How to define a fixed sized array using new in C++

Question

How do you initialize the value of x using array new:

int (*x)[5] = ?

new int[5] doesn't work because it has a type of int*

Do you have to use a C-style cast as follows?

int (*x)[5] = (int( *)[5])(new int[5]);

Are you using an array instead of std:vector because of some constraint of your project? — Jorge Y., Feb 10 '18 at 09:44
You understand `int (*x)[5];` declares a pointer to array of 5 `int`? It needs the address of a valid array of 5-ints, or can serve as the pointer to multiple arrays of 5 ints. — David C. Rankin, Feb 10 '18 at 09:48
Possible duplicate of [Preventing array decay when allocating memory on the heap?](https://stackoverflow.com/questions/48494479/preventing-array-decay-when-allocating-memory-on-the-heap) — xskxzr, Feb 10 '18 at 12:26

score 5 · Answer 1 · answered Feb 10 '18 at 09:46

5

A fixed sized array in C++ would be std::array.

And, guessing from the pointer to array int (*x)[5] this might be either

std::array<int, 5> x;

or

std::array<int, 5> *x = new std::array<int, 5>;

Although the former is preferred, unless there's a real need to put the array on the heap.

If you want a variable number of fixed sized arrays, combine this with a std::vector

std::vector<std::array<int, 5> > x;

answered Feb 10 '18 at 09:46

Olaf Dietsche

1

It's funny. `std::array` is around for a decade now, and it is the first time I see it allocated with `new` :) – YSC Feb 10 '18 at 09:55
1

Now that you mention it, I haven't either. – Olaf Dietsche Feb 10 '18 at 10:00

score 1 · Answer 2 · answered Feb 10 '18 at 10:23

This works

typedef int int5[5];

int main()
{
    int (*x)[5] = new int5[1];
    return 0;
}

There may be a way to do it without a typedef, but I didn't investigate too much.

UPDATE

Somewhat counter intuitively this is also correct

int (*x)[5] = new int[1][5];

And no you shouldn't use a cast, casts are rarely the right solution, especially for beginners.

2 Answers2