Lecture 27012011

8/7/2019 Lecture 27012011

http://slidepdf.com/reader/full/lecture-27012011 1/51

1

ShiftShift--Reduce: The StackReduce: The StackShiftShift--Reduce: The StackReduce: The Stack

Left string can implementedas a stack

Top of the stack is the

Shift pushes a terminal on

the stack

8/7/2019 Lecture 27012011


2





the stack

8/7/2019 Lecture 27012011


3





the stack

8/7/2019 Lecture 27012011


4

int + (int) + (int) $ shift

int + (int) + (int) $ reduce E int E + (int) + (int) $ shift 3 times

E + (int ) + (int) $ reduce E int

int

E

stack

+

(

8/7/2019 Lecture 27012011


5


Reduce pops zero or moresymbols from the stack

(production rhs) and pushesa non-terminal on the stack

(production lhs)

8/7/2019 Lecture 27012011


6

int + (int) + (int) $ shift

int + (int) + (int) $ reduce E int E + (int) + (int) $ shift 3 times

E + (int ) + (int) $ reduce E int

int

E

stack

+

(

E

8/7/2019 Lecture 27012011


7

Discovering HandlesDiscovering HandlesDiscovering HandlesDiscovering Handles

A bottom-up parser buildsthe parse tree starting with

its leaves and workingtoward its root

8/7/2019 Lecture 27012011


8


The upper edge of thispartially contructed parse

tree is called its upper frontier .

8/7/2019 Lecture 27012011


9

int + (int) + (int)

E + (int) + (int)

E + (E) + (int)

E + (int)

E + (E)

int + ( int )int + ( )

E E E

E

8/7/2019 Lecture 27012011


10


At each step, the parser looks for a section of the

upper frontier that matchesright-hand side of some

production.

8/7/2019 Lecture 27012011


11


When it finds a match, theparser builds a new tree

node with the production¶sleft-hand non-terminal thus

extending the frontier upwards towards the root.

8/7/2019 Lecture 27012011


12


The critical step isdeveloping an efficient

mechanism that findsmatches along the tree¶s

current frontier .

8/7/2019 Lecture 27012011


13


Formally, the parser must findsome substring F, of the upper frontier where

1. F is the right-hand side of someproduction A F, and

2. A F is one step in right-mostderivation of input stream

8/7/2019 Lecture 27012011


14





8/7/2019 Lecture 27012011


15





8/7/2019 Lecture 27012011


16


We can represent eachpotential match as a pair

£AF,k ³, where k is theposition on the tree¶s

current frontier of the right-end of F.

8/7/2019 Lecture 27012011


17


The pair £AF,k ³, is calledthe handle of the bottom-up parse.

8/7/2019 Lecture 27012011


18

Handle PruningHandle PruningHandle PruningHandle Pruning

A bottom-up parser operatesby repeatedly locating

handles on the frontier of thepartial parse tree and

performing reductions thatthey specify.

8/7/2019 Lecture 27012011


19

HandlesHandlesHandlesHandles

The bottom-up parser uses a stack to hold the

frontier. The stack simplifies the

parsing algorithm in twoways.

8/7/2019 Lecture 27012011


20


The bottom-up parser uses a stack to hold the

frontier. The stack simplifies the

parsing algorithm in twoways.

8/7/2019 Lecture 27012011


21


First, the stack trivializes theproblem of managing space

for the frontier � To extend the frontier, the

parser simply pushes thecurrent input token onto thetop of the stack

8/7/2019 Lecture 27012011


22


First, the stack trivializes theproblem of managing space

for the frontier � To extend the frontier , the

parser simply pushes thecurrent input token onto thetop of the stack

8/7/2019 Lecture 27012011


23


Second, the stack ensuresthat all handles occur with

their right end at the top of the stack

� eliminates the need torepresent handle¶s position

8/7/2019 Lecture 27012011


24


Second, the stack ensuresthat all handles occur with

their right end at the top of the stack

� eliminates the need torepresent handle¶s position

8/7/2019 Lecture 27012011


25

E + (int) $ shift 3 times

E + (int ) $ reduce E int

E + (E ) $ shift

E + (E) $ red E E+(E)

E $ accept

handle: £Eint ,4³handle: £EE+(E),5³

1 2 3 4

8/7/2019 Lecture 27012011


26

ShiftShift--Reduce Parsing AlgorithmReduce Parsing AlgorithmShiftShift--Reduce Parsing AlgorithmReduce Parsing Algorithm

push $ onto stacksym nextToken()

r epeat unti l (sym == $ and the stack

contains exactly Goal on top of $)if a handle for A F on top of stack

pop |F| symbols off the stack

push A onto the stack

8/7/2019 Lecture 27012011


27

ShiftShift--Reduce Parsing AlgorithmReduce Parsing AlgorithmShiftShift--Reduce Parsing AlgorithmReduce Parsing Algorithm

else if (sym { $ ) push sym onto stack

sym nextToken()

else /* no handle, no input */

report error and halt

8/7/2019 Lecture 27012011


28

Expression Grammar Expression Grammar Expression Grammar Expression Grammar

1 Goal Expr2 Expr Expr + Term3 | Expr ² Term4 | Term

5 Term Term v Fact or6 | Term ¼ Fact or7 | Fact or

8 Fact or num9 | id10 | ( Expr )

8/7/2019 Lecture 27012011


29

The states of the bottom-upparser on input

x ² 2 v y(tokenized as id ² num * id) are

8/7/2019 Lecture 27012011


30

word Stack Handle Action

1 id - none - shift

2 ² id £Fact or id,1³ r educe

3 ² Fact or £Term Fact or,1³ r educe

4 ² Term £Expr Term,1³ r educe

5 ² Expr - none - shift

6 num Expr ² - none - shift

7 v Expr ² num £Fact or num,3³ r educe

8 v Expr ² Fac

t o

r £Term Fac

t o

r,3³

r educe

8/7/2019 Lecture 27012011


31

word Stack Handle Action

9 v Expr ² Term - none - shift

10 id Expr ² Term v - none - shift

11 $ Expr ² Term v id £Fact or id,5³ r educe

12 $ Expr²Term v Fact or £TermTermvFact or,5³ r educe

13 $ Expr ² Term £Expr Expr ² Term,3³ r educe

14 $ Expr £Goal Expr,1³ r educe

15 $ Goal - none - accept

8/7/2019 Lecture 27012011


32


The handle-findingmechanism is the key toeffiecient bottom-up parsing

As it process an input string,the parser must find and

track all potential handles

8/7/2019 Lecture 27012011


33


The handle-findingmechanism is the key toeffiecient bottom-up parsing

As it process an input string,the parser must find and

track all potential handles

8/7/2019 Lecture 27012011


34


For example, every legalinput eventually reduces theentire frontier to grammar¶sgoal symbol

Thus, £Goal Expr,1³ is a

potential handle at the startof every parse

8/7/2019 Lecture 27012011


35


For example, every legalinput eventually reduces theentire frontier to grammar¶sgoal symbol

Thus, £Goal Expr,1³ is a

potential handle at the startof every parse

8/7/2019 Lecture 27012011


36


As the parser builds aderivation, it discovers other

handles At each step, the set of

potential handles representdifferent suffixes that lead to areduction.

8/7/2019 Lecture 27012011


37


As the parser builds aderivation, it discovers other

handles At each step, the set of

potential handles representdifferent suffixes that lead to areduction.

8/7/2019 Lecture 27012011


38


Each potential handlerepresent a string of

grammar symbols that, if seen, would complete the

right-hand side of someproduction.

8/7/2019 Lecture 27012011


39


For the bottom-up parse of the expression grammar

string, we can representthe potential handles thatthe shift-reduce parser should track

8/7/2019 Lecture 27012011


40


Using the placeholder y torepresent top of the stack, thereare nine handles:

8/7/2019 Lecture 27012011


41

Same

Handles

1 £Fact or id y³

2 £Term Fact ory³

3 £Expr Term y³

4 £Fact or numy³5 £Term Fact ory³

6 £Fact or id y³

7£Term Term

vFact or

y³8 £Expr Expr ² Term y³

9 £Goal Expr y³

Same

8/7/2019 Lecture 27012011


42


This notation shows that thesecond and fifth handles areidentical, as are first and sixth

It also create a way torepresent the potential of

discovering a handle infuture.

8/7/2019 Lecture 27012011


43


This notation shows that thesecond and fifth handles areidentical, as are first and sixth

It also create a way torepresent the potential of

discovering a handle infuture.

8/7/2019 Lecture 27012011


44


Consider the parser¶s state in step 6:



8 v Expr ² Fact or £Term Fact or,3³ shift




12 $ Expr²TermvFact or £TermTermvFact or,5³ r educe

13 $ Expr ² Term £ExprExpr²Term,3³ r educe

8/7/2019 Lecture 27012011


45


The parser has recognized Expr ²



8 v Expr ² Fact or £Term Fact or,3³ shift




12 $ Expr²TermvFact or £TermTermvFact or,5³ r educe

13 $ Expr ² Term £ExprExpr²Term,3³ r educe

It needs to recognize, eventually, a Termbefore it can reduce this part of the frontier

8/7/2019 Lecture 27012011


46


Using the stack-relativenotation, we can representthe parser¶s state asExpr Expr ² y Term

The parser has already

recognized an Expr and a ²

8/7/2019 Lecture 27012011


47


Using the stack-relativenotation, we can representthe parser¶s state asExpr Expr ² y Term

The parser has already

recognized an Expr and a ²

8/7/2019 Lecture 27012011


48


If the parser reaches a statewhere it shifts a Term on topof Expr and ±, it willcomplete the handleExpr Expr ² Term y

8/7/2019 Lecture 27012011


49


How many potential handlesmust the parser recognize?

The right-hand side of eachproduction can have aplaceholder at its start, at its endand between any twoconsecutive symbols

8/7/2019 Lecture 27012011


50


How many potential handlesmust the parser recognize?

The right-hand side of eachproduction can have aplaceholder at its start, at its endand between any twoconsecutive symbols

8/7/2019 Lecture 27012011


51


Expr y Expr ²TermExpr Expr y ² Term

Expr Expr ² yTermExpr Expr ² Term y

if the right-hand side hask

symbols,it has k +1 placeholder positions

Lecture 27012011

Documents