Parsing Compiler Baojian Hua [email protected]. Front End source code abstract syntax tree lexical analyzer parser tokens IR semantic analyzer.

Parsing

CompilerBaojian Hua

[email protected]

Front End

source code

abstract syntax

tree

lexical analyzer

parser

tokens

IRsemantic analyzer

Parsing The parser translates the source progr

am into abstract syntax trees Token sequence:

from the lexer abstract syntax trees:

check validity of programs cook compiler internal data structures for pro

grams Must take account the program syntax

Conceptually

token sequence

abstract

syntax tree

parser

language syntax

Syntax: Context-free Grammar

Context-free grammars are (often) given by BNF expressions (Backus-Naur Form) read Dragon 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// Recall the min-ML language in “code3” // (simplified)N = {decs, dec, exp}T = {SEMICOLON, VAL, ID, ASSIGN, NUM}S = decsdecs -> dec SEMICOLON decs |dec -> VAL ID ASSIGN expexp -> ID | NUM

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)

Exampledecs -> dec SEMICOLON decs |dec -> VAL ID ASSIGN expexp -> ID | NUM

val x = 5;val y = x;

derive me

decs -> … (a choice)

Exampledecs -> dec SEMICOLON decs |dec -> VAL ID ASSIGN expexp -> ID | NUM

val x = 5;val y = x;

derive me

decs -> dec SEMICOLON decs -> VAL ID ASSIGN exp SEMICOLON decs -> VAL ID ASSIGN NUM SEMICOLON decs -> VAL ID ASSIGN NUM SEMICOLON dec SEMICOLON decs -> … -> VAL ID ASSIGN NUM SEMICOLON VAL ID ASSIGN ID SEMICOLON decs

Another Way to Derive the same Programdecs -> dec SEMICOLON decs |dec -> VAL ID ASSIGN expexp -> ID | NUM

val x = 5;val y = x;

derive me

decs -> dec SEMICOLON decs -> dec SEMICOLON dec SEMICOLON decs -> …

Derivation For same string, there may exist

many 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 non-

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

how to generate children in the parse tree from the parents

Exampledecs -> dec SEMICOLON decs |dec -> VAL ID ASSIGN expexp -> ID | NUM

val x = 5;val y = x;

derive me

decs

dec SEMI decs

VAL ID = exp

5

dec SEMI decs

similar case

Different Derivations, same Tree

decs -> dec SEMICOLON decs -> VAL ID ASSIGN exp SEMICOLON decs -> …

decs -> dec SEMICOLON decs -> dec SEMICOLON dec SEMICOLON decs -> …

val x = 5;val y = x;

derive me

decs

dec SEMI decs

VAL ID = exp

5

dec SEMI decs

similar case

Parse Tree has Meanings:post-order traversal

decs -> dec SEMICOLON decs -> VAL ID ASSIGN exp SEMICOLON decs -> …

decs -> dec SEMICOLON decs -> dec SEMICOLON dec SEMICOLON decs -> …

val x = 5;val y = x;

derive me

decs

dec SEMI decs

VAL ID = exp

5

dec SEMI decs

similar case

Ambiguous Grammars

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

Exampleexp -> num -> id -> exp + exp -> exp * exp

3+4*5

derive me

exp -> exp + exp -> 3 + exp -> 3 + exp * exp -> 3 + 4 * exp -> 3 + 4 * 5exp -> exp * exp -> exp + exp * exp -> 3 + exp * exp -> 3 + 4 * exp -> 3 + 4 * 5

Exampleexp -> num -> id -> exp + exp -> exp * exp

exp -> exp + exp -> 3 + exp -> 3 + exp * exp -> 3 + 4 * exp -> 3 + 4 * 5exp -> exp * exp -> exp + exp * exp -> 3 + exp * exp -> 3 + 4 * exp -> 3 + 4 * 5

exp

exp + exp

3 exp * exp

54

exp

exp * exp

5exp + exp

43

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

Exampleexp -> num -> id -> exp + exp -> exp * exp

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

Theorists have developed specialized kind of grammar which may be parsed efficiently LL(k) and LR(k)

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

parsing simple to code by hand efficient can parse a large set of grammar

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

Exampledecs -> dec SEMICOLON decs |dec -> VAL ID ASSIGN expexp -> ID | NUM

(* step #1: represent tokens *)datatype token = Val | Id of string | Num of int | Assign | Semicolon | Eof(* step #2: connect with lexer *)token current = ref getToken (); fun advance () = current := getToken ();fun eat (token t) = if !current = t then advance () else error (“want “, t, “but got “, !current)

decs -> dec SEMICOLON decs |dec -> VAL ID ASSIGN expexp -> ID | NUM

(* step #1: represent tokens *)datatype token = Val | Id of string | Num of int | Assign | Semi | Eof(* step #2: connect with lexer *)token current = ref getToken (); fun advance () = current := getToken ();fun eat (token t) = …;(* step #3: build the parser *)fun parseDecs() = case !current of VAL => parseDec (); eat (Semi); parseDecs (); | EOF => () | _ => error (“want VAL or EOF”)fun parseDec () = …fun parseExp () = …

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 ex: exp -> NUM | ID

This motivates the idea of first and follow sets

MoralS -> w1

-> w2

-> …

-> wn

Current nonterminal is S, and the current input token is t if wk starts with t, then choos

e wk, or if wk derives empty string, an

d 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 Dragon sec 4.4.2 and 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

-> a

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 NullableNullable <- {};while (Φ still change) 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

-> a

Round

0 1 2

Φ {}

Example: NullablesZ -> d

-> X Y Z

Y -> c

->

X -> Y

-> a

Round

0 1 2

Φ {} {Y, X}

Example: NullablesZ -> d

-> X Y Z

Y -> c

->

X -> Y

-> a

Round

0 1 2

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 has been computedFirst(X) <- {}; // for each Xwhile (First still change) 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(Y1);∪ …; // Similar as above

Example: FirstZ -> d

-> X Y Z

Y -> c

->

X -> Y

-> a

Round

0 1 2 3

First(Z)

{}

First(Y)

{}

First(X)

{}

Nullable = {X, Y}

Example: FirstZ -> d

-> X Y Z

Y -> c

->

X -> Y

-> a

Round

0 1 2 3

First(Z)

{} {d}

First(Y)

{} {c}

First(X)

{} {c, a}

Nullable = {X, Y}

Example: FirstZ -> d

-> X Y Z

Y -> c

->

X -> Y

-> a

Round

0 1 2 3

First(Z)

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

First(Y)

{} {c} {c}

First(X)

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

Nullable = {X, Y}

Example: FirstZ -> d

-> X Y Z

Y -> c

->

X -> Y

-> a

Round

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)

{c}

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 -> aneither can accept d!

Why?

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)Follow(X) <- {};while (Follow still change) { 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

-> a

Round 0 1 2 3

First(Z)Follow(Z)

{d, c, a}{}

First(Y)Follow(Y)

{c}{}

First(X)Follow(X)

{c, a}{}

Nullable = {X, Y}

Example: FollowZ -> d

-> X Y Z

Y -> c

->

X -> Y

-> a

Round 0 1 2 3

First(Z)Follow(Z)

{d, c, a}{}

{$}

First(Y)Follow(Y)

{c}{} {d, c,

a}

First(X)Follow(X)

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

a}

Nullable = {X, Y}

Example: FollowZ -> d

-> X Y Z

Y -> c

->

X -> Y

-> a

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)

N1 ri

N2 rk

N3 rj

…

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)

N1 r1

N2 rk

N3 ri

…

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

-> a

Nullable = {X, Y}

a c d

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

Y Y-> Y->cY->

Y->

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

-> a

Nullable = {X, Y}

a c d

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

Y Y-> Y->cY->

Y->

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 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

-> a

Round

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

-> a

Nullable = {X, Y}

Q1: What’s reasonable order here?

Q2: How to set this order?

Z

X

Y{c}

{c, a}

{d, c, a}

Order: Y X Z

Reverse Topological Sort Quasi-topological sort the directed gr

aph Quasi: topo-sort general directed graph i

s impossible also known as reverse depth-first orderin

g Reverse: information (First) flows fro

m successors 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:

dragon sec4.3.2, 4.3.3, 4.3.4 tiger sec3.2

Exampleexp -> term exp’exp’ -> + term exp’ -> term -> factor term’term’-> * factor

term’ -> factor -> num -> id

Q: is the right grammar LL(1)? are those two grammars equivalent?

exp -> exp + term -> termterm -> term * factor -> factorfactor -> num -> id

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 and 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!)