std:: endl

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

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See also