what is mean of this declaration int (*a)[2]?

[buddha87]

I saw a declaration like this:

int (*a)[2]

What does this mean? Is this a two dimension array? How should I use it ?

Thanks.

[TarikNeaj]

Read here - http://stackoverflow.com/questions/859634/c-pointer-to-array-array-of-pointers-disambiguation

[giblit]

Probably be easier to just mention it is a pointer to an array of 2 integers and not an array of 2 pointers to integer

[buddha87]

Thanks, both of you.
So generally, if I declare:

int a[2];

now &(a[0]) is the pointer to the array of 2 integers. It is same as "a" in int (*a)[2] ?

Thanks.

[cire]

int a[2];

now &(a[0]) is the pointer to the array of 2 integers. It is same as "a" in int (*a)[2] ?

No. a[0] is an int. Taking the address of an int gives you a pointer-to-int, not a pointer-to-array-of-2-int.

1 2 3 4 5 6 7 8 9 10 11 12 13

// http://ideone.com/ejPK6l #include <iostream> #include <typeinfo> int main() { int a[2]; std::cout << typeid(a).name() << '\n'; std::cout << typeid(a[0]).name() << '\n'; std::cout << typeid(&a[0]).name() << '\n'; std::cout << typeid(&a).name() << '\n'; }

Output from VC++ (GCC tends to be a little more cryptic) int [2] int int * int (*)[2]

[buddha87]

Thanks very much, now I think I understand the difference.