C program sizeof

Question

# include <stdio.h>

void print(int *arr)
{
int n = sizeof(arr)/sizeof(arr[0]);
printf ("%d\n ", sizeof(arr));

printf ("%d\n\r ", sizeof(arr[0]));
//printf ("%d ", n);
int i;
for (i = 0; i < n; i++)
printf("%d ", arr[i]);
}

int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8};
printf ("%d\n ", sizeof(arr));

printf ("%d\n\r ", sizeof(arr[0]));
print(arr);
return 0;
}

Answer 1

Line 5 contains a mistake: when you pass an array you don't pass it's size, so sizeof(arr) will be the size of the pointer type (32 or 64 bits), not of the whole array.

Answer 2

i need explanation about output of this program