C++ How I can get random value from 01 to 12?

Question

Hi How I can get in C++ random value from 01 to 12?

So I will have 03, or 06, or 11?

Answer 1

#include <iomanip>
#include <iostream>
#include <stdlib.h>
#include <time.h>

// initialize random seed
srand( time(NULL) );

// generate random number
int randomNumber = rand() % 12 + 1;

// output, as you seem to wan a '0'
cout << setfill ('0') << setw (2) << randomNumber;

to adress dirkgently's issue maybe something like that would be better?

// generate random number
int randomNumber = rand()>>4; // get rid of the first 4 bits

// get the value
randomNumer = randomNumer % 12 + 1;

edit after mre and dirkgently's comments

Answer 2

Use the following formula:

M + rand() / (RAND_MAX / (N - M + 1) + 1), M = 1, N = 12

and read up on this FAQ.

Edit: Most answers on this question do not take into account the fact that poor PRN generators (typically offered with the library function rand()) are not very random in the low order bits. Hence:

rand() % 12 + 1

is not good enough.

Answer 3

(rand() % 12 + 1).ToString("D2")

Answer 4

You can do this, for example:

#include <cstdlib>
#include <cstdio>
#include <time.h>

int main(int argc, char **argv)
{
    srand(time(0));

    printf("Random number between 1 and 12: %d", (rand() % 12) + 1);
}

The srand function will seed the random number generator with the current time (in seconds) - that's the way it's usually done, but there are also more secure solutions.

rand() % 12 will give you a number between 0 and 11 (% is the modulus operator), so we add 1 here to get to your desired range of 1 to 12.

In order to print 01 02 03 and so on instead of 1 2 3, you can format the output:

printf("Random number between 01 and 12: %02d", (rand() % 12) + 1);

Answer 5

Is there some significance to the leading zero in this case? Do you intend for it to be octal, so the 12 is really 10 (in base 10)?

Getting a random number within a specified range is fairly straightforward:

int rand_lim(int limit) {
/* return a random number between 0 and limit inclusive.
 */

    int divisor = RAND_MAX/(limit+1);
    int retval;

    do { 
        retval = rand() / divisor;
    } while (retval > limit);

    return retval;
}

(The while loop is to prevent skewed results -- some outputs happening more often than others). Skewed results are almost inevitable when/if you use division (or its remainder) directly.

If you want to print it out so even one-digit numbers show two digits (i.e. a leading 0), you can do something like:

std::cout << std::setw(2) << std::setprecision(2) << std::setfill('0') << number;

Answer 6

Probably not terribly speed efficient - but wouldn't the best way to spread evenly across the entire range of RAND_MAX be to simply raise the result to an approriate (fractional) power?

eg. If RAND_MAX is 32768 and you want 0-12 then
RAND_MAX ^ x = 12, so x = ln(12)/ln(32768) = 0.2389

and result = pow(rand(),0.2389)

Answer 7

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <sys/time.h>

int get_random(int n) {
    struct timeval num;
    gettimeofday(&num, NULL);
    srand(num.tv_usec);
    return rand() % n + 1;
}

int main () {
    printf("Random number between 1 and 12: %d \n", get_random(12));
    return 0;
}