C++ pointer to an array

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)?

• Qaz: No, it's a pointer to an array. Arrays are not pointers.
• Carl Norum: This has to be a dupe, right?
• Qaz: @CarlNorum, Oh, let me count the ways.
• SwiftMango: Arrays are pointers with a bit more overhead like array size and padding.

Answer 1

This is an array of five ints:

int a[5];

This is a pointer to an array of five ints:

int (*p)[5] = &a;

This is a pointer to an int:

int * q = a; // same as &a[0]

Answer 2

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.

• Chin: So in the second case, if it's a pointer to an array, what is the type of the pointer? Is it just "array"?
• Sergey Kalinichenko: @Chin Its type is "a pointer to an array of integers of size five" (pretty long name, huh?).
• Sergey Kalinichenko: @Chin Take a look at this link to ideone that illustrates the point from my last comment.
• Chin: Thanks. In my second example, I just want to know from a memory-diagram perspective whether it is a pointer to a pointer. But since it's not a pointer in the first example... I think I get the idea