Why does "printf" not produce any output?

Question

I am learning to program in C. Could you explain why nothing is printed here?

#include <stdio.h>

int main (void)
{
    char a[]="abcde";
    printf ("%s", a);
}

Answer 1

On many systems printf is buffered, i.e. when you call printf the output is placed in a buffer instead of being printed immediately. The buffer will be flushed (aka the output printed) when you print a newline \n.

On all systems, your program will print despite the missing \n as the buffer is flushed when your program ends.

Typically you would still add the \n like:

printf ("%s\n", a);

An alternative way to get the output immediately is to call fflush to flush the buffer. From the man page:

For output streams, fflush() forces a write of all user-space buffered data for the given output or update stream via the stream's underlying write function.

Source: http://man7.org/linux/man-pages/man3/fflush.3.html

EDIT

As pointed out by @Barmar and quoted by @Alter Mann it is required that the buffer is flushed when the program ends.

Quote from @Alter Mann:

If the main function returns to its original caller, or if the exit function is called, all open files are closed (hence all output streams are flushed) before program termination.

See calling main() in main() in c

Answer 2

Strangely enough, it seems that nobody posted the adjusted code where the buffer is flushed yet...:

#include <stdio.h>

int main (void)
{
    char a[]="abcde";
    printf ("%s", a);
    fflush(stdout);
    //On some systems the line above will fail, in that case use: fflush(NULL);
}

Also note that this code probably doesn't do what you actually want to do.
What I assume you really want to do is:

#include <stdio.h>

int main (void)
{
    char a[]="abcde";
    printf ("%s\n", a);
   //The '\n' makes sure the next thing you print will be on the following line
}