Simple c code with sleep gives no output after execution [duplicate] - c

Question

This question already has answers here:
"printf" doesn't print a string immediately [duplicate]
(2 answers)
Closed 5 years ago.
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

Answer 1

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, ...)

Answer 2

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);
}

Answer 3

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;
}