Need to generate 4 random numbers without repetition in C programming. 1 to 4

Question

I want to generate numbers 1 to 4 in a random fashion using C programming. I have made provision to print a[0] directly in a while loop and for any further element the program checks whether the new number from a[1] to a[3] is same as any of the previous elements. A function has been created for the same. int checkarray(int *x, int y).

The function checks current element with previous elements one by one by reducing the passed address. If it matches the value it exits the loop by assigning value zero to the condition variable (int apply).

return apply;

In the main program it matches with the int check if check==1, the number is printed or else the loop is repeated.

Problem faced: The number of random numbers generated is varying between 2 and 4. e.g

2 4
2 4 3 
1 3 3 4

etc

Also repetition is there sometimes.

#include <stdio.h>
#include <conio.h>

int checkarray(int *x, int y);

void main() {
    int a[4], i = 0, check;

    srand(time(0));

    while (i < 4) {
        a[i] = rand() % 4 + 1;
        if (i == 0) {
            printf("%d ", a[i]);
            i++;
            continue;
        } else {
            check = checkarray(&a[i], i);
        }
        if (check == 1) {
            printf("\n%d ", a[i]);
        } else {
            continue;
        }
        i++;                    
    }

    getch();
}

int checkarray(int *x, int y) {
    int arrcnt = y, apply = 1, r = 1;
    while (arrcnt > 0) {
        if (*x == *(x - 2 * r)) {
            apply = 0;
            exit(0);
        } else {
            arrcnt--;
            r++;
            continue;
        }
    }   
    return apply;
}

Your checkarray function ends the program normally when reaching exit(0). I'm not sure that is what you intend. — , Aug 09 '19 at 09:16
There is a far simpler way to do this. Put the four numbers in an array, swap them around randomly, then use them in sequence. — Weather Vane, Aug 09 '19 at 09:22
Hey thank you for putting great effort in the description. But I do not understand what your problem is. Can you please provide a description of the problem you have and not just the task you want to accomplish? Maybe you can also take the [tour] and read up on [mcve] — Kami Kaze, Aug 09 '19 at 09:23

score 1 · Answer 1 · answered Aug 12 '19 at 19:19

Let's look at the checkarray function, which is supposed to check if a number is already present in the array.

It is called this way:

check = checkarray(&a[i], i);

Where a is an array of 4 integers and i is the actual index, so it tries to scan the array backwards looking for any occurrences of a[i]

int checkarray(int *x,int y)
{
    int arrcnt=y,apply=1,r=1;
    while(arrcnt>0)
    {
        if(*x==*(x-2*r))
        //         ^^^    Why? This will cause an out of bounds access.
        {
            apply = 0;
            exit(0);    // <-- This ends the program. It should be a 'break;'
        }
        else
        {
            arrcnt--;
            r++;
            continue;
        }
    }   
    return apply;
}

Without changing the interface (which is error prone, in my opinion) it could be rewritten as

int check_array(int *x, int y)
{
    while ( y )
    {
        if ( *x == *(x - y) )   
            return 0;
        --y;
    }
    return 1;
}

Testable here.

There are many other issues which should be addressed, though, so please, take a look to these Q&A too.

Does "n * (rand() / RAND_MAX)" make a skewed random number distribution?
Why do people say there is modulo bias when using a random number generator?
Fisher Yates shuffling algorithm in C
int main() vs void main() in C
Why can't I find <conio.h> on Linux?

score 0 · Answer 2 · answered Aug 12 '19 at 19:26

Your approach is tedious but can be made to work:

there is no need to special case the first number, just make checkarray() return not found for an empty array.
you should pass different arguments to checkarray(): a pointer to the array, the number of entries to check and the value to search.
you should not use exit(0) to return 0 from checkarray(): it causes the program to terminate immediately.

Here is a modified version:

#include <stdio.h>
#include <conio.h>

int checkarray(int *array, int len, int value) {
    int i;
    for (i = 0; i < len; i++) {
        if (array[i] == value)
            return 0;
    }
    return 1;
}

int main() {
    int a[4], i = 0, value;

    srand(time(0));

    while (i < 4) {
        value = rand() % 4 + 1;
        if (checkarray(a, i, value)) {
            printf("%d ", value);
            a[i++] = value;
        }
    }
    printf("\n");
    getch();
    return 0;
}

Need to generate 4 random numbers without repetition in C programming. 1 to 4

2 Answers2