LEX program to check a Date (dd/mm/yyyy)

Hello dude..   

            I would like to discuss a problem asked by my teacher in our internal lab exam- LEX program to check the validity of a Date. Ahem..In normal programming (i.e, that with C or Java) this is not a big deal!! But in Lex paradigm,it require some more effort..!!

                                                
                                                   

*Check the syntax of Date::

* If we have to check the syntax only ,i.e., dd/mm/yyyy apply the simple logic

[0-9][0-9] \ / [0-9][0-9] \ / [0-9] [0-9] [0-9] [0-9]

* But date should be within range [01-31] ,month be within [01-12] and assume year be within the range [1000-2999]. To apply these constraints, we redefine rules as:

"01-31"----> ([0-2][0-9] | 3 [0-1])
"01-12"----> (0[1-9] | 1[0-2])
"1000-2999"---->  ([1-2][0-9][0-9][0-9])

* Thus we can simply check the syntax of given Date as:

([0-2][0-9] | 3 [0-1])  \ /  (0[1-9] | 1[0-2]) \ / ([1-2][0-9][0-9][0-9])

*To Check the validity of date::

-->  In months 01,03,05,07,08,10 &12 , there will be at most 31 days.

([0-2][0-9]|[3][0-1])\/((0(1|3|5|7|8))|(10|12))\/([1-2][0-9][0-9][-0-9]) 

--> In months 04,06,09 & 11 may have at most 30 days

([0-2][0-9]|30)\/((0(4|6|9))|11)\/([1-2][0-9][0-9][0-9])

-->February has 28 days (in linear and non-linear years)

([0-1][0-9]|2[0-8])\/02\/([1-2][0-9][0-9][0-9])

-->If February has a day 29, check whether it is leap year or not..!!

29\/02\/([1-2][0-9][0-9][0-9])

 {

extract year value;

check whether it is a leap year;

}

-->Extract year value

   1.Iterate upto two "/" in date are over.(i.e.,in dd/mm/yyyy pass over two "/"s to reach at year value.

   2.read all susequent characters (they all are part of year value - yyyy)

while(yytext[i]!='/')i++;   //reach at first "/"
i++;                        // increment pointer
while(yytext[i]!='/')i++;   //reach at next "/"
i++;                        // increment pointer
while(i<yyleng)             // read all characters upto end of the string
yr=(10*yr)+(yytext[i++]-'0');// extract integer value of year

--> Check whether it is Leap year or not:

if(yr%4==0||(yr%100==0&&yr%400!=0))

Well... the complete lex program is as follows..!!

Date.l

%{
#include<stdio.h>
int i=0,yr=0,valid=0;
%}
%%
([0-2][0-9]|[3][0-1])\/((0(1|3|5|7|8))|(10|12))\/([1-2][0-9][0-9][-0-9]) {valid=1;}

([0-2][0-9]|30)\/((0(4|6|9))|11)\/([1-2][0-9][0-9][0-9]) {valid=1;}

([0-1][0-9]|2[0-8])\/02\/([1-2][0-9][0-9][0-9]) {valid=1;}

29\/02\/([1-2][0-9][0-9][0-9]) { while(yytext[i]!='/')i++; i++;while(yytext[i]!='/')i++;i++;while(i<yyleng)yr=(10*yr)+(yytext[i++]-'0'); if(yr%4==0||(yr%100==0&&yr%400!=0))valid=1;}

%%
main()
{
yyin=fopen("vpn.txt","r");
yylex();
if(valid==1) printf("It is a valid date\n");
else printf("It is not a valid date\n");
}
int yywrap()
{
return 1;
}

OUTPUT

Content in input file (here, vpn.txt)	Output
12/12/2011	It is a valid date
32/10/2009	It is not a valid date
29/02/2008	It is a valid date
31/04/1990	It is not a valid date
29/02/2007	It is not a valid date

Hope this post was useful for you. If you have any suggestions or doubts,please do comment here...!!
Thank you