Example UUID in 22 bytes. Linux: `cc -luuid uuid22.c` Darwin: `cc uuid22.c`

uuid22.c

#include <stdio.h>
#include <uuid/uuid.h>
#include <sys/types.h>

u_int64_t bits2uint64(unsigned char const bits[]) {
   return ((u_int64_t)bits[0] << 56)
        | ((u_int64_t)bits[1] << 48)
        | ((u_int64_t)bits[2] << 40)
        | ((u_int64_t)bits[3] << 32)
        | ((u_int64_t)bits[4] << 24)
        | ((u_int64_t)bits[5] << 16)
        | ((u_int64_t)bits[6] <<  8)
        |  (u_int64_t)bits[7];
}

void uuid_init(uuid_t u) {
    // cross platform ifdef here
    uuid_generate(u);
}

// least 22 bytes allocated and owned at call-site
void uuid_fmt22(uuid_t u, char buf[]) {
    static const int len = 11;
    static const int base = 58;
    static const char digits[] =
        "abcdefghijklmnopqrstuvwxyz"
        "ABCDEFGHJKLMNOPQRSTUVWXYZ"
        "2345679"; // Excludes I180

    int i,j;

    // for hi/lo of 128 bits
    for (i=0; i < 2; i++) {
        u_int64_t block = bits2uint64(u+(i*8));

        for (j=0; j < len; j++) {
            buf[j+(i*len)] = digits[block % base];
            block = block / base;
        }
    }
}

int main(int argc, char** argv) {
    uuid_t u;
    char str[23];
    char fmt[37];
    int i;

    for (i = 0; i < 10; i++) {
        uuid_init(u);
        uuid_unparse_lower(u, fmt);
        uuid_fmt22(u, str);
        str[22] = '\0';
        printf("uuid: %s '%s'\n", fmt, str);
    }

    return 0;
}

Can you please explain this comment? I can see they are missing from the digits array, but I'm not sure why.
// Excludes I180

Edit:
Ahh, I think I worked it out, it is encoding with Base58, so we have 62 characters in a-zA-Z0-9 and remove 4 of them to get our Base58 characters.