On Thursday, 2 June 2022 at 08:24:51 UTC, Mike Parker wrote:
And so, `&ts[0]` is the same as `&(*ts.ptr + 0)`, or simply `ts.ptr`.
That should be the same as `&(*(ts.ptr + 0))`!
On Thursday, 2 June 2022 at 08:24:51 UTC, Mike Parker wrote:
And so, `&ts[0]` is the same as `&(*ts.ptr + 0)`, or simply `ts.ptr`.
That should be the same as `&(*(ts.ptr + 0))`!