivalent forms

Eliminating ambiguity In programming language syntax, am

biguity often arises from missing operator precedence or associativity * is of high precedence than + both + and * are left-associative Why or why not?

Rewrite grammar to take account of this

E -> E + T | TT -> T * F | FF -> num | id

Q: is the right grammar ambiguous? Why or why not?

Parser A program to check whether a program is d

erivable from a given grammar expensive in general must be fast

to compile a 2000k lines of kernel even for small application code, speed may be a conce

rn Theorists have developed specialized kind

of grammar which may be parsed efficiently LL(k) and LR(k)

Recursive Decedent Parsing

Predictive parsing A.K.A: Recursive descent parsing, top-down

parsing simple to code by hand efficient can parse a large set of grammar your Tiger compiler will use this

Key idea: one (recursive) function for each nonterminal one clause for each right-hand production rule

Connecting with the lexer(* step #1: represent tokens *)token = ID | IF | NUM | ASSIGN | SEMICOLON | LPAREN | RPAREN | …(* step #2: connect with lexer *)token current_token; /* external var */

void eat (token t) = if (current_token = t) current_token = Lex_nextToken (); else error (“want “, t, “but got”, current_token)

(* step #1: cook a lexer, including tokens *)struct token current_token = lex ();(* step #2: build the parser *)void parse_stm () = switch (current_token) case ID => eat (ID); eat (ASSIGN); parse_exp (); case PRINT => eat (PRINT); eat (LPAREN); parse_exp (); eat (RPAREN); default =>error(“want ID, PRINT”);

void parse_exp () = switch (current_token) case ID: ??? case NUM: ??? // backtracking!!

parse_stm()parse_exp()

How to handle precedence?void parse_stm_all ()

parse_stm();

while (current_token == “;”)

eat (;);

parse_stm ();

void parse_exp_plus ()

parse_exp_times();

while (current_token == “+”)

eat (+);

parse_exp_times();

stm ; stm ; stm ; stm

2+3*4+5*7

Generally: if there are n level of precedence, one may write n parsing functions.

Moral The key point in predicative parsing is

to determine the production rule to use (recursive function to call) must know the “start” symbols of each rule “start” symbol must not overlap e.g.:

exp -> NUM | ID

This motivates the idea of first and follow sets

First & Follow

S -> w1

-> …

For nonterminal S, and current input token t if wk starts with t, then choose wk,

or if wk derives empty string, and the

string follow S starts with t First symbol sets of wi (1<=i<=n)

don’t overlap to avoid backtracking

Nullable, First and Follow sets To use predicative parsing, we must compu

te: Nullable: nonterminals that derive empty string First(ω) : set of terminals that can begin any stri

ng derivable from ω Follow(X): set of terminals that can immediately

follow any string derivable from nonterminal X Read tiger sec 3.2

Fixpoint algorithms

Nullable, First and Follow sets Which symbol X, Y and Z

can derive empty string? What terminals may the

string derived from X, Y and Z begin with?

What terminals may follow X, Y and Z?

Z -> d

-> X Y Z

Y -> c

X -> Y

Nullable If X can derive an empty string, iff:

base case: X ->

inductive case: X -> Y1 … Yn

Y1, …, Yn are n nonterminals and may all derive empty strings

Computing Nullable/* Nullable: a set of nonterminals */Nullable <- {};while (Nullable still changes) for (each production X -> α) switch (α) case : Nullable = {X};∪ break; case Y1 … Yn: if (Y1Nullable && … && YnNullable) Nullable = {X};∪ break;

Example: NullablesZ -> d

-> X Y Z

Y -> c

X -> Y

Round 0 1 2

nullable {}

-> X Y Z

Y -> c

X -> Y

Round 0 1 2

nullable {} {Y, X}

-> X Y Z

Y -> c

X -> Y

Φ {} {Y, X} {Y, X}

First(X) Set of terminals that X begins with:

X => a … Rules

base case: X -> a

First (X) ∪= {a} inductive case:

X -> Y1 Y2 … Yn First (X) ∪= First(Y1) if Y1Nullable, First (X) ∪= First(Y2) if Y1,Y2 Nullable, First (X) ∪= First(Y3) …

Computing First// Suppose Nullable set has been computedforeach (nonterminal X) First(X) <- {}; while (some First set still changes) for (each production X -> α) switch (α) case a: First(X) = {a};∪ break; case Y1 … Yn: First(X) = First(Y1);∪ if (Y1 \not\in Nullable) break; First(X) = First(Y2);∪ …; // Similar as above

Example: FirstZ -> d

-> X Y Z

Y -> c

X -> Y

0 1 2 3

First(Z)

First(Y)

First(X)

Nullable = {X, Y}

-> X Y Z

Y -> c

X -> Y

0 1 2 3

First(Z)

{} {d}

First(Y)

{} {c}

First(X)

{} {c, a}

Nullable = {X, Y}

-> X Y Z

Y -> c

X -> Y

0 1 2 3

First(Z)

{} {d} {d, c, a}

First(Y)

{} {c} {c}

First(X)

{} {c, a} {c, a}

Nullable = {X, Y}

-> X Y Z

Y -> c

X -> Y

0 1 2 3

First(Z)

{} {d} {d, c, a}

{d, c, a}

First(Y)

{} {c} {c} {c}

First(X)

{} {c, a} {c, a} {c, a}

Nullable = {X, Y}

Parsing with FirstZ -> d {d}

-> X Y Z {a, c, d}

Y -> c {c}

X -> Y {c}

-> a {a}

First(Z)

{d, c, a}

First(Y)

First(X)

{c, a}Nullable = {X, Y}

Now consider this string: d

Suppose we choose the production: Z -> X Y Z

But we get stuck at:X -> Y -> aBut neither can accept d!What’s the problem?

Follow(X) Set of terminals that may follow X:

S => … X a … Rules:

Base case: Follow (X) = {}

inductive case: Y -> ω1 X ω2

Follow(X) ∪= Fisrt(ω2) if ω2 is Nullable, Follow(X) ∪= Follow(Y)

Computing Follow(X)foreach (nonterminal X) Follow(X) <- {};while (some Follow still changes) { for (each production Y -> ω1 X ω2 ) Follow(X) = First (∪ ω2); if (ω2 is Nullable) Follow(X) = Follow (Y);∪

Example: FollowZ -> d

-> X Y Z

Y -> c

X -> Y

Round 0 1 2 3

First(Z)Follow(Z)

{d, c, a}{}

First(Y)Follow(Y)

First(X)Follow(X)

{c, a}{}

Nullable = {X, Y}

-> X Y Z

Y -> c

X -> Y

Round 0 1 2 3

First(Z)Follow(Z)

{d, c, a}{}

First(Y)Follow(Y)

{c}{} {d, c,

First(X)Follow(X)

{c, a}{} {d, c,

Nullable = {X, Y}

-> X Y Z

Y -> c

X -> Y

Round 0 1 2 3

First(Z)Follow(Z)

{d, c, a}{}

{$} {$}

First(Y)Follow(Y)

{c}{} {d, c,

a}{d, c, a}

First(X)Follow(X)

{c, a}{} {d, c,

a}{d, c, a}

Nullable = {X, Y}

Predicative Parsing Table

With Nullables, First(), and Follow(), we can make a parsing table P(N,T) each entry contains a set of productions

t1 t2 t3 t4 … $(EOF)

Predicative Parsing Table

For each rule X -> ω for each aFirst(ω), add X -> ω to P(X, a) if X is nullable, add X -> ω to P(X, b) for ea

ch b Follow (X) all other entries are “error”

t1 t2 t3 t4 … $(EOF)

Example: Predicative Parsing Table

First(X)Follow(X)

{c, a}{c, d, a}

First(Y)Follow(Y)

{c}{c, d, a}

First(Z)Follow(Z)

{d, c, a}{$}

Z -> d

-> X Y Z

Y -> c

X -> Y

Nullable = {X, Y}

Z Z->X Y Z Z->X Y Z Z->dZ->X Y Z

Y Y-> Y->cY->

X X->YX->a

X->Y X->Y

Example: Predicative Parsing Table

First(X)Follow(X)

{c, a}{c, d, a}

First(Y)Follow(Y)

{c}{c, d, a}

First(Z)Follow(Z)

{d, c, a}{$}

Z -> d

-> X Y Z

Y -> c

X -> Y

Nullable = {X, Y}

Z Z->X Y Z Z->X Y Z Z->dZ->X Y Z

Y Y-> Y->cY->

X X->YX->a

X->Y X->Y

LL(1) A context-free grammar is called LL(1) if it can be parsed this way: Left-to-right parsing Leftmost derivation 1 token lookahead

This means that in the predicative parsing table, there is at most one production in every entry

Speeding up set Construction

All these sets (Nullable, First, Follow) can be computed simultaneously see Tiger book algorithm 3.13

Order the computation: What’s the optimal order to compute th

ese set?

Example: Speeding up set Construction

Z -> d

-> X Y Z

Y -> c

X -> Y

0 1 2 3

First(Z)

First(Y)

First(X)

Nullable = {X, Y}

Q1: What’s reasonable order here?

Q2: How to set this order?

Directed Graph ModelZ -> d

-> X Y Z

Y -> c

X -> Y

Nullable = {X, Y}

Q1: What’s reasonable order here?

Q2: How to set this order?

{c, a}

{d, c, a}

Order: Y X Z

Reverse Quasi-Topological Sort Quasi-topological sort the directed graph

Quasi: topo-sort general directed graph is impossible

also known as reverse depth-first ordering Reverse: information (here: First) flows fro

m successors backward to predecessors Refer to your favorite algorithm book

Problem

LL(1) can only be used with grammars in which every production rules for a nonterminal start with different terminals

Unfortunately, many grammars don’t have this perfect property

Exampleexp -> NUM -> ID -> exp + exp -> exp * exp

exp -> exp + term -> termterm -> term * factor -> factorfactor -> NUM -> ID

Q: is the right grammar LL(1)? Why or why not?

Solutions

Left-recursion elimination Left-factoring Read:

tiger section 3.2

Example for SLPstm -> id := exp A -> print(exp) AA -> ; stm A ->

Q1: is the right grammar LL(1)?

Q2: are these two grammars equivalent?

stm -> stm ; stm -> id := exp -> print (exp)

LL(k) LL(1) can be further generalized to LL

(k): Left-to-right parsing Leftmost derivation k token lookahead

Q: table size? other problems with this approach?

Summary Context-free grammar is a math tool for spe

cifying language syntax among others…

Writing parsers for general grammar is hard and costly LL(k) and LR(k)

LL(1) grammars can be implemented efficiently table-driven algorithms (again!)

LL(k) Parsing Compiler Baojian Hua bjhua@ustc.edu.cn.

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