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Home/ Questions/Q 597113
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T16:14:41+00:00 2026-05-13T16:14:41+00:00

#include <stdio.h> #include <stdlib.h> #include <unistd.h> int main(void) { int i=0; while(i<10) { printf(%d,

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

int main(void)
{
        int i=0;
        while(i<10)
        {
                printf("%d", i);
                usleep(10000); // or sleep(1)
                i++;
        }
        return 0;
}

I want the program to last 10 secs, i.e. print 1 – wait 1 sec – print 2 – wait 1 sec and so on until the end. But it doesn’t do that – it just waits for all the time (10 secs) and then prints the whole array of numbers together without any time delays between them, it just prints 0123456789 at once.
EDIT: I tried with sleep() instead of usleep but it’s the same
How to fix it ? And why it’s like that ?

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  1. Editorial Team
    Editorial Team
    2026-05-13T16:14:42+00:00Added an answer on May 13, 2026 at 4:14 pm

    Your output buffer is not being flushed. By default, output is written when a new line appears in the stream. Change your printf to this:

    printf("%d\n", i);
    

    or try this:

    printf("%d", i);
    fflush(stdout);
    

    Also, if you want to remove the line-buffering behaviour, you can use setvbuf() and the _IONBUF mode.

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