1. Consider the following program in a PCAT-like language.

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1 CS4XX INTRODUCTION TO COMPILER THEORY MIDTERM EXAM QUESTIONS (Each question carries 20 Points) Total points: Consider the following program in a PCAT-like language. PROCEDURE main; TYPE t = FLOAT; VAR x,y,z:integer; PROCEDURE sub1; TYPE t = STRING; VAR a,y,z : BOOLEAN; PROCEDURE sub2; VAR a,b,z: t; BEGIN (* body of sub2 *) END BEGIN (* body of sub1 *) END BEGIN (* body of main *) END List all the bound identifiers in scope in the body of procedure sub2, and give the kind of entity to which the identifier is bound. If the entity is a variable, also give its type. Make the following assumptions about the language: static scoping is used; all keywords are in upper-case and all identifiers are in lower-case; procedures may be recursive. 2. Consider the following grammar for boolean expressions. E! E or E E! E and E E! not E E! (E) E! true E! false E! id (a) Show that this grammar is ambiguous. 1

2 (b) Rewrite the grammar to remove the ambiguity and enforce the intended precedence order by introducing new non terminals. Make sure that your revised grammar accepts the same language as the original. 3. Consider the following context-free grammar. (It corresponds roughly to the syntax of lists in the programming language LISP.) S is the start symbol, and the terminals are a, ( ). S! ( ) S! a S! ( A ) A! S A! A, S (a) Show precisely why this grammar is not LL(1). (Hint: This will require computing some, but not all, of the FIRST and FOLLOW sets.) (b) Rewrite this grammar to make it suitable for recursive descent parsing. (c) On the basis of your revised grammar from part (b), write a recursive procedure S that parses this grammar by recursive descent. Your procedure may be written in C, C++, Java or pseudocode. You may assume the existence of a global variable token that holds the next input token, a function advance() that reads the next token into token, and a function error that may be called if the input is not in the language generated by the grammar. (d) Rewrite your solution to (c) to remove all tail-recursion and inline functions that are used only once. You should end up with a single function s(). 4. Consider a Unix-shell-like command language for invoking executable files (by naming them), redirecting standard input or output from or to a data file (using < and >), and piping the output from one executable to the input of another (using ). Suppose any number of executables can be piped together, but input can be redirected from a file only to the first executable, and output to a file only from the last executable. Legal commands include simple > outfile < infile and lexer < goodfile.pcat parser ppast > bug.ast. 2

3 (a) Write a plain (not extended) BNF grammar for this command language, assuming the terminals are filename, <, >, and. (b) Draw parse trees for the two commands given above, according to your grammar. 5. Consider the (very artificial) language, over the alphabet of letters and digits and the dollar sign ($), having the following three kinds of tokens: numbers, consisting of one or more consecutive digits; short identifiers, consisting of a single letter; and long identifiers, consisting of one or more letters followed by a single $. a. Write down a regular expression for each of the three token patterns. b. Draw a DFA for each expression. c. Combine the DFA s into a common NFA with a common start state and distinct final states for each token, as described in class. d. Convert your combined NFA into a DFA, using whatever algorithm you like. e. Suppose you use this DFA to perform lexical analysis, backtracking to the most recently encountered final state when necessary, as described in class. Give an example of an input that will require backtracking and 1. re-reading exactly 5 characters. 3

4 CS4XX INTRODUCTION TO COMPILER THEORY MIDTERM EXAM SOLUTIONS 1. Here s the list: Identifier Kind Type main procedure x variable INTEGER sub1 procedure t type (= STRING) y variable BOOLEAN sub2 procedure a variable t (= STRING) b variable t (= STRING) z variable t (= STRING) 2. (a) Here are two leftmost derivations for the same sentence: E => E or E => id or E => id or E and E => id or id and E => id or id and id E => E and E => E or E and E => id or E and E => id or id and E => id or id and id (b) Here s a suitably rewritten grammar: E! E or T E! T T! T and F T! F F! not F F! (E) F! true F! false F! id This problem is completely analogous to arithmetic expressions. Note that in disambiguating, enforcement of precedence order, but also made both and and or left-associative. The alternative with E! T or E T! F and T is also an acceptable answer, since the problem didn t ask for a particular associativity. 4

5 3. (a). A grammar is LL(1) if and only if its predictive parsing table has no multiply-defined entries. Consider the right-hand sides of the first and third productions for S. The terminal ( is in FIRST(()) and also in FIRST((A)). Therefore the table entry for the row labeled S and the column labeled ( will have (at least) two entries for these two productions. So the grammar cannot be LL(1). (Note that there was no need to calculate any FOLLOW() sets after all!) (b) This requires removing left-recursion and left-factoring: S! ( S S! a S! ) S! A ) A! SA A!,SA A! (c) Here s C/Java-like code: void s() { if (token == ( ) { s1(); else if (token == a ) else error(); void s1() { if (token == ) ) else { a(); if (token == ) ) else error(); void a() { s(); a1(); 5

6 void a1() { if (token ==, ) { s(); a1(); (d) First rewrite a1 as a while loop; then inline a1 into a, a into s1, and finally s1 into s. s() { if (token == ( ) { if (token == ) ) else { s(); while (token ==, ) { s(); ; if (token == ) ) else error(); else if (token == a ) else error(); 4.(a) command! filename rdin rdout! filename rdout rdint! filename rdin pipe rdin! < filename! rdout! > filename! 6

7 pipe! filename pipe! filename rdout 7

8 8

9 (e) Example: abcde0 (Only after the 0 is read does the machine discover that it is not reading a long abcde rather than the short a. Characters bcde0 will be rescanned on the next invocation of the lexer.) 9

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