Read an Integer from the User, Part 2

Dave Sinkula

Some issues, such as leading whitespace and trailing characters that cannot be part of a number, were not handled in Read an Integer from the User, Part 1. Here such issues receive lip service.

About the Author
#include <stdio.h>
#include <ctype.h>

int mygeti(int *result)
        char c, buff [ 13 ]; /* signed 32-bit value, extra room for '\n' and '\0' */
        return fgets(buff, sizeof buff, stdin) && !isspace(*buff) &&
        sscanf(buff, "%d%c", result, &c) == 2 && (c == '\n' || c == '\0');

int main(void)
        int value;
        do {
                fputs("Enter an integer: ", stdout);
        } while ( !mygeti(&value) );
        printf("value = %d\n", value);
        return 0;

/* my output
Enter an integer: one
Enter an integer:
Enter an integer: f123
Enter an integer: 123f
Enter an integer:  123
Enter an integer: 123
Enter an integer: 1.23
Enter an integer: -42
value = -42

/* note: this line in the above has a space character following the 123
Enter an integer: 123
drxs33 0 Newbie Poster

Thank you sir dave.. it helped me alot.. thanks!

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