Thank you. I now officially know that I don’t understand. Sorry.

On Sun, Jun 7, 2020 at 7:54 AM Viktor Kojouharov <vkojouha...@gmail.com>
wrote:

> The pointer is being copied via typedmemmove, which itself calls memmove,
> which, according to its documentation, copies bytes from the source to the
> destination. Not sure why that would be impossible, considering it does
> work for some code (the source pointer preserves its data)
>
> Not sure what you mean by "copied via unsafe". Also, the source pointer
> never goes out of scope. It's part of a struct that is passed to the
> function that calls typedmemmove, and its lifetime is more or less static.
> So while the destination pointers go out of scope, the source one never
> does.
>
>
> On Sunday, June 7, 2020 at 4:45:40 PM UTC+3, Michael Jones wrote:
>>
>> Do you mean that you have a problem with the value of the pointer? That
>> is "copying the pointer." This seems impossible.
>>
>> Attempting to access through a pointer copied via unsafe is (generally)
>> inviting doom, and seems highly possible. The instant the last pointer to
>> that data goes out of scope the address range occupied by the formerly
>> pointed to items is formally inaccessible. Using unsafe to keep a shadow
>> copy of the address and then poking around is quite likely to fail, and
>> even when it does not, it is quite likely to be meaningless. (random data
>> from some other use).
>>
>> If I misunderstood, please forgive me.
>>
>> On Sun, Jun 7, 2020 at 6:15 AM Viktor Kojouharov <vkojo...@gmail.com>
>> wrote:
>>
>>> p.s. should such questions be posted in golang-dev, since it deals with
>>> runtime internals?
>>>
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>>
>>
>> --
>>
>> *Michael T. jonesmichae...@gmail.com*
>>
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*Michael T. jonesmichael.jo...@gmail.com <michael.jo...@gmail.com>*

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