C How to Keep a Random Variable From Repeating the Same Number - c

Question

So I'm just learning C and I would like to know how you could prevent a variable randomized with the rand() function from repeating the same number. I have a script which simply randomizes and prints a variable in a for loop 4 times. How could I make it so the variable never gets the same number after each time it uses the rand() function?
#include <stdio.h>
#include <stdlib.h>
int randomInt;
int main()
{
srand(time(0));
for (int i = 0; i < 4; ++i) {
randomInt = rand() % 4;
printf("%d\n", randomInt);
}
return 0;
}

Answer 1

On most machines, int is 32 bits. So after 232 iterations, you are sure that you'll get some repetition (and probably much before).
If you restrict yourself to much less loops, consider e.g. keeping an array of previously met random numbers (or some hash table, or some binary tree, or some other container).
For a loop repeated only 4 times, keeping an array of (at most 4-1) previously emitted numbers is quite simple, and efficient enough.
Read also about the pigeonhole principle.

Answer 2

A slightly different approach.
int set[] = {0, 1, 2, 3 } ;
srand(time(0));
shuffle(set,4);
using the shuffle algorithm given in this question
https://stackoverflow.com/a/6127606/9288531

Answer 3

I'm guessing that you are getting the same numbers because your are running your program multiple times within the same second. If time(0) hasn't changed, you will have the same seed and the same random numbers generated. Unless your program runs extremely quickly, I imagine using a seed based on microseconds instead of seconds would work:
#include <stdio.h>
#include <stdlib.h>
#include <sys/time.h>
int randomInt;
int main()
{
struct timeval my_microtimer;
gettimeofday(&t1, NULL);
srand(t1.tv_sec * my_microtimer.tv_usec);
for (int i = 0; i < 4; ++i) {
randomInt = rand() % 4;
printf("%d\n", randomInt);
}
return 0;
}

Answer 4

What you could do is keeping track of each number you already generated.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int hasMyNumberAlreadyBeenGenerated(int number, int generatedNumbers[], int size){
for(int i = 0; i < size + 1; i++){
//If you already generated the number, it should be present somewhere in your array
if(generatedNumbers[i] == number) return 1;
//If you did not, find the first available space in your array, and put the number you generated into that space
if(generatedNumbers[i] == 0){
generatedNumbers[i] = number;
break; //No need to continue to check the array
}
}
return 0;
}
int main()
{
int randomInt;
int generatedNumbers[4];
//We set "0" in all the array, to be sure that the array doesn't contain unknown datas when we create it
memset(generatedNumbers, 0x0, sizeof(generatedNumbers));
srand(time(0));
for (int i = 0; i < 4; ++i) {
randomInt = rand() % 4 + 1;
//As long as the number you generate has already been generated, generate a new one
while(hasMyNumberAlreadyBeenGenerated(randomInt, generatedNumbers, i) == 1){
randomInt = rand() % 4 + 1;
}
printf("generated : %d\n", randomInt);
}
return 0;
}
The problem with this method is that you can't generate a 0, because if you do you'll endlessly loop.
You can bypass this problem using a dynamic array using malloc() function.
If you want to write clean code you should define how many numbers you want to generate with a #define.

Answer 5

What you seem to be asking is a non-random set of numbers 0 to 3 in a random order. Given that;
int set[] = {0, 1, 2, 3 } ;
int remaining = sizeof(set) / sizeof(*set) ;
while( remaining != 0 )
{
int index = rand() % sizeof(set) / sizeof(*set) ;
if( set[index] > 0 )
{
printf( "%d\n", set[index] ) ;
set[index] = -1 ;
remaining-- ;
}
}
For very large sets, this approach may not be practical - the number of iterations necessary to exhaust the set is non-deterministic.