Re: Working directly on output stream

From:
red floyd <no.spam@here.dude>
Newsgroups:
comp.lang.c++
Date:
Fri, 04 Jan 2008 09:02:18 -0800
Message-ID:
<dGtfj.36177$Pv2.11901@newssvr23.news.prodigy.net>
James Kanze wrote:

On Jan 3, 6:31 pm, red floyd <no.s...@here.dude> wrote:

mathieu wrote:

  I am trying to rewrite this very slow function (due to AFAIK the
extra copy):

void DoAction(std::istream &is, std::ostream &os)
{
      uint16_t c;
      while( is.read((char*)&c,2) )
        {
        c =
          (c >> (16 - 12 - 1)) & pmask;
        os.write((char*)&c, 2 );
        }
}

  I was thinking of doing:

 os.rdbuf( is.rdbuf() )

  and then working directly with a ostreambuf_iterator, but I cannot
read from a ostreambuf_iterator can I ?


Don't reassign the output streambuf. Use transform, or a loop on the
streambuf iterators.

i.e.:

struct convert
{
    uint16_t operator()(const uint16 c) const
    {
       return (c >> shift) & pmask);
    }
    convert(uint16_t& pmask_) : pmask(pmask_) { }
private:
    static const shift = (16 - 12 - 1);
    uint16_t pmask;
};

void DoAction(std::istream& is, std::ostream& os)
{
     std::transform(std::istreambuf_iterator<uint16_t>(is),
                    std::istreambuf_iterator<uint16_t>(),
                    std::ostreambuf_iterator<uint16_t>(os));


And since when can you initialize an
[io]streambuf_iterator<uint16_t> with an [io]stream? [io]stream
is a typedef for basic_[io]stream<char>, which means that it can
only be used with an [io]streambuf_char.


You're right. I was confused. My mistake.

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