How is the working of fflush(stdin) changing the output in below code?

Question

#include <stdio.h>
int main()
{
        int test_no ,count=1,i,n,j;
        scanf("%d",&test_no);
        fflush(stdin);
        int arr1[test_no];
        for(i=0;i<test_no;i++)
        {
                scanf("%d",&n);
                printf("\n");
                int arr[n];
                for(j=0;j<n;j++)
                {
                        fflush(stdin);
                        scanf("%d",&arr[i]);
                }
                for(j=1;j<=n-1;j++)
                {
                        if(arr[j-1]>arr[j])
                        {
                                count++;
                        }
                }
                if(n==1)
                {
                        arr1[i]=1;
                }
                else
                {
                        arr1[i]=count;
                }
                count=1;
        }
        for(i=0;i<test_no;i++)
        {
                printf("%d\n",arr1[i]) ;
        }
        return 0;
}

This solution is to this problem.

I am not getting the desired output for the third case , it's giving me output as 3 or 4 depending on whether I place fflush(stdin) before scanf("%d",arr[i]) or after scanf("%d",arr[i]) , please tell the problem with this code .

Answer 1

In no some magical way.

First of all, fflush(stdin); invokes undefined behavior. Don't use that.

Quoting C11, chapter §7.21.5.2, The fflush function (emphasis mine)

If stream points to an output stream or an update stream in which the most recent operation was not input, the fflush function causes any unwritten data for that stream to be delivered to the host environment to be written to the file; otherwise, the behavior is undefined.

That said,

for(j=0;j<n;j++)
{
    fflush(stdin);
    scanf("%d",&arr[i]);
}

looks pretty wrong to me, arr[i] is not guaranteed to be within bounds. It should rather be

scanf("%d",&arr[j]);