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This lex program is intended to eliminate comments from a C program. This task is simple since in a C program, comments will be associated only with '//' or '/*...*/ only. So our aim is to detect the occurence of these characters and ignore subsequent comments.
%{
#include<stdio.h>
%}
%%
\/\/.* ;
\/\*(.*\n)*.*\*\/ ;
%%
main()
{
yyin=fopen("mypgm.c","r");
yylex();
}
int yywrap()
{
return 1;
}
\/\/.* ; eliminates single line coments. i.e., comments of the form ' //comment.. .. .. ;'
It simply look on '/'immediately followed by another '/' and ignore that line.
\/\*(.*\n)*.*\*\/ ; eliminates multiple comments. i.e, code fragment lies within /*...*/
Have a look on this Example:
Consider mypgm.c
//This is a single comment;
This is not a comment;
/*But..
These are mulptiple comments
*/
Again, It is not a comment
The output will be,
This is not a comment;
Again, It is not a comment
Posted in:
Language processing programs
,
LEX Programs
8 comments :
Please give proper solution...
can anyone tell how does lex program work???
more than 1 multi line comments will result into a greedy match... eliminating the code in between too...
@Anonymousletter [a-zA-z]
digit [0-9]
%%
"if"|"else"|"for"|"while"|"do" printf("KeyWord %s\n",yytext);
"?"/[^:]*":" printf("Ternary %s\n",yytext);
\"[^\"]*\" printf("String Literal %s\n",yytext);
"/*"[^*]*"*/"|"/*"[^/]*"*/" ;
\/\/.* ;
%%
main(int argc,char *argv[])
{
yyin=fopen(argv[1],"r");
yyout=fopen(argv[2],"w+");
yylex();
}
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i dont know
What about a string containing /*.
Eg-
printf("This is a comment - /* comment*/");
This should not be removed.
@Anonymous
+1
can you please explain how the remaining code is printed? where is printing line in main program
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