提高ASIO流缓冲调用消耗后不释放内存？ [英] boost asio streambuf don't release memory after calling consume?

问题描述

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);

在消耗方法被调用时，由流缓冲占用b 内存不会被释放。内存将成长为 async_read_until 被多次调用。是我的使用是否正确？有什么办法释放之前内存streambuf的get指针？

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?

推荐答案

ASIO ::流缓冲基于的std ::载体，生长需要，但从来没有收缩。
因此，消费（）是不应该释放内存，它只是调整内部指针：

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));
}

但每次消费（），并再次阅读（）的时候，内部缓冲区（矢量）被再利用，所以你不需要任何发布

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

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