Generating random string unsigned char in C

Question

I want to generate a random string text of length 100 with the code below, then to verify that I print the length of the variable text but sometimes that is less than 100. How can I fix that?

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>

int main() {
    int i, LEN = 100;
    srandom(time(NULL));
    unsigned char text[LEN];
    memset(text, 1, LEN);
    for (i = 0; i < LEN; i++) {
        text[i] = (unsigned char) rand() & 0xfff;
    }
    printf("plain-text:");
    printf("strlen(text)=%zd\n", strlen(text));

}

Answer 1

Perhaps a random character 0 was added to the string, and then it is considered as the end of string by strlen.

You can generate random characters as (rand() % 255) + 1 to avoid zeros.

And at the end you have to zero-terminate the string.

LEN = 101; // 100 + 1
....
for (i = 0; i < LEN - 1; i++) {
    text[i] = (unsigned char) (rand() % 255 + 1);
}
text[LEN-1] = 0;

Answer 2

I want to generate a random string text of length 100 with the code below, then to verify that I print the length of the variable text but sometimes that is less than 100. How can I fix that?

1. First of all, if you want to generate a string of length 100, you'll need to declare an array of size 101.

   int i, LEN = 101;
   srandom(time(NULL));
   unsigned char text[LEN];
   

2. When you are assigning the characters from the call to rand, make sure that it is not 0, which is usually the null terminator for strings.

   for (i = 0; i < LEN - 1; /* Don't increment i here */) {
       c = (unsigned char) rand() & 0xfff;
       if ( c != '\0' )
       {
          text[i] = c;
   
          // Increment i only for this case.
          ++i
       }
   }
   

   and don't forget to null terminate the string.

   text[LEN-1] = '\0';