Lexical Analysis Questions and Answers

Answer: C

The correct automaton is (C).

In (C), a letter takes the start state to the final state. From there, \(\epsilon\)-moves lead into two branches, one reading a single letter and one reading a single digit, and both rejoin a state with an \(\epsilon\)-move back to the final state. So after the first letter the machine can read any number of letters and digits in any order: exactly letter (letter | digit)*.

(A) has an \(\epsilon\)-move from the start state straight to the final state, so it also accepts the empty string and strings starting with a digit. (B) commits to one branch after the first letter, so it accepts a1 and ab but not a1b, which mixes letters and digits. (D) loops only on the pair 'letter digit', so it rejects ab.

Answer: C

Characters are grouped into tokens by the scanner.

The scanner, or lexical analyser, is the first phase of a compiler. It reads the source program character by character and groups the characters into tokens such as keywords, identifiers, constants and operators, removing white space and comments.

The parser takes the tokens and checks that they form valid syntax. Code optimisation improves intermediate code, and the code generator produces target code; both come much later than token formation.

Answer: C

The variables are yytext, yyleng.

In a LEX-generated scanner, yytext points to the text of the lexeme just matched, and yyleng holds its length. Actions in the rules use them, for example to copy the matched identifier or number.

yylex is the scanning function itself, which returns the next token, and yylval is the variable used to pass a token's value to a YACC parser. lextext, lexval and lexyy are not LEX's global variables (lex.yy.c is the name of the generated file).