2.4 Parsing. Computer Science 332. Compiler Construction. Chapter 2: A Simple One-Pass Compiler : Parsing. Top-Down Parsing

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1 Computer Science 332 Compiler Construction Chapter 2: A Simple One-Pass Compiler : Parsg 2.4 Parsg Parsg : the process of determg whether a strg S is generated by a grammar G Short answer is yes/no To do anythg useful, we need to construct a parse tree Worst case O(N 3 ), where N = S (length of S) In practice, PL's can be parsed O(N) Fundamental distction is top-down vs. bottom-up Easy to construct by hand (Problem Set 1) Not as powerful as bottom-up parsg Basic algorithm: 1) Beg with start symbol as root 2) At node n, labeled with nontermal A, select one of A's productions, and construct children for n usg the symbols the right-hand-side (RHS) of the production 3) Fd the next childless node and goto (2) Production (2) should match current ( lookahead ) symbol S, which advances as we match Example: ML essions from problem set:

2 bdg

3 - - bdg bdg var = var = - - bdg bdg var = var =

4 - - bdg bdg var = var = - - bdg bdg var = var = var

5 - - bdg bdg var = var var = var LOOKAH Note that selection of correct nontermal volved trial and error (backtrackg) Backtrackg is potentially very expensive : exponential depth of parse tree Can we avoid backtrackg? Recursive-Descent / Predictive Parsg Recursive-Descent Parsg : Top-down parsg which each nontermal is handled by a procedure Predictive Parsg: RDP which lookahead symbol unambiguously determes choice of nontermal (so no backtrackg) Sequence of procedure calls implicitly defes parse tree. - private void _() { match(""); (); match(""); (); match(""); }

6 Predictive Parsg Relies on notion of FIRST set: FIRST(A) is set of all first symbols generated from nontermal A. E.g., FIRST() = {, 0, 1,..., 9, a,b,..., f } [why?] Given two productions A and A, predictive parsg will work only if FIRST( ) and FIRST( ) are disjot; i.e., there is no termal that is both FIRST( ) and FIRST( ) [why?] Avoidg Left Recursion Consider grammar rule + term term First part of RHS translates to I.e., loop forever! Can translate to equient What does this remd us of? private void () { (); match( + ); term(); } term R R + term R Avoidg Left Recursion E.g., + term + term + term term + term + term term R term + term R term + term + term R term + term + term More generally, replace A A with A R R R Now prove that they generate the same strg Production Action + term { prt ('+') } - term { prt ('-') } term term 0 { prt ('0') } term 1 { prt ('1') } term 9 { prt ('9') } Translates to term rest rest + - term

7 Problem: Each operand has only one argument: +, - I.e., concrete representation (syntax) now diverges too much from abstract (semantic) representation. We use a different transformation: Gives the modified grammar: -> term rest rest -> + term {prt('+')} rest - term {prt('-')} rest term -> 0 {prt('0')} term -> 1 {prt('1')}... term -> 9 {prt('9')} A A A A R R R R Bottom-Up Parsg term rest term {prt('+')} 2 {prt('2') } + 3 {prt('3')} rest Harder to construct by hand More powerful than top-down parsg So we will use parser generators (yacc, CUP) stead hand-codg Depth-first traversal will produce the ession from this tree