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So if T[] and T[N] are valid, then T[N][M] should also be. This increases

Alexander wrote: the scope greatly but has an incredible difficult questions to answer: What would operator[] return? I don't have a good answer...

Simple questions that already have an answer: The same way shared_ptr<T[N][M]> works today. :)

Its operator[] returns element_type&

So: auto p = allocate_shared<int[5][1024]>(alloc); p[3][675] = 8;

auto q = allocate_unique<double[][2][3][4]>(alloc, size); q[index][1][1][1] = 0.95;

Then: dynamic<int[5][1024], Allocator> a(alloc); a[3][675] = 8;

i.e. Making it work for T[N] also automatically makes it work for T[N][M].

Sure, but shouldn't `Foo<T[N][M]>[]` return `Foo<T[N]>` so you can pass it down like that? Sure returning basically a raw array is possible and easiest, but you (may) loose the goodies for type/range checking and you can't get a lower-D type out of a high-D type for passing that down an interface.