Accessing a std::array via a unique_ptr

Question

This is just for learning purposes, I know I can just use a vector but I have

const int N = 1e3;
auto my_arr = std::make_unique<std::array<int, N>>();

// To access it I have to do this - why [0][0] twice?
my_arr.get()[0][0] = 1;

// Getting the size for fun
std::cout << sizeof(my_arr.get()) << "\n"; // outputs 8
std::cout << sizeof(my_arr.get()[0]) << "\n"; // outputs 800000000
std::cout << sizeof(my_arr.get()[0][0]) << "\n"; // outputs 8

I understand that .get() returns a pointer to the managed object but I don't understand why I need to do my_arr.get()[0][0] twice?

my_arr.get() returns a pointer. Dereferencing a pointer and indexing it at position 0 are the same thing. ptr[0] is defined to be the same at *(ptr + 0) — Richard Critten
– Richard Critten, Commented Oct 11, 2018 at 17:37
my_arr.get()[0] is the same as *(my_arr.get()). You should use the latter because there's no my_arr.get()[1]. — n. m. could be an AI
– n. m. could be an AI, Commented Oct 11, 2018 at 17:38
Are you actually sure about const int N = 1e3;? That looks kind of weird. Also remember that std::unique_ptr is a replacement for a pointer, so strictly speaking you'll get a pointer to a pointer decayed from an array, and you'll end up with more or less sizeof(int). — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Oct 11, 2018 at 17:38

NathanOliver · Accepted Answer · 2018-10-11 17:40:41Z

7

my_arr.get() gives you a std::array<int, N>*. Since you have a pointer doing pointer[0] is the same as *pointer. So you don't actually need my_arr.get()[0][0] and you can instead use

(*my_arr.get())[0]

to show that you are dereferencing the pointer. In fact, you could just use

(*my_arr)[0]

because operator * is overloaded for std::unique_ptr and it will return a reference to the pointed to thing.

answered Oct 11, 2018 at 17:40

NathanOliver

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Peter Ruderman · Accepted Answer · 2018-10-11 17:41:16Z

5

Well, you don't have to do that, but it works.

my_arr.get() returns a std::array<int,N>*

Like any pointer, you can index it like an array.

my_arr.get()[0] returns a reference to the first element in your 'array' of arrays.

You can then use std::array's indexing operator to get an element.

my_arr.get()[0][0] returns a reference to the element you want.

Alternatively, you could write:

my_arr->at(0)
my_arr->operator[](0)
(*my_arr)[0]
*my_arr->data()

answered Oct 11, 2018 at 17:41

Peter Ruderman

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1 Comment

Gar Over a year ago

Thanks for providing those alternatives, they were very helpful

Barry · Accepted Answer · 2018-10-11 18:07:03Z

4

You probably don't want to do this:

auto my_arr = std::make_unique<std::array<int, N>>();

Unless you really specifically want an array<int, N>. If what you just want to do is dynamically allocate N ints and manage that with a unique_ptr, that's:

auto my_arr = std::make_unique<int[]>(N);

The advantages here are:

N can be a runtime value, it doesn't have to be a constant-expression.
unique_ptr<T[]>::operator[] exists and does what you want, so my_arr[2] does index into the array.

answered Oct 11, 2018 at 18:07

Barry

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Michael Deom · Accepted Answer · 2018-10-11 19:36:57Z

3

As you said yourself .get() returns a pointer to the managed object.

So it is std::array<int, N>* what you return and not std::array<int, N>:

std::array<int, N>* array_ptr = my_arr.get();

So either write (*my_arr)[0] or (*my_arr.get())[0]

edited Oct 11, 2018 at 19:36

Michael Deom

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answered Oct 11, 2018 at 17:39

t.niese

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