Why do I need "\n" in c, when using the sleep function?

Question

My code:

#include <stdio.h>
#include <unistd.h>

int main(){
    printf("I sleep\n");
    sleep(3);
    printf("\033[H\033[J");
    return 0;
}

if I don't write "\n" in the printf-function, "I sleep" wil not be displayed, until sleep(3) is done.

Can someone explain? Thanks!

This is because `stdout` is (generally) line buffered. Try `fprintf`ing to `stderr` without `'\n'` instead and it should probably work fine. — mediocrevegetable1, Oct 28 '21 at 11:58
Surely we must have some good duplicate for this? I can't find one though. — Lundin, Oct 28 '21 at 12:58

dbush · Answer 1 · 2021-10-28T14:02:23.153

1

The standard output stream, when not outputting to a file or device, is line-buffered by default on most UNIX-like systems.

That mean that text sent there typically won't actually be printed until a newline character is encountered.

edited Oct 28 '21 at 14:02

answered Oct 28 '21 at 11:59

dbush

Thanks mate, now I understand! – stefanwgn Oct 28 '21 at 12:06
"line-buffered by default. " --> Hmmm, I do not see any C spec that says that. Instead common practice. – chux - Reinstate Monica Oct 28 '21 at 13:56

1 Answers1