Printing randomly from 1 to 10 in C - c

Question

I have written a code for randomly printing a number from 1 to 10 without any repetition, but it isn't working properly sometimes I get the number that is already written.
In short, I'm trying to print numbers from 1-10 randomly with no repetition.
Here is my code :
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<conio.h>
main() {
int no = 0, repeat[100] = { 0 }, i = 0, x = 0, j = 0;
srand(time(NULL));
while (true) {
no = (rand() % 10) + 1;
for (i = 0; i < 100; i++) {
if (no != repeat[i]) {
x = 1;
} else if (no == repeat[i]) {
x = 0;
}
}
if (x == 1) {
repeat[j] = no;
printf("\n%d", repeat[i]);
j = j + 1;
}
getch();
}
}

Answer 1

Don't use rand() to generate the numbers directly, instead fill a sequential array, and then use rand() to shuffle the array, e.g.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
/** shuffle integer array of size 'n'
* (using fisher-yates method)
*/
void shuffle (int *a, int n)
{
int i, tmp;
while (n-- > 1) {
i = rand() % (n + 1);
tmp = a[i];
a[i] = a[n];
a[n] = tmp;
}
}
int main (void) {
int arr[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
srand (time(NULL));
shuffle (arr, 10);
for (int i = 0; i < 10; i++)
printf (" %d", arr[i]);
putchar ('\n');
}
By shuffling the array and swapping random elements within it, you eliminate all possibility of a duplicate number.
Example Use/Output
$ ./bin/shuffle_arr
3 10 7 8 4 5 6 9 1 2

Answer 2

#include <stdio.h>
#define N1 1
#define N2 10
void main() {
int len = N2 - N1 + 1, i, r, temp;
int num[len];
//Fill array with numbers
for (temp = 0, i = N1; temp < len; i++, temp++)
num[temp] = i;
srand(time(NULL));
for (i = len - 1; i > 0; i--) {
r = rand() % i; //pop random number
//swaping
temp = num[i];
num[i] = num[r];
num[r] = temp;
}
/*Random Numbers are stored in Array*/
//print that array
for (i = 0; i < len; i++)
printf("%d\n", num[i]);
}