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copied!<p>It depends on STL implementation. I means, the standard does not prevent to implement <code>std::array</code> using more members, or reserving more memory of that is really necessary (for example, for debugging), but I think is very improbable to found one <code>std::array</code> implementation without just use more of <code>T elem[N];</code> data member.</p> <p>If we assume the std::array implementation includes just one field for store the data and it allocate just the necessary memory (not more), <code>int v[100];</code> and where the data is stored in <code>array<int, 100> v;</code> will have the same layout, since from the standard:</p> <p>[array.overview 23.3.2.1 p1]:</p> <blockquote> <p>The elements of an array are stored contiguously, meaning that if <code>a</code> is an <code>array<T, N></code> then it obeys the identity <code>&a[n] == &a[0] + n</code> for all <code>0 <= n < N</code>.</p> </blockquote> <p>and [class.mem 9.2 p20]:</p> <blockquote> <p>A pointer to a standard-layout struct object, suitably converted using a <code>reinterpret_cast</code>, points to its initial member (or if that member is a bit-field, then to the unit in which it resides) and vice versa. [ Note: There might therefore be unnamed padding within a standard-layout struct object, but not at its beginning, as necessary to achieve appropriate alignment. —end note ]</p> </blockquote> <p>Anyway, that depends on compiler and STL implementation. But the reversed code depends on compiler too. Why are you assuming int <code>a; int b[50];</code> will locate <code>a</code> and then array of <code>b</code> in memory in that order and not in the other if that declarations are not part of a <code>struct</code> or <code>class</code>? The compiler would decide other thing for performance reasons (but I see that is improbable). </p>

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