From hpages at fredhutch.org  Wed Mar 29 20:04:34 2017
From: hpages at fredhutch.org (=?UTF-8?B?SGVydsOpIFBhZ8Oocw==?=)
Date: Wed, 29 Mar 2017 11:04:34 -0700
Subject: [Rd] Transferring ownership of R-managed buffer
In-Reply-To: <CANnL8go0rb+01wbPzygrFO+rEznLsT1K+_buZEzAhAMecNP8RQ@mail.gmail.com>
References: <CANnL8go0rb+01wbPzygrFO+rEznLsT1K+_buZEzAhAMecNP8RQ@mail.gmail.com>
Message-ID: <bd43b699-57fb-96a0-e187-7183bf5f7274@fredhutch.org>

Hi Tim,

On 03/29/2017 08:24 AM, Tim Keitt wrote:
> I have a use case where I would like to create an SEXP around an existing
> buffer that is managed by R, thus avoiding a copy operation.

What to you mean exactly by "an existing buffer managed by R"?

> If I have
> something like:
>
> void *p = (void*) RAW(PROTECT(Rf_allocVector(RAWSXP, n)));
> ... additional maniupulation ...
> SEXP x = somefunc(SXPTYPE, n, p);  // ????
>
> Is there a "placement" constructor available?

What's a "placement" constructor?

>(I have arranged for the
> corresponding UNPROTECT.) I've looked at and experimented with R_allocator
> and allocVector3, but can't quite get it right. I know this is odd, but it
> makes sense for my use case.

Not sure I follow what you are trying to do. Note that an SEXP is a 
pointer to a C struct called SEXPREC. I think that trying to point an
SEXPREC struct to data pointed to by an existing SEXPREC struct is very
likely to lead to a disaster.

Note that if the existing buffer managed by R is an SEXP (e.g. b) and
your code has access to this SEXP then you don't need to create another
SEXP around its data. Since SEXPs are pointers doing

   SEXP x = b;

is fine and doesn't generate any copy.

And if your code only has access to a "naked" pointer to the buffer
managed by R (e.g. to RAW(b) is the buffer is actually in an SEXP)
then why would you need to wrap it inside an SEXP?

H.

>
> Thanks for any pointers.
>
> THK
>
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>
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-- 
Herv? Pag?s

Program in Computational Biology
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[Rd] Transferring ownership of R-managed buffer
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