ECE 468/573 Midterm 1 September 30, 2011

Transcription

1 ECE 468/573 Midterm 1 September 30, 2011 Nme: Purdue emil: Plese sign the following: I ffirm tht the nswers given on this test re mine nd mine lone. I did not receive help from ny person or mteril other thn those explicitly llowed). X Prt Points Score Totl 120

2 Prt 1: Short nswers 10 points) 1) You re writing compiler tht cn produce two different kinds of code: x86 ssembly code or Jv bytecode which will run on ny Jv Virtul Mchine), nd the user cn select which type of code she wnts t compile time. Give scenrio where she would choose to generte ssembly code i.e., give scenrio where generting ssembly code would be better thn producing byecode). ssume tht JVM is vilble on her trget mchine. 4 points). I would rther generte ssembly when it tkes long time to compile for prticulr trget rchitecture, or when the progrm is complex nd my tke long time to compile 2) Give scenrio where the progrmmer would choose to produce bytecode insted of ssembly. ssume the trget mchine hs n x86 rchitecture. 4 points). If the finl progrm is ment to be run on mny rchitectures, or if the behvior of the progrm is very different for different inputs so I my wnt to optimize it differently), I would prefer to generte bytecode. 3) To produce this kind of compiler, nme the compiler psses tht re common for both the x86 nd the bytecode versions. 2 points). Common psses: scnner, lexer, semntic ctions. Code genertion nd some of the optimiztion psses will hve to be different.

3 Prt 2: Regulr expressions, finite utomt nd scnners 15 points) well-prenthesized string follows three rules: i) The only chrcters re nd ). ii) There must be n equl number of nd ) in the string. iii) In ny prefix of the string, there cnnot be more ) thn. 1) I wnt to build finite utomton to cpture well-prenthesized strings. Is this possible? Why or why not? 5 points): Not possible: to stisfy rule ii, we would hve to count how mny there re to mke sure tht we get n equl number of ) s 2) Consider the following NF. Fill in the trnsition tble below with its corresponding DF using the subset construction. 8 points): 2 c b b c b Stte Finl? b c New Nme 1 N N 3 4, N 3, 5 3 4, 5 N 6, , 5 N 3, 5, , 7 Y 6 6 Y 7 3, 5, 7 Y 3, 5, point per line. Prtil credit given. 3) List which sttes should be merged when you reduce the bove DF 2 points): Merge new sttes 6 nd 7. Note tht new sttes 8 nd 5 hve the sme trnsitions, but 8 is finl nd 5 is not, so we cnnot merge them. -1 point for listing 8&5

4 Prt 3: Grmmrs 15 points) Let G be the grmmr: S B$ xb yb λ B z Using this grmmr, nswer the following questions. 1) Wht re the terminls nd non-terminls of this grmmr? 1 point) Terminls: x, y, z lso ccepted $ Non-terminls: S,, B 2) Drw the prse tree for the string xzz 4 point) S B x B z! z 3) Cn the lnguge of this grmmr be cptured by regulr expression? If so, give the regulr expression. If not, give short rgument why not. 5 points) No: this grmmr requires tht if we hve i xs, nd j ys, we must hve i + j + 1) zs, nd we cnʼt do tht with n F. Thus, we cnʼt do this with regulr expression. 4) Give grmmr tht cptures the mtched-prentheses strings from problem points) See the grmmr in prt 5, but with the x rule replced with λ -3 points if ^i)^i grmmr given -1 point if grmmr close i.e., more thn nested prentheses, only little off from correct)

5 Prt 4: LL prsers 25 points) nswer the questions in this prt using the sme grmmr from Prt 3. 1) Give the First sets for ech non-terminl in the grmmr 6 pts) FirstS) = x, y, z First) = x, y, λ FirstB) = z 2 points per set 2) Give the Follow sets for ech non-terminl in the grmmr 6 pts) FollowS) = { } Follow) = {z} FollowB) = {z, $} 2 points per set 3) Fill in the following prse tble 12 pts) x y z $ S B 5 1 point per box 4) Is this grmmr LL1) or not? Why or why not? 1 pts) Yes, this is LL1) there is no conflict in the prse tble.

6 Prt 5: LR0) Prsers 35 points) Use the following grmmr for the next two questions: S $ ) x 1) Fill in the missing informtion for the for the following CFSM 10 points) Stte 0 Stte 1 Stte 6 S! $! )! x Stte 3 x x S! $ Stte 2! x x! ) Stte 5! )! )! x Stte 4! ) )! )! )! x -1 point per missing configurtion 2) List the reduce sttes in the bove CFSM, nd the shift sttes 5 points) Reduce: 2, 6 Shift: 0, 3, 4, 5-1 point if stte 1 listed s reduce stte. OK if listed s shift stte 3) Is this n LR0) grmmr? Why or why not? 4 points) Yes, this is n LR0) grmmr: none of the sttes hve shift/reduce or reduce/reduce conflicts

7 4) Show the ctions tken by the prser s it prses x)x. ctions should either be of the form Shift X or Reduce R goto X) If the prser gets to stte 1, it ccepts. You cn use the following tble to help show your work you my not need ll of the rows) 16 points) Prse stck Remining input ction 0 x)x Shift, x)x Shift, )x Reduce 3 goto 4) )x Shift, x Shift, Reduce 3 goto 6) Reduce 2 goto 1) 0 1 ccept 2 points per ction. Prtil credit given +2 points for ech correct ction, for prtil credit)

8 Prt 6: LR1) Prsers ECE 573 only) 20 points): Consider the following slightly modified grmmr from Prt 5: S $ ) λ 1) Is this grmmr LR0)? Why or why not? 4 points) No, becuse there will be shift/reduce conflict immeditely in Stte 0. 2) Fill in the missing informtion from this prtil LR1) mchine you do not hve to dd ny sttes tht re not lredy shown) 16 points) Stte 0 Stte 1 Stte 2 S! $, {}! ), {$}!!, {$} S! $ Stte 3! ), {$} Stte 4! ), {$}! ), {)}!!, {)}! ), {)}! ), {)}!!, {)} Stte 5! ), {$} ) Stte 6! ), {$}! ), {$}!!, {$} 2 points per configurtion -1 point per wrong lookhed prtil credit given)