printing seconds one by one in loop with sleep() function

Question

for(j = 1; j < 11; j++)
        {
            printf("%ds ",j);
            sleep(1);
        }

i have expected that this code will print "1s 2s 3s ... 10s" where there is 1 second interval between them but instead it prints all the numbers after 10 seconds. how should i fix it? i'm working on ubuntu.

score 2 · Accepted Answer · answered Apr 07 '11 at 19:34

2

The stdout stream is buffered, so if you immediately want your results you need to flush that buffer with fflush():

for (j = 1; j < 11; j++) {
    printf("%ds ", j);
    fflush(stdout);
    sleep(1);
}

answered Apr 07 '11 at 19:34

orlp

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score 0 · Answer 2 · edited May 23 '17 at 10:33

0

add a \n to your printf to flush.

for(j = 1; j < 11; j++)
        {
            printf("%ds\n",j);
            sleep(1);
        }

For more info, see here.

edited May 23 '17 at 10:33

Community

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1

answered Apr 07 '11 at 19:30

NKCSS

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Incorrect. Newlines are not required to flush the stdout buffer and more importantly will break the original asker's format. – orlp Apr 07 '11 at 19:33
It was just a suggestion; I posted the link to a previous question that talks all about fflush() and disabling the buffer. Let's not rehash it all here, let him read the original question. – NKCSS Apr 07 '11 at 19:34
Let's not rehash it all here? You suggested something completely irrelevant: newlines. Only one part of your answer was useful, an other question. – orlp Apr 07 '11 at 19:36
@nightcracker: I might not have formulated it in the best possible way, but what I tried to say was that it didn't print till the end because the buffer wasn't flushed, something that printf automatically does when you put a newline character into your print string. – NKCSS Apr 07 '11 at 19:43
although only when printing to a terminal. – Neil Apr 07 '11 at 19:57

printing seconds one by one in loop with sleep() function

2 Answers2