CODEWITHSUNDEEP
cprintf
Xidobix
Xidobix

Reputation: 2638

printf just before a delay doesn't work in C

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

Upvotes: 6

Views: 5791

Answers (5)

Nick Bedford
Nick Bedford

Reputation: 4435

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);

Upvotes: -1

Roland Rabien
Roland Rabien

Reputation: 8836

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

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

Upvotes: 21

Tordek
Tordek

Reputation: 10872

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();

Upvotes: 1

Greg Hewgill
Greg Hewgill

Reputation: 992937

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.

Upvotes: 8

Ben
Ben

Reputation: 7608

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

try printf ("hi world\n");

Upvotes: 10

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