Trying to put some digits into a char array - c

Question

I'm trying to create a char array made of some letters and numbers (the function was way more complex initially but i kept simplifying it to figure out why it doesn't work properly). So i have a char array in which i put 2 chars, and try to add some numbers to it.
For a reason i can't figure out, the numbers do not get added to the array. It might be really stupid but I'm new to C so here's the simplified code. Any help is much appreciated, thanks!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char some_string[20];
char *make_str() {
some_string[0] = 'a';
some_string[1] = 'x';
int random = 0;
int rand_copy = 0;
random = (rand());
rand_copy = random;
int count = 2;
while ( rand_copy > 0 ) {
rand_copy = rand_copy / 10;
++count;
}
int i=2;
for (i=2; i<count; i++) {
some_string[i] = random%10;
random = random/10;
}
return (some_string);
}
int main(int argc, const char *argv[]) {
printf("the string is: %s\n",make_str());
return 0;
}

Answer 1

You have many problems:
resulting string is not zero-terminated. Add some_string[i] = '\0'; to fix this
character (char) is something like "a letter", but random % 10 produces a number (int) which when converted to character results in control code (ASCII characters 0-9 are control codes). You'd better use some_string[i] = (random % 10) + '0';
you're using fixed length string (20 characters), which may be enough, but it could lead to many problems. If you are a beginner and haven't learn dynamic memory allocation, than that's ok for now. But remember that fixed-length buffers are one of top-10 reasons for buggy C-code. And if you have to use fixed-length buffers (there are legitimate reason for doing this), ALLWAYS check if you are not overrunning the buffer. Use predefined constants for buffer length.
unless the whole point of your excercise is to try converting numbers to strings, use libc function like snprintf for printing anything into a string.
don't use global variable (some_string) and if you do (it's ok for a small example), there is no point in returning this value.
Slightly better version:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define BUF_LENGTH 20
char some_string[BUF_LENGTH];
char *make_str() {
some_string[0] = 'a';
some_string[1] = 'x';
int random = rand();
int rand_copy = random;
int count = 2;
while (rand_copy > 0) {
rand_copy = rand_copy / 10;
++count;
}
int i;
for (i = 2; i < count; i++) {
/* check for buffer overflow. -1 is for terminating zero */
if (i >= BUF_LENGTH - 1) {
printf("error\n");
exit(EXIT_FAILURE);
}
some_string[i] = (random % 10) + '0';
random = random / 10;
}
/* zero-terminate the string */
some_string[i] = '\0';
return some_string;
}
int main(int argc, const char *argv[]) {
printf("the string is: %s\n",make_str());
return 0;
}