boost asio streambuf don't release memory after calling consume?

Question

boost::asio::streambuf b;
...
void handler(const boost::system::error_code& e, std::size_t size)
{
  if (!e)
  {
    std::stringstream sstr(std::string((std::istreambuf_iterator<char>(&b)), 
        std::istreambuf_iterator<char>()));
    b.consume(size);
    ...
  }
}
...
boost::asio::async_read_until(s, b, "END\r\n", handler);

when the consume method is called, the memory occupied by streambuf b is not released. The memory will grow up as async_read_until is called multiple times. Is my usage correct? Is there any way to free the memory before the get pointer of streambuf?

Most likely, you are actually freeing it. The memory will grow up due to fragmentation and the like and will then start leveling off once peak load has been experienced. — David Schwartz

Igor R. Igor R. · Accepted Answer · 2012-03-04T09:54:34

asio::streambuf is based on std::vector that grows as needed, but never shrinks. So, consume() is not supposed to release memory, it just adjusts internal pointers:

void consume(std::size_t n)
{
  if (egptr() < pptr())
    setg(&buffer_[0], gptr(), pptr());
  if (gptr() + n > pptr())
    n = pptr() - gptr();
  gbump(static_cast<int>(n));
}

But each time you consume() and read() again, the internal buffer (vector) is reused, so you don't need to release anything.

boost asio streambuf don't release memory after calling consume?

1 Answers