How to find a missing value in an array? [duplicate]

Question

I am trying to find the smallest missing element of an array using function check, which has two arguments (n and array A). I can't understand why my function check is always returning one and the while loop is never closing.

#include <stdio.h>

bool check(int n, int A[])
{
    for (int i = 0; i < sizeof(A); i++)
    {
        if(A[i] == n) 
        {
            return 1;
        }
    }
    return 0;
}

int main()
{
    int A[] = {1, 3, 6, 4, 1, 2};
    int n = 1;

    while (check(n, A) == 1)
    {
        n++;
    }
        
    printf("%d is missing",n);
}

Answer 1

The compiler adjusts a parameter having an array type to pointer to the array element type.

So this function declaration

bool check(int n, int A[])

is equivalent to

bool check(int n, int *A );

And within the function the expression sizeof(A) is equivalent to the expression sizeof( int * ) and is equal to either 4 or 8 depending on the used system.

Thus this for loop

for (int i = 0; i < sizeof(A); i++)

invokes undefined behavior.

I know but still that's not why the while loop is never stopping.

Answering your above comment it seems that in the used system sizeof( int * ) is equal to 8 and the variable n is placed in memory after the array A as they defined in main

int A[] = {1, 3, 6, 4, 1, 2};
int n = 1;

As a result you get the infinite wile loop because in the for loop within the function the memory occupied by the variable n is checked and n is always equal to itself.

Thus the function always returns 1.

That is in the for loop the array is traversed as it has 8 elements like

int A[] = {1, 3, 6, 4, 1, 2, n, some_indeterminate_value };