Cool. Thanks, Jay.
On Thu, Jun 6, 2019 at 12:25 PM Jay McCarthy wrote:
> On Thu, Jun 6, 2019 at 12:16 PM David Storrs
> wrote:
>
>>
>>
>> On Thu, Jun 6, 2019 at 12:14 PM Jay McCarthy
>> wrote:
>>
>>> Your code is passing bytes by value, but bytes are themselves
>>> pointers, so you are
On Thu, Jun 6, 2019 at 12:16 PM David Storrs wrote:
>
>
> On Thu, Jun 6, 2019 at 12:14 PM Jay McCarthy
> wrote:
>
>> Your code is passing bytes by value, but bytes are themselves
>> pointers, so you are passing copies of the pointer, not copies of the
>> bytes. When you modify it, with
On Thu, Jun 6, 2019 at 12:14 PM Jay McCarthy wrote:
> Your code is passing bytes by value, but bytes are themselves
> pointers, so you are passing copies of the pointer, not copies of the
> bytes. When you modify it, with `bytes-set!` you are modifying the
> underlying structure. When you copy
Your code is passing bytes by value, but bytes are themselves
pointers, so you are passing copies of the pointer, not copies of the
bytes. When you modify it, with `bytes-set!` you are modifying the
underlying structure. When you copy it with `subbytes` or
`bytes-copy`, you are making a new object
On Thu, Jun 6, 2019 at 12:00 PM David Storrs wrote:
> My understanding is that Racket is call by value, not call by reference.
> My application will often be passing around large-ish byte strings; will
> they be copied every time I pass them, or will the interpreter use
> copy-on-write?
>
My understanding is that Racket is call by value, not call by reference.
My application will often be passing around large-ish byte strings; will
they be copied every time I pass them, or will the interpreter use
copy-on-write?
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