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Using Lex

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The Structure of a Lex Program

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(Definition section)

%%(Rules section)

%%

(User subroutines section)

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%{/*

* this sample demonstrates (very) simple recognition:

* a verb/not a verb.

*/

%}

%%

[\t ]+ /* ignore white space */ ;

is |

am |

are |

were |

was |be |

being |

been |

do |

does |

did |

will |would |

should |

can |

could |

has |

have |

had |

go { printf("%s: is a verb\n", yytext); }

[a-zA-Z]+ { printf("%s: is not a verb\n", yytext); }

.|\n { ECHO; /* normal default anyway */ }

%%

main(){

yylex();

}

Example 1-1: Word recognizer ch1-02.l

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The definition section

Lex copies the material between %{ and

%} directly to the generated C file, so you

may write any valid C codes here

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Rules section

Each rule is made up of two parts

A pattern

An action

E.g.

[\t ]+ /* ignore white space */ ;

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Rules section (Contd) E.g.

is |am |are |were |was |be |being |

been |do |does |did |will |would |

should |can |could |has |have |had |go { printf("%s: is a verb\n", yytext); }

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Rules section (Contd)

E.g.

[a-zA-Z]+ { printf("%s: is not a verb\n", yytext); }

.|\n { ECHO; /* normal default anyway */ }

Lex had a set of simple disambiguating rules:1. Lex patterns only match a given input character or

string once

2. Lex executes the action for the longest possiblematch for the current input

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User subroutines section

It can consists of any legal C code

Lex copies it to the C file after the end of

the Lex generated code

%%

main()

{ yylex();

}

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Examples of Regular Expressions [0-9]

[0-9]+

[0-9]*

-?[0-9]+

[0-9]*\.[0-9]+

([0-9]+)|([0-9]*\.[0-9]+)

-?(([0-9]+)|([0-9]*\.[0-9]+))

[eE][-+]?[0-9]+

-?(([0-9]+)|([0-9]*\.[0-9]+))([eE][-+]?[0-9]+)?)

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Example 2-1

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%%

[\n\t ] ;

-?(([0-9]+)|([0-9]*\.[0-9]+)([eE][-+]?[0-9]+)?) { printf("number\n"); }

. ECHO;

%%

main()

{ yylex();

}

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A Word Counting Program

The definition section

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%{unsigned charCount = 0, wordCount = 0, lineCount = 0;

%}

word [^ \t\n]+

eol \n

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A Word Counting Program (Contd)

The rules section

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{word} { wordCount++; charCount += yyleng; }{eol} { charCount++; lineCount++; }

. charCount++;

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A Word Counting Program (Contd) The user subroutines section

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main(argc,argv)

int argc;

char **argv;

{

if (argc > 1) {

FILE *file;

file = fopen(argv[1], "r");

if (!file) {

fprintf(stderr,"could not open %s\n",argv[1]);

exit(1);

}

yyin = file;}

yylex();

printf("%d %d %d\n",charCount, wordCount, lineCount);

return 0;

}

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Another Problem%{

letter [A-Za-z]

digit [0-9]

}%

%%

begin {return (BEGIN);}

end {return (END);}

:= {return (ASGOP);}

{letter} ({letter}|{digit})* {yyval = enter_id();

return(ID);}

{digit}+ {yyval = enter_num();

return (NUM);}

%%

enter_id()

{ /* enter the id in the symbol table and returns entry number */}enter_num()

{ /* enter the number in the constant table and return entry number */}

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