Can someone please explain how stdio buffering works?

Question

I don't understand what the buffer is doing and how it's used. (Also, if you can explain what a buffer normally does) In particular, why do I need fflush in this example?

int main(int argc, char **argv)
{
    int pid, status;
    int newfd;  /* new file descriptor */

    if (argc != 2) {
        fprintf(stderr, "usage: %s output_file\n", argv[0]);
        exit(1);
    }
    if ((newfd = open(argv[1], O_CREAT|O_TRUNC|O_WRONLY, 0644)) < 0) {
        perror(argv[1]);    /* open failed */
        exit(1);
    }
    printf("This goes to the standard output.\n");
    printf("Now the standard output will go to \"%s\".\n", argv[1]);
    fflush(stdout);

    /* this new file will become the standard output */
    /* standard output is file descriptor 1, so we use dup2 to */
    /* to copy the new file descriptor onto file descriptor 1 */
    /* dup2 will close the current standard output */

    dup2(newfd, 1); 

    printf("This goes to the standard output too.\n");
    exit(0);
}

Answer 1

In a UNIX system the stdout buffering happens to improve I/O performance. It would be very expensive to do I/O every time.

If you really don't want to buffer there's some options:

1. Disable buffering calling setvbuf http://www.cplusplus.com/reference/cstdio/setvbuf/

2. Call flush when you want to flush the buffer

3. Output to stderr (that's unbuffered by default)

Here you've more details: http://www.turnkeylinux.org/blog/unix-buffering

I/O is an expensive operation, so to reduce the number of I/O operations the system store the information in a temporary memory location, and delay the I/O operation to a moment when it has a good amount of data.

This way you've a much smaller number of I/O operations, what means, a faster application.