Pointer to Array in C

Question

How can you interpret the following line of code?

int (*arrayABC)[10];

In my textbook, it says that we have a pointer to a pointer to the 0th element of an integer array.

However, I don't quite understand this.

My interpretation: We have some variable, which gets as its value some address. This address is then the address of the 0th element of an UNNAMED integer array. Basically we have a pointer to the 0th element.

Why then to have a pointer TO A POINTER?

Answer 1

This is a pointer to an array. It is not a pointer to a pointer. Arrays and pointers are different. An array has an address, but an array is not an address. An array is a series of contiguous elements.

This pointer points to the whole array and not just the first element, in the same way that a float * points to the whole float and not just the first byte.

If you have for example:

int foo[10];
int (*arrayABC)[10] = &foo;

then the expressions (*arrayABC) and foo are identical. E.g. foo[3] is the same as (*arrayABC)[3].