Scanner wrap-up and Introduction to Parser

Automating Scanner Construction

RENFA (Thompson’s construction) Build an NFA for each term Combine them with -moves

NFA DFA (subset construction) Build the simulation

DFA Minimal DFA (today) Hopcroft’s algorithm

DFA RE (not really part of scanner construction) All pairs, all paths problem Union together paths from s0 to a final state

DFA Minimization

The Big PictureDiscover sets of equivalent statesRepresent each such set with just one

state

DFA Minimization

The Big Picture Discover sets of equivalent states Represent each such set with just one state

Two states are equivalent if and only if: The set of paths leading to them are equivalent , transitions on lead to equivalent states (DFA) -transitions to distinct sets states must be in distinct sets

A partition P of S Each s S is in exactly one set pi P

The algorithm iteratively partitions the DFA’s states

DFA Minimization

Details of the algorithm Group states into maximal size sets, optimistically Iteratively subdivide those sets, as needed States that remain grouped together are equivalent

Initial partition, P0 , has two sets: {F} & {Q-F} (D =(Q,,,q0,F))

Splitting a set (“partitioning a set by a”) Assume qa, & qb s, and (qa,a) = qx, & (qb,a) = qy

If qx & qy are not in the same set, then s must be split

qa has transition on a, qb does not a splits s

One state in the final DFA cannot have two transitions on a

DFA Minimization

The algorithmP { F, {Q-F}}

while ( P is still changing) T { } for each set S P for each partition S by into S1, and S2

T T S1 S2

if T P then P T

Why does this work? Partition P 2Q

Start off with 2 subsets of Q {F} and {Q-F}

While loop takes PiPi+1 by splitting 1 or more sets

Pi+1 is at least one step closer to the partition with |Q| sets

Maximum of |Q | splitsNote that Partitions are never combined Initial partition ensures that final

states are intactThis is a fixed-point

algorithm!

Key Idea: Splitting S around

S

T

R

The algorithm partitions S around

Original set S

Q

S has transitions on to R, Q, & T

Key Idea: Splitting S around

T

R

Original set S

Q

S1

S2

Could we split S2 further?

Yes, but it does not help asymptotically

S2 is everything in S - S1

DFA Minimization

Refining the algorithm As written, it examines every S P on each iteration

This does a lot of unnecessary work Only need to examine S if some T, reachable from S, has split

Reformulate the algorithm using a “worklist” Start worklist with initial partition, F and {Q-F}

When it splits S into S1 and S2, place S2 on worklist

This version looks at each S P many fewer times Well-known, widely used algorithm due to John Hopcroft

Hopcroft's Algorithm

W {F, Q-F}; P {F, Q-F}; // W is the worklist, P the current partition

while ( W is not empty ) do beginselect and remove S from W; // S is a set of states

for each in do begin

let I –1( S ); // I is set of all states that can reach S on

for each R in P such that R Iis not empty

and R is not contained in Ido begin

partition R into R1 and R2 such that R1 R I; R2 R –

R1;

replace R in P with R1 and R2 ;

if R W then replace R with R1 in W and add R2 to W ;

else if || R1 || ≤ || R2 ||

then add add R1 to W ;

else add R2 to W ;

endend

end

A Detailed Example

Remember ( a | b )* abb ?

Applying the subset construction:

Iteration 3 adds nothing to S, so the algorithm halts

a | b

q0 q1 q4 q2 q3

a bb

Iter. State Contains-closure(move(si,a))

-closure(move(si,b))

0 s0 q0, q1 q1, q2 q1

1 s1 q1, q2 q1, q2 q1, q3

s2 q1 q1, q2 q1

2 s3 q1, q3 q1, q2 q1, q4

3 s4 q1, q4 q1, q2 q1

contains q4

(final state)

Our first NFA

A Detailed Example

The DFA for ( a | b )* abb

Not much bigger than the original All transitions are deterministic Use same code skeleton as before

s0

as1

b

s3

bs4

s2

a

b

b

a

a

a

b

a bs0 s1 s2

s1 s1 s3

s2 s1 s2

s3 s1 s4

s4 s1 s2

A Detailed Example

Applying the minimization algorithm to the DFA

Current Partition Worklist s Split on a Split on b

P0 {s4} {s0,s1,s2,s3}{s4}

{s0,s1,s2,s3}{s4}

none{s0, s1, s2}

{s3}

P1 {s4} {s3} {s0,s1,s2}{s0,s1,s2}

{s3}{s3}

none{s0, s2}{s1}

P2 {s4} {s3} {s1} {s0,s2} {s0,s2}{s1} {s1} none none

s0 a s1

b

s3 b s4

s2

a

b

b

a

a

a

b

s0 , s2a s1

b

s3 b s4

b

a

a

a

b

final state

DFA Minimization

What about a ( b | c )* ?

First, the subset construction:

q0 q1 a

q4 q5 b

q6 q7 c

q3 q8 q2 q9

-closure(move(s,*))

NFA states a b c

s0 q0 q1, q2, q3, q4, q6, q9

none none

s1 q1, q2, q3,q4, q6, q9

none q5, q8, q9,q3, q4, q6

q7, q8, q9,q3, q4, q6

s2 q5, q8, q9,q3, q4, q6

none s2 s3

s3 q7, q8, q9,q3, q4, q6

none s2 s3

s3

s2

s0 s1

c

ba

b

b

c

c

Final states

DFA Minimization

Then, apply the minimization algorithm

To produce the minimal DFA

s3

s2

s0 s1

c

ba

b

b

c

c

Split onCurrent Partition a b c

P0 { s1, s2, s3} {s0} none none none

s0 s1

a

b | cMinimizing that DFA produces the one that a human would design!

final states

Limits of Regular Languages

Advantages of Regular Expressions Simple & powerful notation for specifying patterns Automatic construction of fast recognizers Many kinds of syntax can be specified with REs

Example — an expression grammar

Term [a-zA-Z] ([a-zA-z] | [0-9])*

Op + | - | | /

Expr ( Term Op )* Term

Of course, this would generate a DFA …

If REs are so useful …Why not use them for everything?

Limits of Regular Languages

Not all languages are regular

RL’s CFL’s CSL’s

You cannot construct DFA’s to recognize these languages L = { pkqk } (parenthesis languages) L = { wcw r | w *}

Neither of these is a regular language (nor an RE)

But, this is a little subtle. You can construct DFA’s for Strings with alternating 0’s and 1’s

( | 1 ) ( 01 )* ( | 0 ) Strings with and even number of 0’s and 1’s

RE’s can count bounded sets and bounded differences

What can be so hard?

Poor language design can complicate scanning Reserved words are important

if then then then = else; else else = then (PL/I)

Insignificant blanks (Fortran & Algol68)

do 10 i = 1,25

do 10 i = 1.25

String constants with special characters (C, C++, Java, …)

newline, tab, quote, comment delimiters, …

Finite closures (Fortran 66 & Basic) Limited identifier length Adds states to count length

What can be so hard? (Fortran 66/77)

INTEGERFUNCTIONA

PARA METER(A=6,B=2)

IMPLICIT CHARA CTER*(A-B)(A-B)

INTEGER FORMAT(10), IF(10), DO9E1

100 FORMAT(4H)=(3)

200 FORMAT(4 )=(3)

DO9E1=1

DO9E1=1,2

9 IF(X)=1

IF(X)H=1

IF(X)300,200

300 CONTINUE

END

C THIS IS A “COMMENT CARD” $ FILE(1)

END

How does a compiler scan this?

• First pass finds & inserts blanks

• Can add extra words or tags to create a scanable language

• Second pass is normal scanner

Example due to Dr. F.K. Zadeck

Building Faster Scanners from the DFA

Table-driven recognizers waste effort Read (& classify) the next character Find the next state Assign to the state variable Trip through case logic in action() Branch back to the top

We can do better Encode state & actions in the code Do transition tests locally Generate ugly, spaghetti-like code Takes (many) fewer operations per input character

char next character;state s0 ;

call action(state,char);while (char eof) state (state,char); call action(state,char); char next character;

if (state) = final then report acceptance;else report failure;

Building Faster Scanners from the DFA

A direct-coded recognizer for r Digit Digit*

Many fewer operations per character Almost no memory operations Even faster with careful use of fall-through cases

goto s0;s0: word Ø; char next character; if (char = ‘r’) then goto s1; else goto se;s1: word word + char; char next character; if (‘0’ ≤ char ≤ ‘9’) then goto s2; else goto se;

s2: word word + char; char next character; if (‘0’ ≤ char ≤ ‘9’) then goto s2; else if (char = eof) then report success; else goto se;

se: print error message; return failure;

Building Faster Scanners

Hashing keywords versus encoding them directly Some (well-known) compilers recognize keywords as identifiers

and check them in a hash table Encoding keywords in the DFA is a better idea

O(1) cost per transition Avoids hash lookup on each identifier

It is hard to beat a well-implemented DFA scanner

Building Scanners

The point All this technology lets us automate scanner construction Implementer writes down the regular expressions Scanner generator builds NFA, DFA, minimal DFA, and then

writes out the (table-driven or direct-coded) code This reliably produces fast, robust scanners

For most modern language features, this works You should think twice before introducing a feature that defeats a

DFA-based scanner The ones we’ve seen (e.g., insignificant blanks, non-reserved

keywords) have not proven particularly useful or long lasting

Some Points of Disagreement with EAC

Table-driven scanners are not fast EaC doesn’t say they are slow; it says you can do better

Faster code can be generated by embedding scanner in code This was shown for both LR-style parsers and for scanners in

the 1980s

Hashed lookup of keywords is slow EaC doesn’t say it is slow. It says that the effort can be folded

into the scanner so that it has no extra cost. Compilers like GCC use hash lookup. A word must fail in the lookup to be classified as an identifier. With collisions in the table, this can add up. At any rate, the cost is unneeded, since the DFA can do it for O(1) cost per character.

Building Faster Scanners from the DFA

Table-driven recognizers waste a lot of effort Read (& classify) the next character Find the next state Assign to the state variable Trip through case logic in action() Branch back to the top

We can do better Encode state & actions in the code Do transition tests locally Generate ugly, spaghetti-like code Takes (many) fewer operations per input character

char next character;state s0 ;

call action(state,char);while (char eof) state (state,char); call action(state,char); char next character;

if (state) = final then report acceptance;else report failure;

index

index

index

Parsing

The Front End

Parser Checks the stream of words and their parts of speech (produced by

the scanner) for grammatical correctness Determines if the input is syntactically well formed Guides checking at deeper levels than syntax Builds an IR representation of the code

Think of this as the mathematics of diagramming sentences

Sourcecode Scanner

IRParser

Errors

tokens

The Study of Parsing

The process of discovering a derivation for some sentence Need a mathematical model of syntax — a grammar G Need an algorithm for testing membership in L(G) Need to keep in mind that our goal is building parsers, not

studying the mathematics of arbitrary languages

Roadmap1 Context-free grammars and derivations2 Top-down parsing

Hand-coded recursive descent parsers3 Bottom-up parsing

Generated LR(1) parsers

Specifying Syntax with a Grammar

Context-free syntax is specified with a context-free grammar

SheepNoise SheepNoise baa | baa

This CFG defines the set of noises sheep normally make

It is written in a variant of Backus–Naur form

Formally, a grammar is a four tuple, G = (S,N,T,P) S is the start symbol (set of strings in L(G)) N is a set of non-terminal symbols (syntactic variables) T is a set of terminal symbols (words) P is a set of productions or rewrite rules (P : N (N T)+ )

Deriving Syntax

We can use the SheepNoise grammar to create sentences use the productions as rewriting rules

And so on ...

A More Useful Grammar

To explore the uses of CFGs,we need a more complex grammar

Such a sequence of rewrites is called a derivation Process of discovering a derivation is called parsing

We denote this derivation: Expr * id – num * id

Derivations

At each step, we choose a non-terminal to replace Different choices can lead to different derivations

Two derivations are of interest Leftmost derivation — replace leftmost NT at each step Rightmost derivation — replace rightmost NT at each step

These are the two systematic derivations

(We don’t care about randomly-ordered derivations!)

The example on the preceding slide was a leftmost derivation Of course, there is also a rightmost derivation Interestingly, it turns out to be different

The Two Derivations for x – 2 * y

In both cases, Expr * id – num * id The two derivations produce different parse trees The parse trees imply different evaluation orders!

Leftmost derivation Rightmost derivation

Derivations and Parse Trees

Leftmost derivation G

x

E

E Op

–

2

E

E

E

y

Op

*This evaluates as x – ( 2 * y )

Derivations and Parse Trees

Rightmost derivation

x 2

G

E

Op EE

E Op E y

–

*

This evaluates as ( x – 2 ) * y

Derivations and Precedence

These two derivations point out a problem with the grammar:

It has no notion of precedence, or implied order of evaluation

To add precedence Create a non-terminal for each level of precedence Isolate the corresponding part of the grammar Force the parser to recognize high precedence subexpressions first

For algebraic expressions Multiplication and division, first (level one) Subtraction and addition, next (level two)

Derivations and Precedence

Adding the standard algebraic precedence produces:

This grammar is slightly larger

• Takes more rewriting to reach some of the terminal symbols

• Encodes expected precedence

• Produces same parse tree under leftmost & rightmost derivations

Let’s see how it parses x - 2 * y

levelone

leveltwo

Derivations and Precedence

The rightmost derivation

This produces x%2 – ( 2 * y ), along with an appropriate parse tree.

Both the leftmost and rightmost derivations give the same expression, because the grammar directly encodes the desired precedence.

G

E

–E

T

F

<id,x>

T

T

F

F*

<num,2>

<id,y>

Its parse tree

Ambiguous GrammarsOur original expression grammar had other problems

This grammar allows multiple leftmost derivations for x - 2 * y Hard to automate derivation if > 1 choice The grammar is ambiguous

different choice than the first time

Two Leftmost Derivations for x – 2 * y

The Difference: Different productions chosen on the second step

Both derivations succeed in producing x - 2 * yOriginal choice New choice

Ambiguous Grammars

Definitions If a grammar has more than one leftmost derivation for a single

sentential form, the grammar is ambiguous If a grammar has more than one rightmost derivation for a single

sentential form, the grammar is ambiguous The leftmost and rightmost derivations for a sentential form may

differ, even in an unambiguous grammar

Classic example — the if-then-else problem

Stmt if Expr then Stmt | if Expr then Stmt else Stmt | … other stmts …

This ambiguity is entirely grammatical in nature

Ambiguity

This sentential form has two derivations

if Expr1 then if Expr2 then Stmt1 else Stmt2

then

else

if

then

if

E1

E2

S2

S1

production 2, then production 1

then

if

then

if

E1

E2

S1

else

S2

production 1, then production 2

AmbiguityRemoving the ambiguityMust rewrite the grammar to avoid generating the problemMatch each else to innermost unmatched if (common sense rule)

With this grammar, the example has only one derivation

1 Stmt WithElse

2 | NoElse

3 WithElse if Expr then WithElse else WithElse

4 | OtherStmt

5 NoElse if Expr then Stmt

6 | if Expr then WithElse else NoElse

Ambiguity

if Expr1 then if Expr2 then Stmt1 else Stmt2

This binds the else controlling S2 to the inner if

Rule Sentential Form— Stmt2 NoElse5 if Expr then Stmt? if E1 then Stmt1 if E1 then WithElse3 if E1 then if Expr then WithElse else WithElse? if E1 then if E2 then WithElse else WithElse4 if E1 then if E2 then S1 else WithElse4 if E1 then if E2 then S1 else S2

Deeper Ambiguity

Ambiguity usually refers to confusion in the CFG

Overloading can create deeper ambiguity

a = f(17)In many Algol-like languages, f could be either a function or a

subscripted variable

Disambiguating this one requires context Need values of declarations Really an issue of type, not context-free syntax Requires an extra-grammatical solution (not in CFG) Must handle these with a different mechanism

Step outside grammar rather than use a more complex grammar

Ambiguity - the Final Word

Ambiguity arises from two distinct sources Confusion in the context-free syntax (if-then-else) Confusion that requires context to resolve (overloading)

Resolving ambiguity To remove context-free ambiguity, rewrite the grammar To handle context-sensitive ambiguity takes cooperation

Knowledge of declarations, types, … Accept a superset of L(G) & check it by other means†

This is a language design problem

Sometimes, the compiler writer accepts an ambiguous grammar Parsing techniques that “do the right thing” i.e., always select the same derivation