Why does printf() not print anything before sleep(

问题:

I'm just learning C with Kernighan and Ritchie's book; I'm in the basics of the fourth chapter (functions stuff). The other day I became curious about the sleep() function, so tried to use it like this:

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

int main(void)
{
  printf(" I like cows.");
  sleep(5);
  return 0;
}

The problem is the output of the program, it looks like it does the sleep() first and then the printf(), in other words, it waits five seconds and then prints the string. So I thought, maybe the program gets to sleep() so fast that it doesn't let printf() have his work done like I want, that is print the string and then sleep.

How can I show the string and then put the program to sleep? The compiler is GCC 3.3.5 (propolice) in OpenBSD 4.3.

PS I don't know how you put here the preprocessor lines correctly.

回答1:

printf() writes to stdout (the default output stream) which is usually line buffered. The buffer isn't flushed by the time sleep is called so nothing is displayed, when the program exits all streams are automatically flushed which is why it prints right before exiting. Printing a newline will usually cause the stream to be flushed, alternatively you could use the fflush function:

int main(void)
{
  printf(" I like cows.\n");
  sleep(5);
  return 0;
}

or:

int main(void)
{
  printf(" I like cows.");
  fflush(stdout);
  sleep(5);
  return 0;
}

If you are printing to a stream that is not line buffered, as may be the case if stdout is redirected or you are writing to a file, simply printing a newline probably won't work. In such cases you should use fflush if you want the data written immediately.

回答2:

Buffering means that all the output is stored in a place (called buffer) and is output after a certain amount of data is present in it. This is done for efficiency reasons.

Some (most?) implementations clear the buffer after a newline when writing to the console, so you can also try

printf(" I like cows.\n");

instead of the call to fflush()

回答3:

Your problem is that printf (and anything else that uses the stdio library to write to stdout (standard output)) is buffered - line buffered if it goes to the console, and size buffered if it goes to a file. If you do a fflush(stdout); after the printf, it will do what you want. You could try just adding a newline ('\n') to your string, and that would do the right thing as long as you don't redirect standard output to a file.

I'm not 100% sure, but I think stderr isn't buffered, which can cause confusion because you might see output you made to stderr before output you previously made to stdout.

回答4:

I implemented time encounter as following;

for (int i = 1; i <= 60; i++) {
    printf("%02d", i);
    fflush(stdout);
    sleep(1);
    printf("\b\b");
}