Dennis M.Ritche

Question

#include<studio.h>

main()

{        int c, i ,nwhite, nother;

          int ndigit[10];

          nwhite = nother = 0;

         for(i=0; i <10; ++i)

           ndigit[i]= 0;

        while((c= getchar())!=EOF)

              if (c>='0' && c<='9' )

                   ++ndigit[c-'0'];

             elseif ( c== ' ' ll c == ' /n' ll c=='/t ')

                  ++nwhite;

             else

                ++nother;

    printf("digits=");

   for(i= 0; i< 10; ++i)

        printf("%d" , ndigit[I]);

        printf(", white space =%d, other= %d, other =%d", nwhite , nother);

}

Comments

What is the output of this program?

Answer 1

Before posting any "code" please check if the syntax is correct. ( I am a stickler for good written codes. (#include<studio.h>!!! ))

Anyways, I'll dissect this example from K&R.

#include <stdio.h>
   void main()
   {
       int c, i, nwhite, nother;
       int ndigit[10];
       nwhite = nother = 0;
       for (i = 0; i < 10; ++i)
           ndigit[i] = 0;
       while((c = getchar()) != EOF)
           if(c >= '0' && c <= '9')
              ++ndigit[c-'0'];
           else if(c == ' ' || c == '\n' || c == '\t')
               ++nwhite;
           else
               ++nother;
       printf("digits =");
       for (i = 0; i < 10; ++i)
           printf(" %d", ndigit[i]);
       printf(", white space = %d, other = %d\n", nwhite, nother);
}

Let's dissect parts of this. 

for (i = 0; i < 10; ++i)
           ndigit[i] = 0;

Just set all the elements of the array as 0 and using an array it is easier to hold all the 10 digits rather than 10 variables. 

while((c = getchar()) != EOF)
           if(c >= '0' && c <= '9')
              ++ndigit[c-'0']; // implicit char to int conversion, rookie trick
           else if(c == ' ' || c == '\n' || c == '\t')
               ++nwhite;
           else
               ++nother;

Iterate the "text" or "string" or "file" ( user input ) till EOF is reached. While iterating check each character individually and the counter respective to the type in incremented.
(if it's a digit -> ndigit[that digit]++, any of \n, \t or whitespace you would increment nwhite, and if it's something else ( alphabets, special symbols) nother++ )

printf("digits =");
       for (i = 0; i < 10; ++i)
           printf(" %d", ndigit[i]);
       printf(", white space = %d, other = %d\n", nwhite, nother);

This piece of code just prints out the result. 

Let run this!

To compile I'm assuming you have gcc installed or clang will do fine as well.

~$ gcc thisquestion.c -o thisquestion 

~$ ./thisquestion 

I'm wasted for love, and now I shall show my math skills! 1+0 = 1\nEOF

digits = 1 2 0 0 0 0 0 0 0 0, white space = 15, other = 48

Note: '\n' is when I pressed "ENTER" ( or "RETURN" for MacOS ) and EOF for me ( MacOS is Control-D) and it might be different you. To check enter ~$ stty all in your terminal.

So what does the output say? 

First, the digit '0' repeats ONE time, digit '1' repeats TWICE, there are 14 spaces and ONE '\n' character = 14 + 1 = 15 white spaces, and finally 48 alphabets and special characters (+) included which if you really really want go ahead and count. 

We are done? Yep. We are done.

Comments

Compile the program and run it. Type anything. Once you're done you simulate EOF with CTRL+D (for *nix (Any linux distro or OS X) or CTRL+Z  (for Windows) from command line.

You'll get the result for the string/text you entered.

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How to execute this program with myself? Without using any software used. Give any example.

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You mean online? Try Ideone. 
Copy-paste the code above and for input write anything you want. Then compile and keep experimenting.