On Thursday, 2 June 2022 at 08:14:40 UTC, Mike Parker wrote:
More specifically, it points to the starting address of the
allocated block of memory.
I posted too soon.
Given an instance `ts` of type `T[]`, array accesses essentially
are this:
```d
ts[0] == *(ts.ptr + 0);
ts[1] == *(ts.ptr + 1);
ts[2] == *(ts.ptr + 2);
```
Since the size of `T` is known, each addition to the pointer adds
`N * T.sizeof` bytes. If you converted it to a `ubyte` array,
you'd need to handle that yourself.
And so, `&ts[0]` is the same as `&(*ts.ptr + 0)`, or simply
`ts.ptr`.
