printf just before a delay doesn't work in C

Question

Hi!

Does anyone know why if i put a printf just before a delay it waits until the delay is finished before it prints de message?

Code1 with sleep():

int main (void)
{
    printf ("hi world");
    system("sleep 3");    
}

Code2 with a self implemented delay:

void delay(float sec)
{
    time_t start;
    time_t current;
    time(&start);
    do{
        time(&current);
    }while(difftime(current,start) < sec);
}
int main (void)
{
    printf ("hi world");
    delay(3);    
}

And if:

printf ("hi world");
delay(3);    
printf ("hi world");
delay(3);

it waits until the sum of sleeps and then it prints the messages at the same time

Why does this happen?

UPDATE: I writed delay("sleep 3") when i called delay, i meant delay(3). Corrected

Answer 1

the standard output is not flush until you output a '\n' char.

try printf ("hi world\n");

Answer 2

printf buffers it's output until a newline is output.

Add a fflush(stdout); to flush the buffers on demand.

Answer 3

Because printf goes to stdout, which is buffered. The buffer is only flushed after it either fills up (not likely in your case) or when the program exits (after the sleep). Try fprintf to stderr instead. You can also change stdout to be line buffered with setlinebuf(stdout).

Answer 4

Normally, standard output is buffered until you either:

• output a \n character
• call fflush(stdout)

Do one of these things before calling delay() and you should see your output.

Answer 5

When you call printf, you don't print anything until really necessary: until either the buffer fulls up, or you add a new line. Or you explicitly flush it.

So, you can either do

printf("Something\n");
delay();

or

printf("Something");
fflush(stdout);
delay();

Answer 6

Technically that shouldn't even compile. In the delay("sleep 3") call you're trying to convert a const char * to a float. It should be:

void delay (float sec)
{
    // ...
}

delay(3);