Winter 2012-2013 Compiler Principles Exercises on scanning & top-down parsing

1. Winter 2012-2013Compiler PrinciplesExercises on scanning& top-down parsing Roman Manevich Ben-Gurion University

2. Scanning question 1 • Write a regular expression for positive integers where every three consecutive digits, starting from the right, are separated by a _ • Examples: 23 1_000 2_784_153 • Negative examples: 1_0 1_5422

3. Scanning question 1 • Write a regular expression for positive integers where every three consecutive digits, starting from the right, are separated by a _ • Examples: 23 1_000 2_784_153 • Negative examples: 1_0 1_5422 • Answer:D = 0 | 1 | 2 … | 9Small = D | DD | DDDThree = DDDR = Small (_Three)*

4. Parsing question 1 • Consider the grammar belowS  ( L ) | a L  L , S | S • show a parse tree for each of the following • (a, a) • (a, (a, a))

5. Answer S ( L ) S L L L S S S ( a , a ) a ,

6. Parsing question 2 • Consider the grammar GL  L , a | a • What is the language generated by G? • Eliminate the left-recursion in G • Write a non-left-recursive version without ε productions • Construct an LL(1) parser and run it on a, a, a

7. Parsing question 2 • Consider the grammar GL  L , a | a • What is the language generated by G? • Eliminate the left-recursion in G • Write a non-left-recursive version with ε productions • Is the resulting grammar LL(1)? • Answer • The language given by the regular expression (a ,)* a • L  a M M , a M | ε • L  a | a M M , a M | , a • The grammar contains FIRST-FIRST conflicts for both L and M and therefore it is not LL(1)

8. Parsing question 3 Consider the grammar G1 belowX  P PP aP  b Q: Is it ambiguous?

9. Parsing question 3 Consider the grammar G1 belowX  P PP aP  b Q: Is it ambiguous? A: since the only non-terminal with alternatives is P and the first of its two alternatives ({a} and {b}) the grammar is LL(1). All LL(1) grammars are unambiguous

10. Parsing question 4 • Consider the following grammarE  a ( S ) w tE  a ( S ) w bS  S ; c | c • Construct an LL(1) parser, applying any transformations necessary if the grammar is not in LL(1) • Hint: move the ) down to help transformations • Apply the parser to the inputa (c; c) w b • Note: the following solution is over-complicated since there is an implicit constraint – no epsilon productions are allowed. Otherwise the solution is pretty simple.

11. Answer The grammarE  a ( S ) w tE  a ( S ) w bS  S ; c | cwill have a FIRST-FIRST conflict for E, since the first two rules start with ‘a ( S ) w’. We therefore apply left-factoring and obtainE  a ( S ) E’E’  w b | w tS  S ; c | c

12. Answer The grammarE  a ( S ) E’E’  w b | w tS  S ; c | cIs left-recursive on S so we eliminate the left-recursion and obtain the grammar E  a ( S ) E’E’  w b | w tS  c | c ; S

13. Answer Let’s compute FIRST sets forE  a ( S ) E’E’  w b | w tS  c | c ; S FIRST(S  c) = FIRST(c ; S) = {c}FIRST(w b)=FIRST(w t) = {w}so there are FIRST-FIRST conflict. We apply left-factoring to both E’ and S and obtainE  a ( S ) E’E’  w E’’ E’’  b | tS  c S’ S’  ε | ; S Now if we apply substitution for S’ we will just get back to a left-recursive grammar.Time to use the hint

14. Answer Starting fromE  a ( S ) E’E’  w E’’ E’’  b | t S  c | c ; SLet’s move ) from E to S E  a ( S E’E’  w E’’ E’’  b | t S  c ) | c ; S Now let’s apply left-factoring to S and obtain:E  a ( S E’E’  w E’’ E’’  b | t S  c S’S’  ) | ; S

15. Answer Let’s compute FIRST sets for (1) E  a (SE’(2) E’  w E’’ (3) E’’  b(4) E’’  t (5) S  c S’(6) S’  )(7) S’  ; S FIRST(E) = {a}FIRST(E’) = {w}FIRST(E’’) = {b, t}FIRST(S) = {c}FIRST(S’) = {), ;} There are no conflicts so the grammar is in LL(1)

16. Parsing table

17. Running on a ( c; c ) w b

18. Running on a ( c; c ) w b Since the input has been read and the stack is empty – the parsing was successful and the input is accepted