Re: Working directly on output stream: second try
On Jan 4, 11:10 am, mathieu <mathieu.malate...@gmail.com> wrote:
Because my first post was not clear, I'll try again to explain what I
am looking for. Let's assume I had access directly to the memory, I
would do:
void DoAction(const char *rawinput, unsigned long l)
{
uint16_t *deb = (uint16_t*)rawinput;
for(unsigned long i = 0; i<l/2; i++)
{
*deb = (*deb >> (16 - 12 - 1)) & pmask;
deb++;
}
}
Well for the design of my IO library I decided that it would be more
convenient (more flexible) to design it with stream, since the stream
could hide a subclass and could hide a compressed stream.
Streams aren't particularly designed for modifying memory in
place. As others have mentionned, what you're looking for is
probably std::transform.
You can use stream_iterators with std::transform, but if the
file is binary, you'll have to design some sort of binary stream
for them, and define stream_iterators for it.
So I am stuck with an API which is:
void DoAction(std::istream& is, std::ostream& os);
My previous design (with raw pointer) is a factor two faster than the
current one (with stream interface), and that's a real issue.
Certainly: direct in place or memory to memory will almost
always be faster than stream to stream.
In this *very* particular case I know I will not be reusing
the input buffer (input stream) and it has all the good
properties so that I can directly work on the input buffer.
I'm not sure I understand. Do you mean that you know that the
transformation will be in place and in memory?
Is there a way to run a filter in-place on the is.rdbuf, and then
simply pass is.rdbuf() to os ?
Well, you can always use a filtering streambuf. But if your
goal is an in place transformation in memory, any introduction
of streams can only be considered obfuscation.
Hopefully this time it does make sense,
Not really. I doesn't make sense to use streams in the
interface if all you want is an in place transformation.
--
James Kanze (GABI Software) email:james.kanze@gmail.com
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