Question

int my_array[5] = {0};
int *my_pointer = 0;

my_pointer = &my_array;  // compiler error
my_pointer = my_array;   // ok

If my_array is address of array then what does &my_array gives me?

I get the following compiler error:

error: cannot convert ‘int (*)[5]’ to ‘int*’ in assignment

Answer 1

my_array is the name of an array of 5 integers. The compiler will happily convert it to a pointer to a single integer.

&my_array is a pointer to an array of 5 integers. The compiler will not treat an array of integers as a single integer, thus it refuses to make the conversion.