Simple c code with sleep gives no output after execution

Question

I wrote the following piece of code to simulate a callback. But this got stuck and not even prints "inside main". I tried this on my unix machine and also on online compiler but same behavior. Have I missed anything ?

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

void sleep_50sec()
{
    printf("inside sleep_50sec");

    sleep(50);

}

int main()
{
    int i;
    printf("inside main");
    sleep(1);

    for( i =0; i < 100;i++)
    {   printf("Loop %d",i);
        sleep_50sec();
    }
return 0;
}

Output

root@xyz> ./a.out

`printf` is flushed to `stdout` when it sees a newline character `\n`. Append that to your `printf` calls and you should see some printouts. — yano, Jun 13 '17 at 15:07
@yano assuming you don't inherit a `stdout` set to `_IOFBF`. — , Jun 13 '17 at 15:09

arminb · Accepted Answer · 2017-06-13T15:12:03.187

3

Add \n when printing to stdout.

printf("inside main\n");

The reason a \n is needed is because printf flushes data to stdout after \n was reached.

You can print to stderr to see your data immediately, without using \n:

fprintf(stderr, "test");

By the way, when calling printf() output will be written to stdout. So printf(...) is the same as fprintf(stdout, ...)

edited Jun 13 '17 at 15:12

answered Jun 13 '17 at 15:07

arminb

1,788
3
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2

with *some* explanation, this could be a good answer. – Jun 13 '17 at 15:08

score 1 · Answer 2 · answered Jun 13 '17 at 15:11

1

You have to flush the stdout. It can be done by the '\n' character inside the format string or by the fflush command:

void sleep_50sec() {
    printf("inside sleep_50sec");
    fflush(stdout);
    sleep(50);
}

answered Jun 13 '17 at 15:11

Akira

4,100
3
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Ishay Peled · Answer 3 · 2017-06-14T07:34:14.617

0

The reason you're not seeing anything printed is because the crt (C runtime) buffers your outputs and will only output to the screen after K characters have been accumulated. One solution will be to use '\n' - this hints crt that a complete line has been written, and that it should print it to the screen. Another solution will be to implicitly flush stdout like in my example below:

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

void sleep_50sec()
{
    printf("inside sleep_50sec");
    fflush(stdout);

    sleep(50);

}

int main()
{
    int i;
    printf("inside main");
    fflush(stdout);
    sleep(1);

    for( i =0; i < 100;i++)
    {   printf("Loop %d",i);
        fflush(stdout);
        sleep_50sec();
    }
return 0;
}

edited Jun 14 '17 at 07:34

answered Jun 13 '17 at 15:11

Ishay Peled

2,600
1
19
34

It's your C runtime that controls the buffering. – Jun 13 '17 at 15:13
@FelixPalmen correct, fixed – Ishay Peled Jun 13 '17 at 15:14
@IshayPeled How do you "flush" the CRT? I think you mean `stdout` and that is explicit flushing not implicit flushing. – Ajay Brahmakshatriya Jun 14 '17 at 07:19
@AjayBrahmakshatriya Of course you're right (: corrected – Ishay Peled Jun 14 '17 at 07:34

Simple c code with sleep gives no output after execution

3 Answers3