printf just before a delay doesn't work in C

Question

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          


        

Answer 1

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

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

Answer 2

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 3

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 4

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);

Answer 5

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

try printf ("hi world\n");