I understand there are a lot of questions around even recursion here, but I don’t know in C wether its my logic or my syntax that’s wrong (or presumably both) because from what I know I may need pointers. Especially when I’m using function in C, so I can harldy find any question that I can understand as a freshman in cs undergrads.

So my assignment was to simply create an array with even numbers in range determined in input, if input n = 10, then the array needs to contain [0, 2, 4, 6, 8, 10], if n = 9 then the array would simply be [0, 2, 4, 6, 8]. But I have to do it with recursion and I don’t know what’s wrong with my code since the output is somewhat weird.

#include <stdio.h>

int even(int x, int arrEven[]) {

    if (x == 0) {
        arrEven[0] = 0;
        return 0;
    }
    else if (x > 0) {    
        arrEven[x - 1] = even(x - 2, arrEven) + 1;
        return arrEven[x - 1];
    }
}

int main(void) {

    printf("n = ");
    int n;
    scanf("%d", &n);
    n = (n / 2) + 1;
    int arrEven[n];
    even(n, arrEven);

    for (int i = 0; i < (n - 1); i++) {
        printf("%d, ", arrEven[i]);
    }
    printf("%d", arrEven[n - 1]);
}

When I input 10 the output was

0, 1, -731908444, 2, 781232032, 3

instead of

0, 2, 4, 6, 8, 10

>Solution :

You mix the usage of one variable for two purposes.
In your code, n is used as index in the array and as value to be put into the array at the same time.

There are tons of options to do this. One could looke like this:

#include <stdio.h>

void even(int *arrEven, int value, int limit) {

    *arrEven = value;

    if (value < limit) {
        even(arrEven+1, value+2, limit);
    }
}

int main (void) {
    int limit;
    int count;

    printf("n = ");
    scanf("%d", &limit);

    count = (limit / 2) + 1;
    int arrEven[count];

    even(arrEven, 0, limit);

    for (int i = 0; i < count-1; i++){
        printf("%d, ", arrEven[i]);
    }
    printf("%d", arrEven[count-1]);
}