Why is it a compile error to assign the address of an array to a pointer “my_pointer = &my_array”?

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 a

Answer 1

This would be a fascinating topic for research on neuroscience, semantics, and software development. Even though we can explain the difference between my_array and &my_array, and even though we can repeat the mantra that "arrays are (or are not) pointers", this distinction is still confusing.

Usually when we take the address of something with the "&" operation, we arrive at a completely different value.

int x;
x=5;
cout <

First of all, let's challenge this intuition. A pointer can be accidentally equal to the value it is pointing at:

int* x;
x=(int*)(&x);
cout <<"same: "<

So in some contexts semantically different things can evaluate to the same thing. But just because two expressions are equal in the current context does not mean that they are interchangeable.

int w=3;
int* ip=&w;
void* vp=&w;
cout <<"Equal: "<

The same happens with pointers to arrays. A pointer to an array is often technically "equal" to an array itself, but since they have different semantics, there are situations in which they are not interchangeable. Try this one:

int my_array[5] = {5,6,7,8,9};
cout <