Allocating static array with new

Question

No matter what syntax I try, this code is always a compile error:

int (*foo)[3] = new ???;

I've tried

int (*foo)[3] = new (int[3]);
int (*foo)[3] = new (int(*)[3]);
int bar[3];
int (*foo)[3] = new decltype(bar);

https://stackoverflow.com/questions/6111565/now-that-we-have-stdarray-what-uses-are-left-for-c-style-arrays — πάντα ῥεῖ, Feb 20 '19 at 21:58
Why not `std::vector` or `std::array` as this is C++ and not C? — tadman, Feb 20 '19 at 21:58
`Cannot convert from int* to int(*)[3]`, `Cannot convert from int(**)[3] to int(*)[3]`, `Cannot convert from int* to int(*)[3]` — Artikash-Reinstate Monica, Feb 20 '19 at 22:05
Use an alias to help clean things up: `typedef int myArrType[3];` or `typedef std::array myArrType;` or `using myArrType = int[3];` or `using myArrType = std::array;`. Then you can do `myArrType *foo = new myArrType; ... delete foo;` or `std::unique_ptr foo(new myArrType);` or `auto foo = std::make_unique();` — Remy Lebeau, Feb 21 '19 at 00:48

score 3 · Accepted Answer · answered Feb 20 '19 at 22:02

3

Well, it's easy.

int (*foo)[3] = new int[1][3];

answered Feb 20 '19 at 22:02

KamilCuk

1

Ah, of course. Is it possible to make it deletable with `delete` instead of `delete[]`? – Artikash-Reinstate Monica Feb 20 '19 at 22:04
@Artikash Anything allocated with `new` is "deletable". The key here is to figure out how to delete it. – Code-Apprentice Feb 20 '19 at 22:10
I am trying `int (*f)[3] = new int[3];` but the `new int[3]` decays from array to pointer. I don't thing using `delete` is possible - even if you can allocate memory, you would allocate an array of 3 integers and store the pointer to it in `foo`. That's still an array. – KamilCuk Feb 20 '19 at 22:10
You can allocate the memory for pointer itself `int (**foo)[3] = new (int(*)[3]);` that would free with delete. – KamilCuk Feb 20 '19 at 22:13
My guts tells me `int (*foo)[3] = reinterpret_cast(new int[3])` should be ok, because it's the same as `int mem[3]; int (*foo)[3] = reinterpret_cast(mem);` and `&mem` and `mem` have the same value (different type)... – KamilCuk Feb 20 '19 at 22:20

1 Answers1