C Programming - Library Functions - Discussion

In my case, it is printing %.0#s IndiaBIX.

Please, anyone, help me by explaining this clearly.

Why we used the #symbol in printf statement? Please explain to me.

It will print %.0#s IndiaBIX.

It shows following error.

In function \'main\':
demo6.c:6:12: warning: unknown conversion type character \'#\' in format [-Wformat=]
printf("%.#s %2s", str, str);
^
demo6.c:6:12: warning: too many arguments for format [-Wformat-extra-args]

Can anyone help me to get it?

Thank you all for explaining it

Hello, I didn't get the last line that %2s adds space before printing str.

Please explain.

This program is actually really very simple to understand. Let's just go step wise.

char str[] = "IndiaBIX"; this judt defines str.

printf ("%. #s %2s", str, str);

It seems a bit complex lets break it into two halves.

printf ("%. #s", str); this statement I syntactically correct i will tell you how.

This statement when executed prints %. #s in turbo C while %. 0's in GCC you must know that we have certain formatters like %. (integer)'s eg: % 5s this means the first five char are printed.

So when the compiler reads this statement after reading %. It expects an integer but it does not get that it reads # which is not an integer so it does not know what to do so in turbo c it simply treats this as a string not a formatter and simply prints %. #s while in GCC it does consider it as a string but itself adds 0 between. And # and then simply prints it.

Now coming onto the second halve i.e 5.

printf ("%2s", str); this is again for formatting one of the formatter it just adds 2 spaces before printing str.

Hope it is clear now.

char str[] = "IndiaBIX"; :- Declaration of character array.

printf("%.#s %2s", str, str); :- Its print %.#s due any symbol as u

print like anything in printf function and its print IndiaBix due %s, str here one more str which does not error in the program.

It prints %.#s IndiaBix . can any one explain tracing of this programme.

Why dont print the %.#s ?