Good evening to you, please I need a line by line explanation on these lines of Code. Got it from you tube, but the explanation given was not clear. Thanks in advance


Input file in C
#include<stdio.h>
#include<conio.h>
void main()
{

int i,j=1,k=2,flag;
for(i=0;<=4;i++)
{
k=i+j*k;
}
flag=55;
if(flag==10)
printf(“Hello All”);
getch();

}

Please, I need help on the explanation of this code in C language

From a functional perspective, the code does nothing.
It does some processing on some variables (i, j, k) which it never uses.
Now 'Hello All' will never be displayed because flag can never be 10.
Finally getchar() just keeps the program on till you enter a character and press enter at which the program terminates.

So it does nothing.

I disagree with the post above that it does nothing. The fact that it doesn't display any output given the state of the variables doesn't mean it does nothing.

The code assigns a new value to variable k, checks if the flag variable is equal to 10 (and since it doesn't, nothing is displayed) and waits for input from user to terminate the program execution

valzey:
From a functional perspective, the code does nothing.
It does some processing on some variables (i, j, k) which it never uses.
Now 'Hello All' will never be displayed because flag can never be 10.
Finally getchar() just keeps the program on till you enter a character and press enter at which the program terminates.

So it does nothing.

Ok, I appreciate your opinion. I actually got the code from you tube. I am to do a presentation on the implementation of a Lexical Analyzer in C Language so the code is meant to be my working code. But I couldn't really decode what's all about. Thanks for your insight.

airsaylongcon:
I disagree with the post above that it does nothing. The fact that it doesn't display any output given the state of the variables doesn't mean it does nothing.

The code assigns a new value to variable k, checks if the flag variable is equal to 10 (and since it doesn't, nothing is displayed) and waits for input from user to terminate the program execution

Ok. Another good insight. So, what name can you give the above program? Or do you have any small program in C that could be used to show how a Lexical Analyzer works. That will be fine

Of a quick Google search this is what I got. And it's a bit intuitive to analeand understand it.

60

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<ctype.h>

 

int isKeyword(char buffer[]){

char keywords[32][10] ={"auto","break","case","char","const","continue","default", "do","double","else","enum","extern","float","for","goto", "if","int","long","register","return","short","signed", "sizeof","static","struct","switch","typedef","union", "unsigned","void","volatile","while"};

int i, flag = 0;

for(i = 0; i < 32; ++i){

if(strcmp(keywords[i], buffer) == 0){

flag = 1;

break;

}

}



return flag;

}

 

int main(){

char ch, buffer[15], operators[] = "+-*/%=";

FILE *fp;

int i,j=0;



fp = fopen("program.txt","r" ;


if(fp == NULL){

printf("error while opening the file\n" ;

exit(0);

}



while((ch = fgetc(fp)) != EOF){

   for(i = 0; i < 6; ++i){

   if(ch == operators[i])

   printf("%c is operator\n", ch) ;

   }

  

   if(isalnum(ch)){

   buffer[j++] = ch;

   }

   else if((ch == ' ' || ch == '\n') && (j != 0)){

   buffer[j] = '\0';

   j = 0;

     

   if(isKeyword(buffer) == 1)
printf("%s is keyword\n", buffer) ;

   else

   printf("%s is indentifier\n", buffer) ;

   }

  

}



fclose(fp) ;



return 0;

}

airsaylongcon:
Of a quick Google search this is what I got. And it's a bit intuitive to analeand understand it.

60

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<ctype.h>

 

int isKeyword(char buffer[]){


char keywords[32][10] ={"auto","break","case","char","const","continue","default", "do","double","else","enum","extern","float","for","goto", "if","int","long","register","return","short","signed", "sizeof","static","struct","switch","typedef","union", "unsigned","void","volatile","while"};

int i, flag = 0;

for(i = 0; i < 32; ++i){

if(strcmp(keywords[i], buffer) == 0){

flag = 1;

break;

}

}



return flag;

}

 

int main(){

char ch, buffer[15], operators[] = "+-*/%=";

FILE *fp;

int i,j=0;



fp = fopen("program.txt","r" ;


if(fp == NULL){

printf("error while opening the file\n" ;

exit(0);

}



while((ch = fgetc(fp)) != EOF){

   for(i = 0; i < 6; ++i){

   if(ch == operators[i])

   printf("%c is operator\n", ch) ;

   }

  

   if(isalnum(ch)){

   buffer[j++] = ch;

   }

   else if((ch == ' ' || ch == '\n') && (j != 0)){

   buffer[j] = '\0';

   j = 0;

     

   if(isKeyword(buffer) == 1)
printf("%s is keyword\n", buffer) ;

   else

   printf("%s is indentifier\n", buffer) ;

   }

  

}



fclose(fp) ;



return 0;

}

?

octaflav:

?

Don't be afraid. Its a simple program that:
Opens a file called "program.txt"
Goes through every string in the file:
If it finds an operator (+,-,/,*) it prints its an operator.
If it finds a keyword it prints it is a keyword
If it is none it prints it is an identifier.

octaflav:

?

Some questions for you.

1) Can you program? Can you program in C (or C++)
2) Do you know what a lexical analyser does?
3) If you can program, do you actually have a C compiler and IDE installed on your system where you actively write, compile and debug programs?

I haven't touched programming in a while but I can understand what the bit of code I shared with you is doing. It takes an external file called PROGRAM.TXT and analyses the content of the file to identify tokens within the file. The tokens are identified as either constants, keyword (C language keywords) or operators (such as +, -, ÷, x). Its a very straightforward program despite its seemingly complicated look

valzey:

Don't be afraid. Its a simple program that:
Opens a file called "program.txt"
Goes through every string in the file:
If it finds an operator (+,-,/,*) it prints its an operator.
If it finds a keyword it prints it is a keyword
If it is none it prints it is an identifier.

Thank you my brother... I guess its the way NL has scattered the code that is scaring him. In a proper IDE it's very easy to "parse" with the human eye.

PS: The keywords being identified are C language keywords which have been stored in the KEYWORDS array.

valzey:

Don't be afraid. Its a simple program that:
Opens a file called "program.txt"
Goes through every string in the file:
If it finds an operator (+,-,/,*) it prints its an operator.
If it finds a keyword it prints it is a keyword
If it is none it prints it is an identifier.

Thanks for your time in shedding more light. I am to explain or teach how a lexical analzer works in C. I can explain the concepts and its rudiments but I need a short program like [b]sum of the first fifty odd numbers [/b]to be used as my working example within the time allotted to me

airsaylongcon:


Some questions for you.

1) Can you program? Can you program in C (or C++)
2) Do you know what a lexical analyser does?
3) If you can program, do you actually have a C compiler and IDE installed on your system where you actively write, compile and debug programs?

I haven't touched programming in a while but I can understand what the bit of code I shared with you is doing. It takes an external file called PROGRAM.TXT and analyses the content of the file to identify tokens within the file. The tokens are identified as either constants, keyword (C language keywords) or operators (such as +, -, ÷, x). Its a very straightforward program despite its seemingly complicated look

I'm learning C++ and python. And yes I know how a lexical analyzer works because it's part of the contents in one of my courses Compiler Construction. Thanks for your input

octaflav:


Thanks for your time in shedding more light. I am to explain or teach how a lexical analzer works in C. I can explain the concepts and its rudiments but I need a short program like [b]sum of the first fifty odd numbers [/b]to be used as my working example within the time allotted to me

That program is good for the use case, just provide a valid simple C code in 'program.txt' and watch the results. Then modify as you see fit. Remember anything more than this is entering syntax analysis.

I wonder if you've even run it yet.

octaflav:


Ok, I appreciate your opinion. I actually got the code from you tube. I am to do a presentation on the implementation of a Lexical Analyzer in C Language so the code is meant to be my working code. But I couldn't really decode what's all about. Thanks for your insight.

Are u writing a compiler? Or is it
a final project ?

octaflav:


Ok. Another good insight. So, what name can you give the above program? Or do you have any small program in C that could be used to show how a Lexical Analyzer works. That will be fine

Many open source programs show how [b]lexers [\b] work. I assume you mean parsing a program into tokens and executing based on the interpretation of the tokens. But simpler examples could have been used for the case study e.g a simple calculator. The parser (or lexers) would break the script into numbers and operators (functions). It's easier to digest than a full blown language