CMPSC 470: Compiler Construction

Transcription

1 CMPSC 47: Compiler Construction Plese complete the following: Midterm (Type A) Nme Instruction: Mke sure you hve ll pges including this cover nd lnk pge t the end. Answer ech question in the spce provided. If you need more spce thn provided, use the cks of the pges nd e sure to write the Answer continued on next pge. Prolem Mx. Points Received Points Totl

2 . ( points) True/Flse Questions. True () True / Flse Compiler is progrm tht red source progrm in one lnguge nd trnslte it into n equivlent trget progrm in nother lnguge. True () True / Flse Three ddress code is one of the forms for intermedite code, which is n ssemly like instruction hving three opernds per opertion. Flse (c) True / Flse Front-end of compiler performs lexicl nlysis, syntx nlysis, nd mchine dependent code genertion. True (d) True / Flse Lexicl nlyzer dd lexemes into symol tle when it is new identifier found in input strem, produce tokens. Flse (e) True / Flse Linker comines reloctle mchine codes with other reloctle oject files or lirry files, nd generte intermedite code.

3 2. (8 points) You hve the following regulr expression tht hs two lphets: nd. Convert the regulr expression to n NFA using McNughton-Ymd-Thompson lgorithm (the method discussed in clss). You don t need to drw its prse tree. (* )* strt

4 3. (5 points) You hve the following set of ptterns of tokens nd token vlues to define BNF. Pttern of tokens for BNF (ckus norml form) Return Exmples vlue :== :== Nonterminl whose nme is composed of more thn one lower cse letters nd _ without white spce, nd enclosed with < nd >. <expr>, <zip_code>, etc. Terminl, ny string enclosed with doule quottion mrks 2 "", "<", "=", ".", ",", "Jr.", etc. Write JFlex progrm under the ox tht looks for the ove ptterns, nd returns the specified vlue whenever one of these ptterns is found. Any input tht is not prt of one of these ptterns should e ignored. Hint: \c specifies one chrcter symol. For exmple, \" indictes doule quottion mrk. import jv.io.*; %% %clss Lexer %yccj %int %{ pulic sttic void min(string[] rgs) throws IOException { if(rgs.length < ) return; Lexer lexer = new Lexer(new jv.io.filereder(rgs[])); while(token!= ) { int token = lexer.yylex(); } } %} DQ = \" ASSIGN = ":==" NONTER = "<"[-z_]+">" TERMIN = DQ[^DQ]*DQ %% ASSIGN { return ; } NONTER { return ; } TERMIN { return 2; } [^] { /* skip */ }

5 4. (8 points) Convert the right NFA to DFA using suset construction lgorithm (the method discussed in clss), nd show the finl DFA grph or its trnsition tle. Show your work for prtil points. strt A=e-clo()=36 DTrns[A,] = e-clo(move(36,)) = e-clo(2,4,6) = 2346 = B DTrns[A,] = e-clo(move(36,)) = e-clo(7) = 7 = C DTrns[B,] = e-clo(move(2346,)) = e-clo(4,6) = 46 = D DTrns[B,] = e-clo(move(2346,)) = e-clo(5,7) = 57 = E DTrns[C,] = e-clo(move(7,)) = e-clo() = - DTrns[C,] = e-clo(move(7,)) = e-clo(7) = C DTrns[D,] = e-clo(move(46,)) = e-clo(6) = 6 = F DTrns[D,] = e-clo(move(46,)) = e-clo(5,7) = E DTrns[E,] = e-clo(move(57,)) = e-clo() = - DTrns[E,] = e-clo(move(57,)) = e-clo(7) = C DTrns[F,] = e-clo(move(6,)) = e-clo(6) = F DTrns[F,] = e-clo(move(6,)) = e-clo(7) = C ======================== A: :B 7:C B: :D 57:E C:7-7:C D:46 6:F 57:E E:57-7:C F:6 6:F 7:C strt A:36 B:2346 D:46 F:6 C:7 E:57

6 5. (7 points) Show tht the following grmmr is miguous. Upper cse letters represent nonterminls, nd underlined lowercse letters represent tokens. SS TTUU TTUU TT SS cc UU TT cc Sentence ccc hs 2 prse trees since it hs 2 left most derivtion. SS TT ccccuu cccctt ccccss cccctt ccccccccuu cccccccccc SS TT SS TT ccccuu ccccccccuu cccccccccc

7 6. (6 points) The following grmmr is miguous. SS SS = SS SS + SS SS SS (SS) iiii Rewrite the ove grmmr to remove its miguity, nd to ccept the following rules: nd + opertors re left-ssocitive = opertor is e right-ssocitive. The order of precedence is: (lowest precedence) =, +,, prentheses (highest precedence) SS EE = SS EE EE EE + TT TT TT TT FF FF FF (SS) iiii or FF (SS) SS iiii

8 7. ( points) Remove ll left recursion from the following grmmr. Upper cse letters represent nonterminls, nd underlined lowercse letters represent tokens. Show your work nd your finl grmmr to get full points. SS TT TT TT SScc Step Remove immedite left recursion in SS. Nothing cn e done. Step 2 Sustituting SS in TT SScc yields TT TT cc cc = TT cc cc Now we hve, TT TT TT cc cc. Removing immedite left recursion yields TT cc TT TT TT cc TT Finlly, we hve the following grmmr. SS TT TT cc TT TT TT cc TT

9 8. (6 points) In the following CFG, EE, BB nd TT re nonterminls, EE is the strting symol, nd (, ),, oooo, iiii, nd nnnnnn re tokens. Find first nd follow of ll nonterminls: FFFFFFFFFF(EE), FFFFFFFFFF(BB), FFFFFFFFFF(TT), FFFFFFFFFFFF(EE), FFFFFFFFFFFF(BB) nd FFFFFFFFFFFF(TT). (Note: if you solve without, you will get 2 points. If you solve with, you will get 6 points) EE EE BB EE nnnnnn EE TT BB TT (EE) iiii 4 points: FFFFFFFFFF(EE) =, (, nnnnnn, iiii, EE nnnnnn EE EE TT (EE) iiii EE EEEEEE BBBB 2 points: FFFFFFFFFF(BB) = 2 points: FFFFFFFFFF(TT) = (, iiii 2 points: FFFFFFFFFFFF(EE) = $, ), EE is strting symol EE EEEEEE TT (EE) 4 points: FFFFFFFFFFFF(BB) = $, (, ),, iiii, nnnnnn EE EEEEEE EE EEEEEE EEEEEE EEEE 2 points: FFFFFFFFFFFF(TT) = $, ), EE TT nnnnnn, FFFFFFFFFF(EE) (, iiii FFFFFFFFFF(EE) FFFFFFFFFF(EE) $ FFFFFFFFFFFF(EE) FFFFFFFFFF(BB) = FFFFFFFFFFFF(EE) ) FFFFFFFFFFFF(EE) FFFFFFFFFF(EE) = nnnnnn,, (, iiii, FFFFFFFFFFFF(BB) FFFFFFFFFFFF(EE) = $, ), FFFFFFFFFFFF(BB) FFFFFFFFFFFF(EE) = $, ), FFFFFFFFFFFF(TT) Without, 2 points: FFFFFFFFFF(EE) = (, nnnnnn, iiii, (, nnnnnn, iiii, with 2 points: FFFFFFFFFF(BB) = 2 points: FFFFFFFFFF(TT) = (, iiii 2 points: FFFFFFFFFFFF(EE) = $, ), EE is strting symol $ FFFFFFFFFFFF(EE) EE EEEEEE FFFFFFFFFF(BB) = FFFFFFFFFFFF(EE) TT (EE) ) FFFFFFFFFFFF(EE) 2 points: FFFFFFFFFFFF(BB) = (, nnnnnn, iiii $, (, ),, iiii, nnnnnn with EE EEEEEE FFFFFFFFFF(EE) = (, nnnnnn, iiii FFFFFFFFFFFF(BB) 2 points: FFFFFFFFFFFF(TT) = $, ), EE TT FFFFFFFFFFFF(EE) = $, ), FFFFFFFFFFFF(TT)

10 (extr 5 points) Minimize the following DFA using Moore's lgorithm (the method discussed in clss). Note: show your work to get this extr points.. Initilly, there re two sets{c,d,e} nd {,,f,g} 2. The second itertion, {c,d,e} {c,d,e}, {,} {,,f,g}, {c,d,e} {,,f,g}, {,} {c,d,e}, strt c {f,g} {f,g}, {f,g} {f,g}, Since the ehviors of {,} nd {f,g} re different, split them, nd we hve the following sets {{,}, {c,d,e}, {f,g}} d,, g e f 3. The third itertion, {c,d,e} {c.d.e}, {} {,}, {} {f,g}, {c,d,e} {f,g}, {} {c,d,e}, {} {c,d,e}, {f,g} {f,g}, {f,g} {f,g}, Since the ehviors of {} nd {} re different, split them, nd we hve the following sets {{}, {}, {c,d,e}, {f,g}} 4. The fourth itertion, {c,d,e} {c,d,e}, {} {}, {} {f,g}, {f,g} {f,g}, we hve the following sets {{}, {}, {c,d,e}, {f,g}} {c,d,e} {fg}, {} {f,g}, {} {c,d,e}, {f,g} {f,g}, 5. no chnge strt c,d,e f,g,