Re: ostream::operator<<() and sputn()

[Lewis Hyatt <lhyatt@gmail.com>]

On Wed, Jul 14, 2021 at 5:31 PM Jonathan Wakely <jwakely.gcc@gmail.com> wrote:
>
> On Wed, 14 Jul 2021 at 22:26, Lewis Hyatt via Libstdc++
> <libstdc++@gcc.gnu.org> wrote:
> >
> > Hello-
> >
> > I noticed that libstdc++'s implementation of ostream::operator<<() prefers
> > to call sputn() on the underlying streambuf for all char, char*, and string
> > output operations, including single characters, rather than manipulate the
> > buffer directly. I am curious why it works this way, it feels perhaps
> > suboptimal to me because sputn() is mandated to call the virtual function
> > xsputn() on every call, while e.g. sputc() simply manipulates the buffer and
> > only needs a virtual call when the buffer is full. I always thought that the
> > buffer abstraction and the resulting avoidance of virtual calls for the
> > majority of operations was the main point of streambuf's design, and that
> > sputn() was meant for cases when the output would be large enough to
> > overflow the buffer anyway, if it may be possible to skip the buffer and
> > flush directly instead?
> >
> > It seems to me that for most typical use cases, xsputn() is still going to
> > want to use the buffer if the output fits into it; libstdc++ does this in
> > basic_filebuf, for example. So then it would seem to be beneficial to try
> > the buffer prior to making the virtual function call, instead of after --
> > especially because the typical char instantiation of __ostream_insert that
> > makes this call for operator<<() is hidden inside the .so, and is not
> > inlined or eligible for devirtualization optimizations.
> >
> > FWIW, here is a small test case.
> >
> > ---------
> > #include <ostream>
> > #include <iostream>
> > #include <fstream>
> > #include <sstream>
> > #include <chrono>
> > #include <random>
> > using namespace std;
> >
> > int main() {
> >     constexpr size_t N = 500000000;
> >     string s(N, 'x');
> >
> >     ofstream of{"/dev/null"};
> >     ostringstream os;
> >     ostream* streams[] = {&of, &os};
> >     mt19937 rng{random_device{}()};
> >
> >     const auto timed_run = [&](const char* label, auto&& callback) {
> >         const auto t1 = chrono::steady_clock::now();
> >         for(char c: s) callback(*streams[rng() % 2], c);
> >         const auto t2 = chrono::steady_clock::now();
> >         cout << label << " took: "
> >              << chrono::duration<double>(t2-t1).count()
> >              << " seconds" << endl;
> >     };
> >
> >     timed_run("insert with put()", [](ostream& o, char c) {o.put(c);});
> >     timed_run("insert with op<< ", [](ostream& o, char c) {o << c;});
> > }
> > ---------
> >
> > This is what I get with the current trunk:
> > ---------
> > insert with put() took: 6.12152 seconds
> > insert with op<<  took: 13.4437 seconds
> > ---------
> >
> > And this is what I get with the attached patch:
> > ---------
> > insert with put() took: 6.08313 seconds
> > insert with op<<  took: 8.24565 seconds
> > ---------
> >
> > So the overhead of calling operator<< vs calling put() was reduced by more
> > than 3X.
> >
> > The prototype patch calls an internal alternate to sputn(), which tries the
> > buffer prior to calling xsputn().
>
> This won't work if a user provides an explicit specialization of
> basic_streambuf<char, MyTraits>. std::basic_ostream<char, MyTraits>
> will still try to call your new function, but it won't be present in
> the user's specialization, so will fail to compile. The basic_ostream
> primary template can only use the standard API of basic_streambuf. The
> std::basic_ostream<char> specialization can use non-standard members
> of std::basic_streambuf<char> because we know users can't specialize
> that.

Thanks, makes sense, this was more just a quick proof of concept. I
guess a real version could work around this, well it could be
implemented purely in terms of sputc() too. Am curious if you think
the overall idea is worthwhile though? Partly I am trying to
understand it better, like it was a bit surprising to me that the
standard says that sputn() *must* call xsputn(). Feels like calling
it, only if a call to overflow() would otherwise be necessary, makes
more sense to me...


-Lewis