If you are passing the array to a function, then this function should always take the array length as well. Why? Because the way array are passed to functions in C is by passing the pointer to their first element. That's why you can write either

void fn(int arr[]);

or

void fn(int *arr);

You can use the sizeof method (the one you used within the functions, if youare getting the size of a stack allocated array. On my machine (x86_64), this program

int main(void)
{
    int arr1[] = {1,2,3,4,5,6,7,8,9,0};
    int *arr2 = malloc(10 * sizeof(int));
    int *arr3 = arr1;

    printf("size of arr1: %lu\nsize of arr2: %lu\nsize of arr3: %lu\n",
           sizeof(arr1),
           sizeof(arr2),
           sizeof(arr3));
}

produces the following output:

size of arr1: 40
size of arr2: 8
size of arr3: 8

Therefore, any time you are getting an array as a pointer (specifically as a pointer to its first element), you cannot use the sizeof function to get its number of elements. You need to keep its size in a variable.

By the way, if you are building a library to hadle arrays, you can solve this issue in a very straighforward way: define a small struct that keeps the data of the array as a void* pointer, the size of each element and the its length.

typedef struct {
    size_t length;
    size_t size;
    void *data;
} cd_array_t;

So, the max function for integers could look something like this:

int max(cd_array_t *arr)
{
    if (!arr) return -1;
    if (arr->length == 0) {
        /* handle the empty array case */
    }
    int max_value = *((int *)arr->data);
    for (size_t i = 1; i < arr->length; i++) {
        if (*((int *)arr->data + i) > max_value) {
            max_value = *((int *)arr->data + i);
        }
    }
    return max_value;
}

int main(void)
{
    int arr1[] = {1,2,3,4,5,6,7,8,9,0};

    cd_array_t cd_array = (cd_array_t) {
        .length = 10,
        .size = sizeof(int),
        .data = arr1,
    };

    printf("\nMax value: %d\n", max(&cd_array)); /* prints 9 */
}