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Test 1 Post Mortem

Test 1 Post Mortem. CSCE 531 Compiler Construction. Topics Questions and points Grade Distribution Answers. February 20, 2006. Questions and Point Evaluations. 5 Recursive Descent parsing Transition diagram 7pts recursive parsing 7pts LL(1) table construction pts Modifications 1 pt

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Test 1 Post Mortem

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  1. Test 1 Post Mortem CSCE 531 Compiler Construction • Topics • Questions and points • Grade Distribution • Answers February 20, 2006

  2. Questions and Point Evaluations • 5 Recursive Descent parsing • Transition diagram 7pts • recursive parsing 7pts • LL(1) table construction pts • Modifications 1 pt • FIRST 4 pts • FOLLOW 4 pts • Table 5 pts • LR(0) sets of items 14 pts • 1a. Reg exprNFA 14 pts • 2. Subset Construction 14 pts • Grammars, derivations • Flex/Lex patterns 4 pts • L( (a|b)*aa(a|b) ) 5 pts • Reg expr a’s before b’s 5pts • Left-recusion • Left recursion 4 pts • Left factor 4 pts • LL(1) 2pts • M[A,a] 4pts

  3. Grade Distribution

  4. 1. Thompson Construction Reg Expr  NFA • (a | b)(ab)*a solution build from innermost out ε a ε ε ε ε a ε b ε ε a ε b ε ε

  5. 2. Subset Construction NFA  DFA • Accepting states those that contain s6:

  6. 3. Regular Languages • 3a. What do the following Lex/Flex patterns match? • x? --- zero or one x’s (spaces on the dynamically bound question) • [^abc] --- any single character not ‘a’, ‘b’ or ‘c’ • ab+ --- a single ‘a’ followed by one or more ‘b’s’ • ab{3,5} --- a single ‘a’ followed by 3, 4, or 5 b’s • 3b. Describe the language denoted by the regular expression (a|b)*aa(a|b) • Any string of a’s and b’s such that the third and second to last characters are a’s • Or strings of a’s and b’s that end in aaa or aab • 3c. Regular expression for strings from {a,b,c}* all a’s before any b • (a | c)* (b | c)*

  7. 3a. Mistakes; imprecise answers, etc. • x? --- zero or one x’s (spaces ) • An optional match • Matches anything that is not an “a”, “b” or “c” • [^abc] --- any single character not ‘a’, ‘b’ or ‘c’ • Negated character class any thing other than [abc] • Any string other than abc • Any string that does not contain the characters ‘a’, ‘b’ and ‘c’ • ab+ --- a single ‘a’ followed by one or more ‘b’s’ • ab{3,5} --- a single ‘a’ followed by 3, 4, or 5 b’s • Matches 3 to 5 b’s

  8. 3b. Mistakes • A string of a’s and b’s which contain two consecutive a’s (a|b)*aa(a|b)* • 3c. Mistakes, etc. Notations • L3c – the language specified in 3c • La – the language specified by the student’s answer • Mistakes • a*b*c* | c* a* b* | a* c* b* --- acacb is not in La . • (a* | c*) b* c* -- close but acbcb is not in La . • ((a* | c*) b c*)* -- close but abcabc is in La .

  9. 4. Points(4/4/2/4) Left Recursion • S Sa | bL • L  ScL | Sd | a • 4a. Equivalent non left recursive grammar • The only left recursive production is S Sa • The problem would have been more interesting if SLb were to replace SbL because then we would have some non-direct recursion to eliminate • To handle SSα | β (α=a, β=bL) replace with S βS’ and S’ αS’ | Є, so • S  bLS’ • S’  aS’ | Є • L  ScL | Sd | a

  10. Common error: rewrite Eliminate L • S Sa | bL • L  ScL | Sd | a • The idea some students used is to substitute for L the right habd sides of the L productions to obtain • S Sa | bScL | bSd | ba • But what’s wrong with this? • We still have an L!!! • This is because the L productions are recursive; so no matter how many times you rewrite it you can never get rid of the L.

  11. 4b Left Factor this resulting grammar • S  bLS’ • S’  aS’ | Є • L  ScL | Sd | a • The only portion needing factoring is LScL | Sd • So • L  SL’ | a • L’  cL | d • With the S and S’ productions from above • S  bLS’ • S’  aS’ | Є

  12. 4. continued • 4c. (2pts) a grammar is LL(1) if in constructing the predictive parse table there are no multiply defined entries • There is an alternate definition in the text page 192. • G is LL(1) if and only if if Aα | β are two productions then the following conditions hold • FIRST*(α) ∩ FIRST*(β) = Φ • At most one of α and β can derive the empty string Є • If β  Є then α does not derive any string beginning with a token in FOLLOW(A) • 4d. (4 pts) Pop A off the stack and push X Y and Z in reverse order

  13. 5 Recursive Descent Parsing • 5a. Construct a transition diagram for X given the productions X  aAB | bB | Є Є a A B b B

  14. 5b. Recursive Descent Parsing • 5b. Construct a recursive parsing routine for X assuming routines for A and B. • Reference Transition diagrams in fig 4.12 used to generate parsing routine in Chapter 2, page 75. • ParseX (){ • if (current_token == ‘a’){ • parseA(); • parseB(); • }else if (current_token == ‘b’){ • parseB(); • }else return; /* epsilon*/ • }

  15. 6 FIRST, FOLLOW Predictive Parsing • 6a. Non required • 6b. FIRST - FIRST (tα) ={t} for any token t • Considering productions • Zd add d to FIRST(Z) • Yc add c to FIRST(Y) • YЄ add Є to FIRST(Y) • Xa add a to FIRST(X) • XY add FIRST*(Y) to FIRST(X) • Then considering • ZXYZ add FIRST*(X) to FIRST(Z) • Since XYЄ add to FIRST*(Y) to FIRST(Z) • And since XЄ and YЄ add FIRST(Z) to FIRST(Z) duh!

  16. 6c. FOLLOW • 6c. Add $ to FOLLOW(Z) • Here we will use FIRST*(N) to denote FIRST(N) – {Є} • Considering productions • ZXYZ add • FIRST*(Y) to FOLLOW(X) • FIRST*(Z) to FOLLOW(Y) • Since Z XYZ  XЄZ • FIRST*(Z) to FOLLOW(X) • And considering • XY add FOLLOW(X) to FOLLOW(Y)

  17. 6d. Predictive Parse Table

  18. 6d. Predictive Parse Table Analysis • Consider Productions in order and add them to M[N,t] according to algorithm 4.4 • ZXYZ • Using rule 2 add ZXYZ for each a in FIRST*(XYZ) • But since Є is in FIRST(X) and FIRST(Y) • FIRST*(XYZ) = FIRST*(X) U FIRST*(Y) U FIRST*(Z) = {a, c, d } • Rule 3 does not apply as XYZ does not derive Є • Z  d • Using rule 2 again add Zd for M[Z, d]  double entry  not LL(1) • Yc similarly adds Yc to M[Y, c] • And X a adds X a to M[X, a] • YЄ, using rule3 add to M[Y, b] foreach token b in FOLLOW(Y)={a,c,d} • XY using rule 2 add to a to M[X, a] for each a in FIRST*(Y)={c} • XY, using rule3 add to M[X, b] for each token b in FOLLOW(X)={a,c,d}

  19. 7. LR(0) Sets of Items • D  T L ; | ЄTokens = {‘;’, id, ‘,’, int, float} • L  L , id | id • T  int | float • Augment with S’  D • J0 = closure({S’  ● D}) • = { [S’  ● D], [D ● T L ;], [D●], [T●int], [T ●float]} • Now consider GOTO(J0, X) for each X just to right of “dot” • J1 = GOTO(J0, D) = closure({[S’  D ● ]} = {[S’  D ● ]} • J2 = GOTO(J0, T) = closure({[D T ● L ;]} • = {[D T ● L ;], [L  ● L , id], [L  ● id]}

  20. 7. LR(0) Sets of Items • J3 = GOTO(J0, int) = closure({[T int ● ]} = {[T int ● ]} • J4 = GOTO(J0, float) = closure({[T float ●]} = {[Tfloat ● ]} • J5 = GOTO(J2, L) = closure({[D T L ● ;], [L  L ● , id]}) • = ({[D T L ● ;], [L  L ● , id]}) • J6 = GOTO(J2, id) = closure({[L  id ● ]}) = {[L  id ● ]} • J7 = GOTO(J5, ; ) = closure( {[L  T L ; ● ]}) = {[L  T L ; ● ]} • J8 = GOTO(J5, , ) = closure( {[L  L , ● id]}) = {[L  L , ● id]} • J9 = GOTO(J8, id ) = closure( {[L  L , id ●]}) = {[L  L , id●]}

  21. What I should have also asked . • Leftmost derivations, parse trees • Ambiguous grammars

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