lex程序上没有注释行数 [英] lex program on counting no of comment lines

问题描述

这里程序不计算注释行，单行注释和多行注释的数目，并给出以file.txt作为输入的总注释输出

here the program counts the no of comment lines, single line comments and multi line comments and gives a total comments output with a file.txt as input

file.txt

//hellow world
/*hello world1*/
/*hello world2
*/
/*hello world3
hello world3.1*/
#include<>

count.l

    %{
    #include<stdio.h>
    #include<stdlib.h>
    int a=0,b=0,c=0,d;
%}
%%
"//".* {a++;}
"/*" {b++;}
.*"*/" {b--;c++;}
%%
void main(int argc,char *argv[]){
    yyin=fopen(argv[1],"r");
    yylex();
    printf("single line %d \nmultiline %d \n",a,c);
    d=a+c;
    printf("total %d \n",d);
}

我得到的输出是

./a.out file.txt

./a.out file.txt

hello world2 

hello world3



#include<>
single line 1 
multiline 3 
total 4 

我需要获得的输出仅仅是

the output that i need to get is just

#include<>
single line 1 
multiline 3 
total 4 

我也这样尝试过，我在"/*"的前面添加了.*，就像这样"/*".*，然后它将删除该行中的"*/"，并给我多行注释计数为2. 我尝试了各种方法，但我有点卡住了.

I tried this way too i added .* infront of "/*" like this "/*".* then it would delete the "*/" in that line and give me the multi line comment count as 2. i tried various ways but im kinda stuck.

推荐答案

添加了精确的逻辑以使其更好地工作.

added precise logic to make it work better.

    %{
    #include<stdio.h>
    #include<stdlib.h>
    int a=0,c=0,d,e=0;
%}
%%
"/*" {if(e==0)e++;}
"*/" {if(e==1)e=0;c++;}
"//".* {if(e==0)a++;}
. {if(e==0)ECHO;}
%%
void main(int argc,char *argv[]){
    yyin=fopen(argv[1],"r");
    yyout=fopen(argv[2],"w");
    yylex();
    printf("single line %d \nmultiline %d \n",a,c);
    d=a+c;
    printf("total %d \n",d);
}

这篇关于lex程序上没有注释行数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！