CODEWITHSUNDEEP
coutput
woodfordb
woodfordb

Reputation: 73

No output in basic C program

I have been reading the first chapter of K&R, but I am having trouble with one of the very first problems.

#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);

}

I have zero output in the terminal from this. I have tried many different variations of this, but can not seem to get it to output and numbers.

Any help is very appreciated!

Upvotes: 0

Views: 406

Answers (3)

John Hascall
John Hascall

Reputation: 9416

BTW, in the years since K&R came out, we have learned a lot about what makes code more reliable/maintainable. Some examples:

#include <stdio.h>

/* count digits, white space, others */

int main ( void ) {                 /* type explicitly */

        int c;                      /* declare 1 variable per line */
        int i;
        int nwhite     = 0;         /* initialize variables when possible */
        int nother     = 0;
        int ndigit[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};

        while ((c = getchar()) != EOF) {          /* use brackets */
                if ((c >= '0') && (c <= '9')) {   /* explicit precedence is clearer */
                        ++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]); /* 1 liner is ok w/{} */
        printf(", white space = %d, other = %d\n", nwhite, nother);
        return 0;                            /* don't forget a return value */
}

Upvotes: 0

qwertz
qwertz

Reputation: 14792

The problem is, that you do not normally input EOF from the console. As stated in Tony Ruths answer, it works by redirecting a file to stdin of the program, because files are terminated by EOF. Since the entire file is then passed into stdin, which is read by the program via getch(), the "end of the input" gets recognized the correct way.

You just don't get to input EOF via the console, thus not getting any output. (actually there might be some special shortcut for EOF)

Upvotes: 0

Tony Ruth
Tony Ruth

Reputation: 1408

This program requires an input file to be passed to it, but it works as is.

 gcc test.c -o test
 echo "12345" > data
 ./test < data

 digits = 0 1 1 1 1 1 0 0 0 0, white space = 1, other = 0

Here is another output:

 echo "111 111 222" > data
 ./test < data

 digits = 0 6 3 0 0 0 0 0 0 0, white space = 3, other = 0

Upvotes: 2

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