/*
* This implementation of the above algorithm has some distinct
* limitations, although I believe they are not a serious problem.
* This is intended for processing short strings of digits (10 or so),
* not arbitrarily long digit strings. Each of the substrings that are
* broken out by the code below are assumed to be a value that can
* be assigned to a long when converted to an integral type. This
* limitation requires that the input string not be more than about 20
* characters in length, which is more than adequate for the intended
* application. I'm assuming that a long is at least 32 bits.
* Further, it assumes a character set where the digits' collating
* sequence mimics that of ASCII, so that things like:
* x = val - '0';
* will evaluate to an integer value of 3 in the case where val == '3'.
*/
#include <stdio.h>
#define MOD10 10
#define MOD11 11
/*
* -------------- - check_digit() -------- - *
* Calculates a check digit according to the algorithm given above.
* NOTE
* that there are no checks for array bounds overflow. Beware.
*/
check_digit (string, mod)
char * string;
int mod;
{
char tmpbuf[20];
int i, j, sum;
unsigned long odd, even;
/* extract the odd digits */
j = strlen (string);
for ( j-, i = 0 ; j >= 0 ; j -= 2, i++ )
{
tmpbuf[i] = string[j];
}
tmpbuf[i] = '\0';
odd = atol (tmpbuf);
/* extract the even digits */
j = strlen (string);
for ( j -= 2, i = 0 ; j >= 0 ; j -= 2, i++ )
{
tmpbuf[i] = string[j];
}
tmpbuf[i] = '\0';
even = atol (tmpbuf);
/* now make the composite string & sum the digits */
sprintf (tmpbuf, "%lu%lu", even, 2L * odd);
i = strlen (tmpbuf);
for ( sum = j = 0 ; j < i ; j++)
{
sum += tmpbuf[j] - '0';
}
/* now do the mod10 or mod11 operation */
i = (mod - (sum % mod));
if (i == mod)
i = 0;
return (i);
}
#ifdef TEST
main ()
{
char buf[30];
while (1)
{
gets (buf);
if (strlen (buf) == 0)
break;
printf ("string: %s, mod10 checkdigit: %d,"
"mod11 checkdigit %d:n",
buf, check_digit (buf, MOD10),
check_digit (buf, MOD11));
}
}
#endif /* TEST */
/* End of File */