文件名称：编译原理龙书答案
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龙书答案 完整性高
编译原理龙书答案 完整性高 第二章 2.2 Exercises for Section 2.2 2.2.1 Consider the context-free grammar: S -> S S + | S S * | a Show how the string aa+a* can be generated by this grammar. Construct a parse tree for this string. What language does this grammar generate? Justify your answer. answer S -> S S * -> S S + S * -> a S + S * -> a a + S * -> a a + a * L = {Postfix expression consisting of digits, plus and multiple signs} 2.2.2 What language is generated by the following grammars? In each case justify your answer. S -> 0 S 1 | 0 1 S -> + S S | - S S | a S -> S ( S ) S | ε S -> a S b S | b S a S | ε ⧗ S -> a | S + S | S S | S * | ( S ) answer L = {0n1n | n>=1} L = {Prefix expression consisting of plus and minus signs} L = {Matched brackets of arbitrary arrangement and nesting, includes ε} L = {String has the same amount of a and b, includes ε} ？ 2.2.3 Which of the grammars in Exercise 2.2.2 are ambiguous answer No No Yes Yes Yes 2.2.4 Construct unambiguous context-free grammars for each of the following languages. In each case show that your grammar is correct. Arithmetic expressions in postfix notation. Left-associative lists of identifiers separated by commas. Right-associative lists of identifiers separated by commas. Arithmetic expressions of integers and identifiers with the four binary operators +, - , *, /. answer 1. E -> E E op | num 2. list -> list , id | id 3. list -> id , list | id 4. expr -> expr + term | expr - term | term term -> term * factor | term / factor | factor factor -> id | num | (expr) 5. expr -> expr + term | expr - term | term term -> term * unary | term / unary | unary unary -> + factor | - factor factor - > id | num | (expr) 2.2.5 Show that all binary strings generated by the following grammar have values divisible by 3. Hint. Use induction on the number of nodes in a parse tree. num -> 11 | 1001 | num 0 | num num Does the grammar generate all binary strings with values divisible by 3? answer prove any string derived from the grammar can be considered to be a sequence consisting of 11, 1001 and 0, and not prefixed with 0. the sum of this string is: sum = Σn (21 + 20) * 2 n + Σm (23 + 20) * 2m = Σn 3 * 2 n + Σm 9 * 2m It is obviously can divisible by 3. No. Consider string "10101", it is divisible by 3, but cannot derived from the grammar. Question: any general prove? 2.2.6 Construct a context-free grammar for roman numerals. Note: we just consider a subset of roman numerals which is less than 4k. answer wikipedia: Roman_numerals via wikipedia, we can categorize the single noman numerals into 4 groups: I, II, III | I V | V, V I, V II, V III | I X then get the production: digit -> smallDigit | I V | V smallDigit | I X smallDigit -> I | II | III | ε and we can find a simple way to map roman to arabic numerals. For example: XII => X, II => 10 + 2 => 12 CXCIX => C, XC, IX => 100 + 90 + 9 => 199 MDCCCLXXX => M, DCCC, LXXX => 1000 + 800 + 80 => 1880 via the upper two rules, we can derive the production: romanNum -> thousand hundred ten digit thousand -> M | MM | MMM | ε hundred -> smallHundred | C D | D smallHundred | C M smallHundred -> C | CC | CCC | ε ten -> smallTen | X L | L smallTen | X C smallTen -> X | XX | XXX | ε digit -> smallDigit | I V | V smallDigit | I X smallDigit -> I | II | III | ε 2.3 Exercises for Section 2.3 2.3.1 Construct a syntax-directed translation scheme that trans lates arithmetic expressions from infix notation into prefix notation in which an operator appears before its operands; e.g. , -xy is the prefix notation for x - y . Give annotated parse trees for the inputs 9-5+2 and 9-5*2.。 answer productions: expr -> expr + term | expr - term | term term -> term * factor | term / factor | factor factor -> digit | (expr) translation schemes: expr -> {print("+")} expr + term | {print("-")} expr - term | term term -> {print("*")} term * factor | {print("/")} term / factor | factor factor -> digit {print(digit)} | (expr) 2.3.2 Construct a syntax-directed translation scheme that trans lates arithmetic expressions from postfix notation into infix notation. Give annotated parse trees for the inputs 95-2* and 952*-. answer productions: expr -> expr expr + | expr expr - | expr expr * | expr expr / | digit translation schemes: expr -> expr {print("+")} expr + | expr {print("-")} expr - | {print("(")} expr {print(")*(")} expr {print(")")} * | {print("(")} expr {print(")/(")} expr {print(")")} / | digit {print(digit)} Another reference answer E -> {print("(")} E {print(op)} E {print(")"}} op | digit {print(digit)} 2.3.3 Construct a syntax-directed translation scheme that trans lates integers into roman numerals answer assistant function: repeat(sign, times) // repeat('a',2) = 'aa' translation schemes: num -> thousand hundred ten digit { num.roman = thousand.roman || hundred.roman || ten.roman || digit.roman; print(num.roman)} thousand -> low {thousand.roman = repeat('M', low.v)} hundred -> low {hundred.roman = repeat('C', low.v)} | 4 {hundred.roman = 'CD'} | high {hundred.roman = 'D' || repeat('X', high.v - 5)} | 9 {hundred.roman = 'CM'} ten -> low {ten.roman = repeat('X', low.v)} | 4 {ten.roman = 'XL'} | high {ten.roman = 'L' || repeat('X', high.v - 5)} | 9 {ten.roman = 'XC'} digit -> low {digit.roman = repeat('I', low.v)} | 4 {digit.roman = 'IV'} | high {digit.roman = 'V' || repeat('I', high.v - 5)} | 9 {digit.roman = 'IX'} low -> 0 {low.v = 0} | 1 {low.v = 1} | 2 {low.v = 2} | 3 {low.v = 3} high -> 5 {high.v = 5} | 6 {high.v = 6} | 7 {high.v = 7} | 8 {high.v = 8} 2.3.4 Construct a syntax-directed translation scheme that trans lates roman numerals into integers. answer productions: romanNum -> thousand hundred ten digit thousand -> M | MM | MMM | ε hundred -> smallHundred | C D | D smallHundred | C M smallHundred -> C | CC | CCC | ε ten -> smallTen | X L | L smallTen | X C smallTen -> X | XX | XXX | ε digit -> smallDigit | I V | V smallDigit | I X smallDigit -> I | II | III | ε translation schemes: romanNum -> thousand hundred ten digit {romanNum.v = thousand.v || hundred.v || ten.v || digit.v; print(romanNun.v)} thousand -> M {thousand.v = 1} | MM {thousand.v = 2} | MMM {thousand.v = 3} | ε {thousand.v = 0} hundred -> smallHundred {hundred.v = smallHundred.v} | C D {hundred.v = smallHundred.v} | D smallHundred {hundred.v = 5 + smallHundred.v} | C M {hundred.v = 9} smallHundred -> C {smallHundred.v = 1} | CC {smallHundred.v = 2} | CCC {smallHundred.v = 3} | ε {hundred.v = 0} ten -> smallTen {ten.v = smallTen.v} | X L {ten.v = 4} | L smallTen {ten.v = 5 + smallTen.v} | X C {ten.v = 9} smallTen -> X {smallTen.v = 1} | XX {smallTen.v = 2} | XXX {smallTen.v = 3} | ε {smallTen.v = 0} digit -> smallDigit {digit.v = smallDigit.v} | I V {digit.v = 4} | V smallDigit {digit.v = 5 + smallDigit.v} | I X {digit.v = 9} smallDigit -> I {smallDigit.v = 1} | II {smallDigit.v = 2} | III {smallDigit.v = 3} | ε {smallDigit.v = 0} 2.3.5 Construct a syntax-directed translation scheme that trans lates postfix arithmetic expressions into equivalent prefix arithmetic expressions. answer production: expr -> expr expr op | digit translation scheme: expr -> {print(op)} expr expr op | digit {print(digit)} Exercises for Section 2.4 2.4.1 Construct recursive-descent parsers, starting with the follow ing grammars: S -> + S S | - S S | a S -> S ( S ) S | ε S -> 0 S 1 | 0 1 Answer 1） S -> + S S | - S S | a void S(){ switch(lookahead){ case "+": match("+"); S(); S(); break; case "-": match("-"); S(); S(); break; case "a": match("a"); break; default: throw new SyntaxException(); } } void match(Terminal t){ if(lookahead = t){ lookahead = nextTerminal(); }else{ throw new SyntaxException() } } 2） S -> S ( S ) S | ε void S(){ if(lookahead == "("){ S(); match("("); S(); match(")"); S(); } } 3） S -> 0 S 1 | 0 1 void S(){ switch(lookahead){ case "0": match("0"); S(); match("1"); break; case "1": // match(epsilon); break; default: throw new SyntaxException(); } } Exercises for Section 2.6 2.6.1 Extend the lexical analyzer in Section 2.6.5 to remove com ments, defined as follows: A comment begins with // and includes all characters until the end of that line. A comment begins with /* and includes all characters through the next occurrence of the character sequence */. 2.6.2 Extend the lexical analyzer in Section 2.6.5 to recognize the relational operators <, <=, ==, ! =, >=, >. 2.6.3 Extend the lexical analyzer in Section 2.6.5 to recognize float ing point numbers such as 2., 3.14, and . 5. Answer Source code: commit 8dd1a9a Code snippet(src/lexer/Lexer.java)： public Token scan() throws IOException, SyntaxException{ for(;;peek = (char)stream.read()){ if(peek == ' ' || peek == '\t'){ continue; }else if(peek == '\n'){ line = line + 1; }else{ break; } } // handle comment if(peek == '/'){ peek = (char) stream.read(); if(peek == '/'){ // single line comment for(;;peek = (char)stream.read()){ if(peek == '\n'){ break; } } }else if(peek == '*'){ // block comment char prevPeek = ' '; for(;;prevPeek = peek, peek = (char)stream.read()){ if(prevPeek == '*' && peek == '/'){ break; } } }else{ throw new SyntaxException(); } } // handle relation sign if("<=!>".indexOf(peek) > -1){ StringBuffer b = new StringBuffer(); b.append(peek); peek = (char)stream.read(); if(peek == '='){ b.append(peek); } return new Rel(b.toString()); } // handle number, no type sensitive if(Character.isDigit(peek) || peek == '.'){ Boolean isDotExist = false; StringBuffer b = new StringBuffer(); do{ if(peek == '.'){ isDotExist = true; } b.append(peek); peek = (char)stream.read(); }while(isDotExist == true ? Character.isDigit(peek) : Character.isDigit(peek) || peek == '.'); return new Num(new Float(b.toString())); } // handle word if(Character.isLetter(peek)){ StringBuffer b = new StringBuffer(); do{ b.append(peek); peek = (char)stream.read(); }while(Character.isLetterOrDigit(peek)); String s = b.toString(); Word w = words.get(s); if(w == null){ w = new Word(Tag.ID, s); words.put(s, w); } return w; } Token t = new Token(peek); peek = ' '; return t; } Exercises for Section 2.8 2.8.1 For-statements in C and Java have the form: for ( exprl ; expr2 ; expr3 ) stmt The first expression is executed before the loop; it is typically used for initializ ing the loop index. The second expression is a test made before each iteration of the loop; the loop is exited if the expression becomes O. The loop itself can be thought of as the statement {stmt expr3 ; }. The third expression is executed at the end of each iteration; it is typically used to increment the loop index. The meaning of the for-statement is similar to expr1 ; while ( expr2 ) {stmt expr3 ; } Define a class For for for-statements, similar to class If in Fig. 2.43. Answer class For extends Stmt{ Expr E1; Expr E2; Expr E3; Stmt S; public For(Expr expr1, Expr expr2, Expr expr3, Stmt stmt){ E1 = expr1; E2 = expr2; E3 = expr3; S = stmt; } public void gen(){ E1.gen(); Label start = new Lable(); Lalel end = new Lable(); emit("ifFalse " + E2.rvalue().toString() + " goto " + end); S.gen(); E3.gen(); emit("goto " + start); emit(end + ":") } } 2.8.2 The programming language C does not have a boolean type. Show how a C compiler might translate an if-statement into three-address code. Answer Replace emit("isFalse " + E.rvalue().toString() + " goto " + after); with emit("ifNotEqual " + E.rvalue().toString() + " 0 goto " + after); or emit("isNotEqualZero " + E.rvalue().toString() + " goto " + after);
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----第7章.doc(215KB)
----第3章.doc(344KB)
----第4章.doc(420KB)
----第5章.doc(115KB)
----第8章.doc(277KB)
----第2章.doc(183KB)
----第6章.doc(278KB)
----编译原理(龙书)答案第三章.pdf(524KB)