Question

I know that in C++, the name of an array is just the pointer to the first element of the array. For example:

int foo[5];

=> foo is a pointer of type int, and it points to the first element in the array, which is a[0]

But then, we have this:

int (*foo)[5]

Here, foo is a pointer to an array of type. So is foo a pointer to a pointer (of type int)?

Answer 1

in C++, the name of an array is just the pointer to the first element of the array

That’s not correct: a name of an array can be converted to a corresponding pointer type “for free”, but it is definitely not a pointer. The easiest way you can tell is by comparing sizes of a pointer and of an array:

int foo[5];
int *ptr;
cout << sizeof(foo) << endl;
cout << sizeof(ptr) << endl;

Your second example is a pointer to an array of ints of size 5.

cout << sizeof(foo) << endl;
cout << sizeof(*foo) << endl;

In a 32-bit environment this prints 4 and 20, as expected.