Why Does `System.out.println` and `System.err.println` Sometimes Produce Out-of-Order Output in Java?

Out-of-Order Output from System.out.println and System.err.println

In Java, System.out.println() writes to the standard output stream, while System.err.println() writes to the standard error stream. Typically, these streams are printed to the console in sequence. However, under certain circumstances, this behavior may not be consistent.

Consider the following code snippet:

public static void main(String[] args) {
    for (int i = 0; i < 5; i++) {
        System.out.println("out");
        System.err.println("err");
    }
}

When you run this program, it produces the following output:

out
out
out
out
out
err
err
err
err
err

Instead of alternating between "out" and "err," the output shows all "out" messages followed by all "err" messages. This discrepancy is explained by the different nature of the standard output and error streams.

Output streams in Java are cached, meaning that data is not immediately written to the console. Instead, it is stored in an internal buffer and flushed periodically. The flushing process is triggered by various criteria, such as a certain period of inactivity or the buffer reaching a specific size.

In the case of the code snippet above, both standard output and error streams are writing to their respective buffers. Since the writes are not synchronized, it is possible for the error stream's buffer to fill up first and be flushed, even though some "out" messages still remain in the output stream's buffer. This results in the observed out-of-order output.

To resolve this issue, you can ensure that both output streams are flushed after each write. The following code includes calls to System.out.flush() and System.err.flush() within the loop:

public static void main(String[] args) {
    for (int i = 0; i < 5; i++) {
        System.out.println("out");
        System.out.flush();
        System.err.println("err");
        System.err.flush();
    }
}

With this modification, the output will be printed in the expected alternating order:

out
err
out
err
out
err
out
err
out
err

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