How does subscripting std::array multiple times work?

Question

How does subscripting multiple times work for std::array even though all operator[]returns is a reference, without using any proxy-objects (as shown here)?

Example:

#include <iostream>
#include <array>

using namespace std;

int main()
{

    array<array<int, 3>, 4> structure;
    structure[2][2] = 2;
    cout << structure[2][2] << endl;

    return 0;
}

How/why does this work?

When performing operations on references, you perform them on objects they refer to. — W.B., Sep 10 '14 at 15:11
The example you link to shows how to use a proxy object so that you can subscript an object multiple times when you have a single object that _models_ a multidimensional array. With `array, M>` you literally have an array of arrays, so subscripting the outer object returns a reference to an array, which you can then subscript. No proxy needed, because this case is much simpler! — Jonathan Wakely, Sep 10 '14 at 15:11
When you have a pointer `int*p`, you can compute `p+6+2`. Same thing. — Marc Glisse, Sep 10 '14 at 15:14

Baum mit Augen · Accepted Answer · 2014-09-10T15:21:41.857

7

You simply call structure.operator[](2).operator[](2), where the first operator[] returns a reference to the third array in structure to which the second operator[] is applied.

Note that a reference to an object can be used exactly like the object itself.

edited Sep 10 '14 at 15:21

answered Sep 10 '14 at 15:12

Baum mit Augen

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score 6 · Answer 2 · answered Sep 10 '14 at 15:11

As you say, structure[2] gives a reference to an element of the outer array. Being a reference, it can be treated exactly as if it were an object of the same type.

That element is itself an array (array<int,3>), to which a further [] can be applied. That gives a reference to an element of the inner array, of type int.

How does subscripting std::array multiple times work?

2 Answers2