What is this construction in C and C++

Question

what kind of declaration is this in C and C++?

int (*array1)[10]

Also, after resolving this, since it is some kind of array, how to access those members?
If this is a function, which returns a pointer to an array of 10, pointer, than is it possible to declare it without parenthesis? What arguments does it take?

EDIT:
As you answered it is pointer to an array of 10 integers, what is the difference between these two?

int (*array1)[10]
int array1[10]

EDIT:
How to assign that pointer? I try

int array[10] __attribute__((used)) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
int (*array_ptr)[10] = array;

And it give error:

cannot convert 'int*' to 'int (*)[10]' in initialization

There is a websit that can tell you [See cdecl](http://cdecl.org/). It says *declare array1 as pointer to array 10 of int*. — Iharob Al Asimi, Jul 25 '16 at 11:28
The second is an array with 10 elements of type `int*`, the first is a *pointer* to an array with 10 elements of type `int`. — Rudy Velthuis, Jul 25 '16 at 11:48
Buy a good book that teaches you C++. stackoverflow.com is not a C++ tutorial site. — Sam Varshavchik, Jul 25 '16 at 12:36

score 0 · Answer 1 · 2016-07-25T11:35:08.070

0

it is a declaration of array1 as pointer to array 10 of int.

Access to it`s member is as usual, array1[0], array1[1] ...

edited Jul 25 '16 at 11:35

answered Jul 25 '16 at 11:31

1

If there was a trailing semicolon, it would be a **definition** too (unless there's an `extern` in there somewhere. – Martin Bonner supports Monica Jul 25 '16 at 11:34

What is this construction in C and C++

1 Answers1