Parsing III (Top-down parsing: recursive descent & LL(1) )

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Parsing III (Top-down parsing: recursive descent & LL(1) )

Roadmap (Where are we?) We set out to study parsing Specifying syntax  Context-free grammars   Ambiguity  Top-down parsers  Algorithm & its problem with left recursion   Left-recursion removal  Predictive top-down parsing  The LL(1) condition  Simple recursive descent parsers  Table-driven LL(1) parsers

Picking the “Right” Production If it picks the wrong production, a top-down parser may backtrack Alternative is to look ahead in input & use context to pick correctly How much lookahead is needed? In general, an arbitrarily large amount Use the Cocke-Younger, Kasami algorithm or Earley’s algorithm Fortunately, Large subclasses of CFGs can be parsed with limited lookahead Most programming language constructs fall in those subclasses Among the interesting subclasses are LL(1) and LR(1) grammars

Predictive Parsing Basic idea Given A    , the parser should be able to choose between  &  F IRST sets For some rhs   G, define F IRST (  ) as the set of tokens that appear as the first symbol in some string that derives from  That is, x  F IRST (  ) iff   * x , for some  We will defer the problem of how to compute F IRST sets until we look at the LR(1) table construction algorithm

Predictive Parsing Basic idea Given A    , the parser should be able to choose between  &  F IRST sets For some rhs   G, define F IRST (  ) as the set of tokens that appear as the first symbol in some string that derives from  That is, x  F IRST (  ) iff   * x , for some  The LL(1) Property If A   and A   both appear in the grammar, we would like F IRST (  )  F IRST (  ) =  This would allow the parser to make a correct choice with a lookahead of exactly one symbol ! This is almost correct See the next slide

Predictive Parsing What about  -productions?  They complicate the definition of LL(1) If A   and A   and   F IRST (  ), then we need to ensure that F IRST (  ) is disjoint from F OLLOW (  ), too Define F IRST + (  ) as F IRST (  )  F OLLOW (  ), if   F IRST (  ) F IRST (  ), otherwise Then, a grammar is LL(1) iff A   and A   implies F IRST + (  )  F IRST + (  ) =  F OLLOW (  ) is the set of all words in the grammar that can legally appear immediately after an 

Predictive Parsing Given a grammar that has the LL(1) property Can write a simple routine to recognize each lhs Code is both simple & fast Consider A   1 |  2 |  3, with F IRST + (  1 )  F IRST + (  2 )  F IRST + (  3 ) =  /* find an A */ if (current_word  F IRST (  1 )) find a  1 and return true else if (current_word  F IRST (  2 )) find a  2 and return true else if (current_word  F IRST (  3 )) find a  3 and return true else report an error and return false Of course, there is more detail to “find a  i ” ( § in EAC ) Grammars with the LL(1) property are called predictive grammars because the parser can “predict” the correct expansion at each point in the parse. Parsers that capitalize on the LL(1) property are called predictive parsers. One kind of predictive parser is the recursive descent parser.

Recursive Descent Parsing Recall the expression grammar, after transformation This produces a parser with six mutually recursive routines: Goal Expr EPrime Term TPrime Factor Each recognizes one NT or T The term descent refers to the direction in which the parse tree is built.

Recursive Descent Parsing (Procedural) A couple of routines from the expression parser Goal( ) token  next_token( ); if (Expr( ) = true & token = EOF) then next compilation step; else report syntax error; return false; Expr( ) if (Term( ) = false) then return false; else return Eprime( ); Factor( ) if (token = Number) then token  next_token( ); return true; else if (token = Identifier) then token  next_token( ); return true; else report syntax error; return false; EPrime, Term, & TPrime follow the same basic lines (Figure 3.7, EAC) looking for EOF, found token looking for Number or Identifier, found token instead

Recursive Descent Parsing To build a parse tree: Augment parsing routines to build nodes Pass nodes between routines using a stack Node for each symbol on rhs Action is to pop rhs nodes, make them children of lhs node, and push this subtree To build an abstract syntax tree Build fewer nodes Put them together in a different order Expr( ) result  true; if (Term( ) = false) then return false; else if (EPrime( ) = false) then result  false; else build an Expr node pop EPrime node pop Term node make EPrime & Term children of Expr push Expr node return result; This is a preview of Chapter 4 Success  build a piece of the parse tree

Left Factoring What if my grammar does not have the LL(1) property?  Sometimes, we can transform the grammar The Algorithm  A  NT, find the longest prefix  that occurs in two or more right-hand sides of A if  ≠  then replace all of the A productions, A   1 |  2 | … |  n | , with A   Z |  Z   1 |  2 | … |  n where Z is a new element of NT Repeat until no common prefixes remain

A graphical explanation for the same idea becomes … Left Factoring A   1 |  2 |  3 A   Z Z   1 |  2 |  n A  1  3  2 ZZ 11 33 22 A

Left Factoring (An example) Consider the following fragment of the expression grammar After left factoring, it becomes This form has the same syntax, with the LL(1) property F IRST (rhs 1 ) = { Identifier } F IRST (rhs 2 ) = { Identifier } F IRS T(rhs 3 ) = { Identifier } F IRST (rhs 1 ) = { Identifier } F IRS T(rhs 2 ) = { [ } F IRST (rhs 3 ) = { ( } F IRST (rhs 4 ) = F OLLOW (Factor)  It has the LL(1) property

Graphically becomes … Left Factoring Factor Identifier []ExprList () FactorIdentifier[]ExprList ()  No basis for choice Word determines correct choice

Question By eliminating left recursion and left factoring, can we transform an arbitrary CFG to a form where it meets the LL(1) condition? (and can be parsed predictively with a single token lookahead?) Answer Given a CFG that doesn’t meet the LL(1) condition, it is undecidable whether or not an equivalent LL(1) grammar exists. Example {a n 0 b n | n  1}  {a n 1 b 2n | n  1} has no LL(1) grammar Left Factoring (Generality)

Language that Cannot Be LL(1) Example {a n 0 b n | n  1}  {a n 1 b 2n | n  1} has no LL(1) grammar G  aAb | aBbb A  aAb | 0 B  aBbb | 1 Problem: need an unbounded number of a characters before you can determine whether you are in the A group or the B group.

Recursive Descent (Summary) 1.Build F IRST (and F OLLOW ) sets 2.Massage grammar to have LL(1) condition a.Remove left recursion b.Left factor it 3.Define a procedure for each non-terminal a.Implement a case for each right-hand side b.Call procedures as needed for non-terminals 4.Add extra code, as needed a.Perform context-sensitive checking b.Build an IR to record the code Can we automate this process?

F IRST and F OLLOW Sets F IRST (  ) For some   T  NT, define F IRST (  ) as the set of tokens that appear as the first symbol in some string that derives from  That is, x  F IRST (  ) iff   * x , for some  F OLLOW (  ) For some   NT, define F OLLOW (  ) as the set of symbols that can occur immediately after  in a valid sentence. F OLLOW (S) = { EOF }, where S is the start symbol To build F IRST sets, we need F OLLOW sets …

Computing F IRST Sets Define F IRST as If   * a , a  T,   (T  NT)*, then a  F IRST (  ) If   * , then   F IRST (  ) Note: if  = X , F IRST (  ) = F IRST ( X) Terminal: a,b,c,  Non-terminal: L,R,Q,R’,Q’, L’ First(a) ={a}, First(b) ={b}, First(c)={c}, First{  } ={  } First(L) ={a,b,c}; First(R) ={a,c}, First(Q)={b} First(R’) = {b,  }, First(Q’) = {b,c}, First(L’) = {b,c}

Computing F OLLOW Sets FOLLOW(S)  {EOF } for each A  NT, FOLLOW(A)  Ø while (FOLLOW sets are still changing) for each p  P, of the form A  1  2 …  k FOLLOW(  k )  FOLLOW(  k )  FOLLOW(A) TRAILER  FOLLOW(A) for i  k down to 2 if   FIRST(  i ) then FOLLOW(  i-1 )  FOLLOW(  i-1 )  { FIRST(  i ) – {  } }  TRAILER else FOLLOW(  i-1 )  FOLLOW(  i-1 )  FIRST(  i ) TRAILER  Ø FOLLOW(R’) = {a}

To Combine First(alpha) and FOLLOW(alpha) FIRST + First+(L) =First(L) = {a,b,c}; First+(R) = First(R) = {a,c}, First+(Q)=First(Q) ={b} First+(R’) = First(R’) U Follow(R’) = {b,a  }, First+(Q’) = First(Q)= {b,c}, First+(L’)= First(L’) = {b,c} Table:abcEOF L R Q R’ Q’ L’

Building Top-down Parsers Given an LL(1) grammar, and its F IRST & F OLLOW sets … Emit a routine for each non-terminal  Nest of if-then-else statements to check alternate rhs’s  Each returns true on success and throws an error on false  Simple, working (, perhaps ugly,) code This automatically constructs a recursive-descent parser Improving matters Nest of if-then-else statements may be slow  Good case statement implementation would be better What about a table to encode the options?  Interpret the table with a skeleton, as we did in scanning I don’t know of a system that does this …

Building Top-down Parsers Strategy Encode knowledge in a table Use a standard “skeleton” parser to interpret the table Example The non-terminal Factor has three expansions  ( Expr ) or Identifier or Number Table might look like: +-*/Id.Num. EOF Factor————1011— Reduce by rule 10 on `+ ’ Error on `+ ’ Terminal Symbols Non-terminal Symbols

Building Top Down Parsers Building the complete table Need a row for every NT & a column for every T Need a table-driven interpreter for the table

LL(1) Skeleton Parser token  next_token() push EOF onto Stack push the start symbol, S, onto Stack TOS  top of Stack loop forever if TOS = EOF and token = EOF then break & report success else if TOS is a terminal then if TOS matches token then pop Stack// recognized TOS token  next_token() else report error looking for TOS else // TOS is a non-terminal if TABLE[TOS,token] is A  B 1 B 2 …B k then pop Stack // get rid of A push B k, B k-1, …, B 1 // in that order else report error expanding TOS TOS  top of Stack exit on success ababca R’ a EOF TOS L -> abaR’a

Building Top Down Parsers Building the complete table Need a row for every NT & a column for every T Need an algorithm to build the table Filling in TABLE[X,y], X  NT, y  T 1. entry is the rule X  , if y  F IRST (  ) 2. entry is the rule X   if y  F OLLOW (X ) and X    G 3. entry is error if neither 1 nor 2 define it If any entry is defined multiple times, G is not LL(1) This is the LL(1) table construction algorithm