Sample Midterm Solutions COMS W4115 Programming Languages and Translators Monday, October 12, 2009

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1 Deprtment of Computer cience Columbi University mple Midterm olutions COM W4115 Progrmming Lnguges nd Trnsltors Mondy, October 12, 2009 Closed book, no ids. ch question is worth 20 points. Question 5(c) is extr credit, 10 points. 1. Briefly explin the essentil difference between ) cll-by-vlue nd cll-by-reference. How re prmeters pssed in C nd Jv? In cll-by-vlue, the ctul prmeter is evluted or copied; its vlue is plced in the loction of the corresponding forml prmeter of the clled procedure. In cll-by-reference, the ddress of the ctul prmeter is pssed to the cllee s the vlue of the corresponding forml prmeter. C nd Jv use cll-by-vlue. b) sttic scope nd dynmic scope. How is scoping done in C nd Jv? cope specifies the textul region of progrm in which there is n ctive ssocition (binding) between nme nd the object it represents. ttic scoping ssocites the use of nme with the closest lexiclly enclosing declrtion. Dynmic scoping chooses the most recent ctive declrtion t runtime. C nd Jv use sttic scoping. 2. Jv compiltion. ) Drw block digrm showing how progrms re compiled nd executed in Jv. 1

2 source progrm Trnsltor Intermedite progrm input Jv Virtul Mchine output Jv Compiler b) Wht is Jv just-in-time compiler? The intermedite progrm from the Jv trnsltor is sequence of rchitecturlly neutrl bytecodes tht re interpreted by the Jv virtul mchine. A Jv just-in-time compiler trnsltes the bytecodes into n equivlent sequence of ntive code for the trget mchine in order to chieve fster run-time performnce. 3. Let L be the set of strings of the form bxb where x is string of s, b s, nd c s tht does not contin b s substring. ) Write regulr expression for L. Let R = b( b*c)*b+. b) how how your regulr expression genertes the string bbcb. The prefix b of R genertes the prefix b of the string. Then ( b*c)* genertes bc. Finlly b+ genertes b nd the finl of R genertes the finl of the string. 2

3 c) Construct deterministic finite utomton for L.,c b b b c All unspecified trnsitions re to ded stte. d) how how your utomton processes the input bbcb. b b c b Consider the context-free grmmr G: + *. ) how tht G is mbiguous by constructing ll prse trees for + *. + * * + 3

4 b) Construct n unmbiguous grmmr for L(G) in which + is left ssocitive, * is nonssocitive nd of higher precedence thn +. (1) + T (2) T (3) T * (4) T Drw the prse tree in your grmmr for the input string + *. + T T * 5. yntx-directed trnsltion. ) Construct n DT tht mps postfix expressions contining the digits 0, 1,, 9 nd the binry opertors nd % into equivlent infix expressions. 4

5 Here is DT using synthesized ttribute.v of type string for the nonterminl. In the semntic rules, we hve used juxtposition s the conctention opertor {.v = "(" 1.v " - " 2.v ")"; } 1 2 % {.v = "(" 1.v " % " 2.v ")"; } 0 {.v = "0"; } 1 {.v = "1"; } 2 {.v = "2"; } 3 {.v = "3"; } 4 {.v = "4"; } 5 {.v = "5"; } 6 {.v = "6"; } 7 {.v = "7"; } 8 {.v = "8"; } 9 {.v = "9"; } b) how how your DT trnsltes the expression 123 %. Here is n nnotted prse tree for 123 % with the vlue of.v shown t ech node. The output is the infix expression (1%(2-3)), the vlue of.v t the root of the tree...v=(1%(2-3)) %.v = 1.v = (2-3) 1.v = 2.v = c) Modify your DT so tht it uses the fewest possible number of prentheses in the output. 5

6 Here is modified DT tht uses two synthesized ttributes,.v nd.p, for the nonterminl in the productions..v is the infix string ssocited with the nonterminl nd.p is n integer giving the precedence level of the opertor ssocited with. We ssume the precedence level of % is 2, nd is 1. For convenience, we set the precedence level of digit to 3. To determine whether we need to put prentheses round subexpression opernd, we use the following rule. uppose we hve the prse tree node: 1 2 op Then we put prentheses round 1.v if 1.p is less thn the precedence level of op; we put prentheses round 2.v if 2.p is less thn or equl to the precedence level of op. Otherwise, we do not dd prentheses. The prentheses re there to mke sure we evlute the infix expression in the sme order s the postfix expression. Here is the modified DT: { if ( 2.p == 1) 2.v = "(" 2.v ")";.v = 1.v "-" 2.v;.p = 1; } 1 2 % { if ( 1.p == 1) 1.v = "(" 1.v ")"; if ( 2.p 2) 2.v = "(" 2.v ")";.v = 1.v "%" 2.v;.p = 2; } 0 {.v = "0";.p = 3; } 1 {.v = "1";.p = 3; } 2 {.v = "2";.p = 3; } 3 {.v = "3";.p = 3; } 4 {.v = "4";.p = 3; } 5 {.v = "5";.p = 3; } 6 {.v = "6";.p = 3; } 7 {.v = "7";.p = 3; } 8 {.v = "8";.p = 3; } 9 {.v = "9";.p = 3; } 6

7 Here is the nnotted prse tree for the input 123-% using the modified DT:.v = 1%(2-3).p = 2.v = 1.v = 2-3.p = 3.p = 1 % 1.v = 2.v = 3.p = 3.p =