Randomize 4 different numbers - c

Question

I'm trying to randomize 4 different numbers in C and trying the next code:
{
int num1 = 0, num2 = 0, num3 = 0, num4 = 0;
int i = 0;
while (num1 == num2 && num1 == num3 && num1 == num4 && num2 == num3 && num2 == num4 && num3 == num4 && num3 == num2)
{
num1 = rand() % 7;
num2 = rand() % 7;
num3 = rand() % 7;
num4 = rand() % 7;
}
printf("%d %d %d %d\n", num1, num2, num3, num4);
}
The code suppose to check if the numbers are not equal and if they are equal, it needs to generate new numbers until they are all distinct.
But for some reason, it's not working well and even right numbers it puts them as wrong and it becomes and endless loop.
What am I missing?

Answer 1

This code will pick 4 different numbers in the range 0 .. 6, it works by creating an array of available numbers, as each is picked it is removed from the list.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define RANGE 7 // how many different numbers
#define PICK 4 // how many to pick
int main(void) {
int pool[RANGE];
int size, n, i;
for (size=0; size<RANGE; size++) {
pool[size] = size; // create number pool 0..6
}
srand((unsigned)time(NULL));
// pick different numbers
for(i=0; i<PICK; i++) {
n = rand() % size; // random array index
printf("%d ", pool[n]); // select number from pool
pool[n] = pool[--size]; // remove from pool
}
printf("\n");
return 0;
}

Answer 2

Try like this
void
get_random_values(int *values)
{
int source[] = {0, 1, 2, 3, 4, 5, 6};
for (int i = 0 ; i < 7 ; ++i)
{
int saved;
int j = rand() % 7;
int k = rand() % 7;
saved = source[j];
source[j] = source[k];
source[k] = saved;
}
values[0] = source[0];
values[1] = source[1];
values[2] = source[2];
values[3] = source[3];
}
int
main(void)
{
int values[4];
srand(time(NULL));
get_random_values(values);
for (int i = 0 ; i < 4 ; ++i)
fprintf(stderr, "%d ", values[i]);
fprintf(stderr, "\n");
return 0;
}
Don't forget to set the random seed srand() at the program startup or you will get the same sequence always.

Answer 3

To get a random, unbiased sequence:
#include <stdio.h>
#include <stdlib.h>
#define N 7
int main(int argc, char **argv)
{
// seed random number generator with first argument for easier testing
if (argc > 1) {
srand(atoi(argv[1]));
}
int array[N] = {0,1,2,3,4,5,6};
// Fisher–Yates shuffle:
// https://en.wikipedia.org/w/index.php?oldid=697311634#The_modern_algorithm
for (unsigned i = 0; i < N - 1; ++i) {
unsigned modulo = N - i;
// unbiased rand() % modulo:
// https://stackoverflow.com/a/10989061/416224
unsigned j;
do {
j = rand();
} while (j >= RAND_MAX - (RAND_MAX % modulo));
j %= modulo;
if (j > 0) {
int tmp = array[i];
array[i] = array[i + j];
array[i + j] = tmp;
}
}
for (unsigned i = 0; i < N; ++i) {
printf("%u. %d\n", i + 1, array[i]);
}
return 0;
}
Please follow the referenced links in the code.

Answer 4

I would suggest universal solution:
array of any size
lower and upper bounds sent as parameters.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void random(int* values, int amount, int lower_bound, int upper_bound)
{
int i=0, j=0, temp;
if(amount > upper_bound - lower_bound + 1)
return; // if there are more numbers than given bound
for(i=0; i<amount; )
{
temp = rand() % ( upper_bound - lower_bound + 1 ) + lower_bound;
for(j=i-1; j>=0; --j)
if(temp==values[j])
break;
if(temp==values[j])
continue;
values[i]=temp;
++i;
}
}
int main()
{
srand(time(NULL));
int arr[4]={0,0,0,0};
random(arr, 4, 5, 10);
for(int i=0; i<4; ++i)
printf("%d\n", arr[i]);
}
With this you can have (for example) 10 numbers of original values from -6 to 7.