Though you are getting desired output, here in your code you are not exactly generating characters at all indices/positions of the char pass[12]; array. let me explain
consider the following for loop in your code:
for (i = 0; i < 4; i++) {
pass[i] = rand() % 10; //10 instead of 9 to produce digits from 0 to 9
char capLetter = 'A' + (rand() % 26);
pass[i + 2] = capLetter;
char letter = 'a' + (rand() % 26);
pass[i + 3] = letter;
printf("%d%c%c", pass[i], pass[i + 2], pass[i + 3]);
}
consider the maximum cases i.e, when i = 3 and pass[i + 3], then pass[i + 3] is equivalent to pass[6] so the maximum index value that you can access is 6. So you are not generating random value at all indices of the pass[] array but only within 0 to 6 positions and printing the generated characters in each iteration.
Instead try using for loop this way to generate random character at all the indices:
for (i = 0; i < 4; i++)
{
pass[ 3 * i ] = rand() % 10;
char capLetter = 'A' + (rand() % 26);
pass[(3 * i) + 1] = capLetter;
char letter = 'a' + (rand() % 26);
pass[(3 * i) + 2] = letter;
printf("%d%c%c", pass[3 * i], pass[(3 * i) + 1], pass[(3 * i) + 2]);
}
here you can access all the 12 indices i.e, 0 to 11
further, instead of using pass[i] = rand() % 9; to generate a random character and using %d, you can do pass[i] = '0' + (rand() % 10); (use 10 instead of 9 if not 9 will never be printed) to generate a numeric character and use %c. Since you have a character array, you can append the array with a null terminating character ('\0') and make it a string and print the generated password this way:
#include <stdio.h>
#include <stdlib.h>
#include <math.h> //not required
#include <time.h>
int main(void) //int main() is not a valid signature in C
{
srand((unsigned int)(time(NULL)));
int i;
char pass[13]; //extra byte for null terminating character
printf("Press enter to get a twelve-character password\n");
getchar();
for (i = 0; i < 4; i++)
{
//revised logic to generate random characters at all positions (0 - 11)
pass[ 3 * i ] = '0' + (rand() % 10); //generating numeric character
char capLetter = 'A' + (rand() % 26); //generating upper case alpha character
pass[(3 * i) + 1] = capLetter;
char letter = 'a' + (rand() % 26); //generating lower case alpha character
pass[(3 * i) + 2] = letter;
}
pass[3 * i] = '\0'; //placing null terminating character at the end
printf("generated password : %s\n\n",pass); //printing the string
printf("\n\n");
}
I want to make it more efficient if I can, and perhaps add some
punctuation, which does not seem that easy, since the ASCII codes are
quite spread out.
Here is a way to generate passwords which includes punctuation without caring about ASCII code at all! Right now, you are following a fixed pattern of printing uppercase letter followed by number and then a lowercase letter, and you repeat it for three more times to generate a 12 character password. Instead you could randomly generate the password without any such pattern and not using ASCII codes at all!
- create a string of all permissible characters
- use the string to print 12 character from any 12 random positions
here's the program which follows the above two steps to generate a password and you can add all the permissible characters in the string which I have not mentioned
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main(void)
{
srand((unsigned int)(time(NULL)));
int index = 0;
//step 1
char characters[] = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789/,.-+=~`<>:";
//I just added a few punctuations characters for explanatory purpose
//you can add all the additional punctuations which are required
//step 2
for(index = 0; index < 12; index++)
{
printf("%c", characters[rand() % (sizeof characters - 1)]);
}
}
I'd say generating the password is much more advantageous as you can only randomly select from the characters that provide in the string. All you need to do is just put in all the permissible characters in the string and you are bound to get a randomly generated password :)
rand()%10for digits in the range 0 to 9. \$\endgroup\$rand() % 9. Thanks. \$\endgroup\$