Question

printf("/*something else*/"); /*note that:without using \n in printf*/

I know printf() uses a buffer which prints whatever it contains when, in the line buffer, “\n” is seen by the buffer function. So when we forget to use “\n” in printf(), rarely, line buffer will not be emptied. Therefore, printf() wont do its job. Am I wrong?

Answer 1

The example you gave above is safe as there are no variable arguments to printf. However it is possible to specify a format string and supply variables that do not match up with the format, which can deliver unexpected (and unsafe) results. Some compilers are taking a more proactive approach with printf use case analysis, but even then one should be very, very careful when printf is used.