How to make a simple C++ program in which std::cout is not flushed

Question

In an effort to better understand buffered streams in C++, I would like to write a simple program in which the std::cout buffer is NOT flushed before termination. Since I have read that std::cout is flushed on normal termination, I tried throwing a runtime error. I also avoided using std::endl, as I understand that forces a flush. First attempt:

//file noflush.cpp
#include <iostream>

int main() {
    std::cout << "Don't write me to the console!";
    throw 0;
}

Compile with g++, call from terminal:

$ ./noflush
libc++abi.dylib: terminating with uncaught exception of type int
Don't write me to the console!Abort trap: 6

Even when I force a runtime error, it seems the buffer still gets flushed on termination. Is it possible to "strand" some data in the buffer, leaving it unwritten to the device?

Answer 1

This is not standard c++, but in POSIX, you can send a "kill" signal to kill the running process. This will stop the execution without cleanup such as flushing buffers.

Edit: I realized that signals are not only POSIX but actually part of C standard library (and included in the C++ standard library).

#include <csignal>
// ...
std::cout << "Don't write me to the console!";
std::raise(SIGKILL);

Answer 2

As far as I can tell, there is no standard compliant and clean way to avoid std::cout to flush() before program termination (but, of course, you can use unclean methods, e.g. raising a signal either directly or indirectly). According to cppreference, the actual buffer type controlled by std::cout is implementation defined but derived from std::streambuf, which does not appear to allow public access in a way that would emulate silent swallowing of the buffer.

Further, as I noted in a comment, even abnormal program termination (via std::terminate() or std::abort() may or may not close open resources, so this is again implementation defined.