std::endl 和 '\n' 之间的区别用于流缓冲区实现

发布于 2024-11-15 05:12:08 字数 706 浏览 5 评论 0原文

我目前正在尝试实现 stringbuf 的子类，以允许缓冲区标记特定字符（在我的例子中为“\n”），并在出现此字符时采取操作（将消息转储到记录器，然后在我的记录器中清除缓冲区）案件）。为了实现这个目标，我覆盖了 sputc （以实现对 '\n' 的监视）和 xsputn （确实使用 sputc，因为 GCC 实现默认情况下似乎不这样做）。出于调试目的，我让 sputc 将传递给它的每个字符写出到 stdout。

现在这是我的问题：如果我使用像

mystream << "Some text" << std::endl;

sputc 这样的东西接收除了 '\n' 之外的每个字符，该字符应该由 std::endl 引入，所以预期的操作不会完成，因为 '\n' 没有过去了。如果我使用类似的东西

mystream << "Some text" << '\n';

，甚至

mystream << "Some text" << "\n" << std::flush;

一切都按预期工作，并且我的 sputc 实现得到 '\n' 字符。

所以我的问题是：两个代码行是否应该对后面的 stringbuf 执行完全相同的操作，如果不是，我必须重写哪些其他方法才能获取 '\n'？

原文

I'm currently trying to implement a subclass of stringbuf to allow the buffer to tokenize for specific chars ('\n' in my case) and undertake an action if this char occurs (dump the message to a logger and clear buffer afterwards in my case).
To achieve this goal, I overrode sputc (to implement watching out for the '\n') and xsputn (to use sputc indeed, as the GCC implementation doesn't seem to do this by default).
For debugging purposes, I let sputc write out each character that is passed to it to stdout.

Now this is my question: If I use something like

mystream << "Some text" << std::endl;

sputc receives each character except of the '\n' which should be inducted by std::endl, so the action that is expected is not done because the '\n' isn't passed on.
If I use something like

mystream << "Some text" << '\n';

or even

mystream << "Some text" << "\n" << std::flush;

everything works as expected and my sputc implementation gets the '\n' char.

So my question is: Shouldn't both code lines do exactly the same concerning the stringbuf behind, and if not, which other methods do I have to override to get the '\n'?

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