I really hope someone can give a well explained example. I've been searching everywhere but can't find a proper solution.
I am taking an introduction to C Programming class, and our last assignment is to write a program which validates a 10 digit ISBN with dashes... The ISBN is inputted as a string in a CHAR array. From there I need to separate each digit and convert them into an integer, so I can calculated the validity of the ISBN. On top of that, the dashes need to be ignored..
My thought process was to create an INT array and then use a loop to store each character into the array, and pass it through the atoi() function. I also tried using an IF statement to check each part of the CHAR array to see if it found a dash. If it did find one, it would skip to the next spot in the array. It looked something like this:
int num[12], i = 0, j = 0, count = 0;
char isbn[12];
printf ("Enter an ISBN to validate: ");
scanf ("%13[0-9Xx-]%*c", &isbn);
do {
if (isbn[i] == '-') {
i++;
j++;
}
else {
num[i]= atoi(isbn[j]);
i++;
j++;
}
count++;
} while (count != 10);
But that creates a segmentation fault, so I can't even tell if my IF statement has actually filtered the dashes....
If someone could try and solve this I'd really appreciate that. The Assignment was due Dec 4th, however I got an extension until Dec 7th, so I'm pressed for time.
Please write out the code in your explanation. I'm a visual learner, and need to see step by step.
There's obviously a lot more that needs to be coded, but I can't move ahead until I get over this obstacle.
Thanks in advance!
First of all, your definition of isbn
is not sufficient to hold 13 characters; it should therefore be 14 chars long (to also store the terminating '\0'
).
Second, your loop is overly complicated; three loop variables that maintain the same value is redundant.
Third, the loop is not safe, because a string might be as short as one character, but your code happily loops 10 times.
Lastly, converting a char that holds the ascii value of a digit can be converted by simply subtracting '0'
from it.
This is the code after above improvements have been made.
#include <stdio.h>
int main(void)
{
int num[14], i;
char isbn[14], *p;
printf("Enter an ISBN to validate: ");
scanf("%13[0-9Xx-]%*c", &isbn);
// p iterates over each character of isbn
// *p evaluates the value of each character
// the loop stops when the end-of-string is reached, i.e. '\0'
for (p = isbn, i = 0; *p; ++p) {
if (*p == '-' || *p == 'X' || *p == 'x') {
continue;
}
// it's definitely a digit now
num[i++] = *p - '0';
}
// post: i holds number of digits in num
// post: num[x] is the digit value, for 0 <= x < i
return 0;
}