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C++ - Most efficient way to remove an element from a Message's repeated array
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jaypipes  
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 More options Oct 5 2009, 1:33 pm
From: jaypipes <jaypi...@gmail.com>
Date: Mon, 5 Oct 2009 10:33:07 -0700 (PDT)
Local: Mon, Oct 5 2009 1:33 pm
Subject: C++ - Most efficient way to remove an element from a Message's repeated array
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Hi all,

I'm wondering what is the most efficient method to remove an element
from a Message's vector of repeated elements?  I looked at the
RepeatedFieldPtr API but would appreciate some example code on how to
do this in the most efficient way.  I gather from the API that the
underlying storage is a vector and so does not support efficient index-
based, random-access deletion from the vector.  Is this correct?

Help is much appreciated,

Jay


 
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Kenton Varda  
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 More options Oct 5 2009, 3:56 pm
From: Kenton Varda <ken...@google.com>
Date: Mon, 5 Oct 2009 12:56:17 -0700
Local: Mon, Oct 5 2009 3:56 pm
Subject: Re: C++ - Most efficient way to remove an element from a Message's repeated array
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On Mon, Oct 5, 2009 at 10:33 AM, jaypipes <jaypi...@gmail.com> wrote:
> I'm wondering what is the most efficient method to remove an element
> from a Message's vector of repeated elements?  I looked at the
> RepeatedFieldPtr API but would appreciate some example code on how to
> do this in the most efficient way.  I gather from the API that the
> underlying storage is a vector and so does not support efficient index-
> based, random-access deletion from the vector.  Is this correct?

Yes, that is correct.

The most efficient solution is to permute the contents so that the things
you want to remove are at the end, then use RemoveLast().

Note that the underlying storage is an array of *pointers*, so re-arranging
pointers doesn't require copying whole message objects.


 
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jaypipes  
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 More options Oct 5 2009, 4:14 pm
From: jaypipes <jaypi...@gmail.com>
Date: Mon, 5 Oct 2009 13:14:28 -0700 (PDT)
Local: Mon, Oct 5 2009 4:14 pm
Subject: Re: C++ - Most efficient way to remove an element from a Message's repeated array
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Gotcha. Thanks for the assistance, Kenton!

 
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