different output with exit() and _exit()?

问题

why it showing different output???can anyone explain me in depth.

1.

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

int main (void)
{
printf ("Using exit ... \ n");
printf ("This is the content in buffer");
exit (0);
}

Output:

Using exit ... This is the content in buffer

2.

# Include <unistd.h>
# Include <stdio.h>

int main (void)
{
printf ("Using exit ... \ n");
printf ("This is the content in buffer");
_exit (0);
}

Only output:

Using exit ...

回答1:

If we read _exit()'s documentation, we note:

Causes normal program termination to occur without completely cleaning the resources.

This presumably would include flushing stdout.

回答2:

First thing you should know is STDOUT is line buffered, means it flushes the memory after getting '\n'. Second thing is exit() flushes the stdio buffer while _exit() won't.

Now in your case, in first example exit() flushes the stdio buffer, So it is printing both printf output while in _exit() no flushing takes place so it is not printing both printf statement.

If you want to get correct output on the second statement, disable the buffering by putting

int main (void)
{
setbuf(stdout, NULL);
printf ("Using exit ... \ n");
printf ("This is the content in buffer");
_exit (0);
}

来源：https://stackoverflow.com/questions/25222617/different-output-with-exit-and-exit