Ben-Gurion University Winter 2012-2013 Compiler Principles Exercises on scanning & top-down parsing Roman Manevich Ben-Gurion University

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

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 … | 9 Small = D | DD | DDD Three = DDD R = Small (_Three)*

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

Answer S ( L ) S L L L S S S a , ( a , a )

Parsing question 2 Consider the grammar G L  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

Parsing question 2 Consider the grammar G L  L , a | a Answer 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)

Parsing question 3 Consider the grammar G1 below X  P P P  a P  b Q: Is it ambiguous?

Parsing question 3 Consider the grammar G1 below X  P P P  a P  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

Parsing question 4 Consider the following grammar E  a ( S ) w t E  a ( S ) w b S  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 input a (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.

Answer The grammar E  a ( S ) w t E  a ( S ) w b S  S ; c | c will have a FIRST-FIRST conflict for E, since the first two rules start with ‘a ( S ) w’. We therefore apply left-factoring and obtain E  a ( S ) E’ E’  w b | w t S  S ; c | c

Answer (with bugs) The grammar E  a ( S ) E’ E’  w b | w t S  S ; c | c Is left-recursive on S so we eliminate the left-recursion and obtain the grammar E  a ( S ) E’ E’  w b | w t S  c | c ; S

Answer Let’s compute FIRST sets for E  a ( S ) E’ E’  w b | w t S  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 obtain E  a ( S ) E’ E’  w E’’ E’’  b | t S  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

Answer Starting from E  a ( S ) E’ E’  w E’’ E’’  b | t S  c | c ; S Let’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

Answer Let’s compute FIRST sets for (1) E  a (S E’ (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)

Parsing table ; ) c t b w ( a 1 E 2 E’ 4 3 E’’ 5 S 7 6 S’

Running on a ( c; c ) w b Input suffix Stack content Move predict(E, a) = E  a ( S E’ a ( S E’ $ match(a, a) ( c; c ) w b $ ( S E’ $ match( ‘(‘, ‘(‘) c; c ) w b $ S E’ $ predict(S, c) = S  c S’ c S’ E’ $ match(c, c) ; c ) w b $ S’ E’ $ predict(S’, ;) = S’  ; S ; S E’ $ match( ‘(;, ‘;‘) c ) w b $ ) w b $ predict(S’, ‘)’) = S’  ) ) E’ $ match( ‘)‘, ‘)‘) w b $ E’ $ predict(E’, w) = E’  w E’’ w E’’ $ match(w, w) b $ E’’ $ predict(E’’, b) = E’’  b match(b, b) $

Running on a ( c; c ) w b Input suffix Stack content Move a ( c; c ) w b$ E$ predict(E, a) = E  a ( S E’ a ( S E’ $ match(a, a) ( c; c ) w b $ ( S E’ $ match( ‘(‘, ‘(‘) c; c ) w b $ S E’ $ predict(S, c) = S  c S’ c S’ E’ $ match(c, c) ; c ) w b $ S’ E’ $ predict(S’, ;) = S’  ; S ; S E’ $ match( ‘(;, ‘;‘) c ) w b $ ) w b $ predict(S’, ‘)’) = S’  ) ) E’ $ match( ‘)‘, ‘)‘) w b $ E’ $ predict(E’, w) = E’  w E’’ w E’’ $ match(w, w) b $ E’’ $ predict(E’’, b) = E’’  b match(b, b) $ Since the input has been read and the stack is empty – the parsing was successful and the input is accepted