Compiler Principles Winter 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 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 Answer: D = 0 | 1 | 2 … | 9 Small = D | DD | DDD Three = DDD R = Small (_Three)* 3

Parsing question 1 Consider the grammar below S  ( L ) | a L  L, S | S 1.show a parse tree for each of the following 1.(a, a) 2.(a, (a, a)) 4

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

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 6

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 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) 7

Parsing question 3 Consider the grammar G1 below X  P P P  a P  b Q: Is it ambiguous? 8

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 9

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. 10

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 11

Answer 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 12

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 13

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 14

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) 15

Parsing table ;)ctbw(a 1E 2E’ 43E’’ 5S 76S’ 16

Running on a ( c; c ) w b 17 Input suffixStack contentMove a ( c; c ) w b $E$ predict(E, a) = E  a ( S E ’ a ( c; c ) w b $ 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; c ) w b $ c S ’ E ’ $ match(c, c) ; c ) w b $ S ’ E ’ $ predict(S’, ;) = S ’  ; S ; c ) w b $ ; S E ’ $ match( ‘(;, ‘;‘) c ) w b $ S E ’ $ predict(S, c) = S  c S ’ c ) w b $ c S’ E ’ $ match(c, c) ) w b $ S’ E ’ $ predict(S’, ‘)’) = S ’  ) ) w b $ ) E’ $match( ‘)‘, ‘)‘) w b $E’ $ predict(E’, w) = E’  w E ’’ w b $ w E ’’ $ match(w, w) b $E’’ $ predict(E’’, b) = E ’’  b b $ match(b, b) $$

Running on a ( c; c ) w b 18 Input suffixStack contentMove a ( c; c ) w b $E$ predict(E, a) = E  a ( S E ’ a ( c; c ) w b $ 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; c ) w b $ c S ’ E ’ $ match(c, c) ; c ) w b $ S ’ E ’ $ predict(S’, ;) = S ’  ; S ; c ) w b $ ; S E ’ $ match( ‘(;, ‘;‘) c ) w b $ S E ’ $ predict(S, c) = S  c S ’ c ) w b $ c S’ E ’ $ match(c, c) ) w b $ S’ E ’ $ predict(S’, ‘)’) = S ’  ) ) w b $ ) E’ $match( ‘)‘, ‘)‘) w b $E’ $ predict(E’, w) = E’  w E ’’ w b $ w E ’’ $ match(w, w) b $E’’ $ predict(E’’, b) = E ’’  b b $ match(b, b) $$ Since the input has been read and the stack is empty – the parsing was successful and the input is accepted