Difference between revisions of "cpp/io/manip/endl"

Revision as of 20:14, 18 November 2017

Inserts a newline character into the output sequence os and flushes it as if by calling os.put(os.widen('\n')) followed by os.flush().

This is an output-only I/O manipulator, it may be called with an expression such as out << std::endl for any out of type std::basic_ostream.

Notes

This manipulator may be used to produce a line of output immediately, e.g. when displaying output from a long-running process, logging activity of multiple threads or logging activity of a program that may crash unexpectedly. An explicit flush of std::cout is also necessary before a call to std::system, if the spawned process performs any screen I/O. In most other usual interactive I/O scenarios, std::endl is redundant when used with std::cout because any input from std::cin, output to std::cerr, or program termination forces a call to std::cout.flush(). Use of std::endl in place of '\n', encouraged by some sources, may significantly degrade output performance.

In many implementations, standard output is line-buffered, and writing '\n' causes a flush anyway, unless std::ios::sync_with_stdio(false) was executed. In those situations, unnecessary endl only degrades the performance of file output, not standard output.

The code samples on this wiki follow Bjarne Stroustrup and The C++ Core Guidelines in flushing the standard output only where necessary.

When an incomplete line of output needs to be flushed, the std::flush manipulator may be used.

When every character of output needs to be flushed, the std::unitbuf manipulator may be used.

Parameters

Return value

os (reference to the stream after manipulation)

Example

#include <iostream>
#include <chrono>
 
template<typename Diff>
void log_progress(Diff d)
{
    std::cout << std::chrono::duration_cast<std::chrono::milliseconds>(d).count()
              << " ms passed" << std::endl;
}
 
int main()
{
    std::cout.sync_with_stdio(false); // on some platforms, stdout flushes on \n
    volatile int sink = 0;
 
    auto t1 = std::chrono::high_resolution_clock::now();
    for (int j=0; j<5; ++j)
    {
        for (int n=0; n<10000; ++n)
            for (int m=0; m<20000; ++m)
                sink += m*n; // do some work
        auto now = std::chrono::high_resolution_clock::now();
        log_progress(now - t1);
    }
}

487 ms passed
974 ms passed
1470 ms passed
1965 ms passed
2455 ms passed

See also