printf does not work out of loop K&R array example

Code:

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#include <stdio.h>
/* count digits, white space, others */
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);
}


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this code an example for an array in K&R 2 but it is not executed unless printf inside the while loop is there a specific reason for that or may be something wrong with my mac ?

You need to manually trigger an end-of-file indication on stdin when you run the program (after supplying all the input you intend to be counted). The way you do that depends on your operating system, but definitely requires action by you (the user).

If you don't do that, getchar() will never return the value EOF, so the while loop will never terminate, and the printf() statements will not be reached.

Thank you very much! I found that control+d

Or you could just replace the EOF with

Code:

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while ((c = getchar()) != '0')

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and the program should stop with the loop.

Quote:

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Originally Posted by Dakshin

Or you could just replace the EOF with

Code:

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while ((c = getchar()) != '0')

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and the program should stop with the loop.

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It'll certainly stop the loop, but given that the loop should accept any character and not just digits, your suggestion breaks the algorithm.

Try creating a text file that contains your input. Then run the program like: ./foo < sample.txt, or "cat sample.txt | ./foo"

Quote:

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Originally Posted by Prelude

It'll certainly stop the loop, but given that the loop should accept any character and not just digits, your suggestion breaks the algorithm.

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I don`t see why... the 0 is in quotes, isn`t it? The program is checking a character, not integer

Quote:

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The program is checking a character, not integer

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Exactly. The program should be able to accept a '0' as acceptable input, not to stop the loop.