why use of %n in printf() is not printing the number of variable occurrence before %n In C?

Question

#include <stdio.h>
int main ()
{ 
   int c;
   printf ("the value of %nc :  
   ", &c);

   return 0;
}

Output : the value of 0

Answer 1

Per C 2018 7.21.6.1 8, for the conversion specifier n :

The argument shall be a pointer to signed integer into which is written the number of characters written to the output stream so far by this call to fprintf [or printf ].…

Thus, the effect of printf ("the value of %nc : ", &c); is to write the characters “the value of c : ” to output and to put the number of characters in “the value of ” in c , which is 13.

Answer 2

For starters you may not split a string literal with an intermediate new line character. So this call

   printf ("the value of %nc :  
   ", &c);

is syntactically invalid. Either write

   printf ("the value of %nc :  \n   ", &c);

or write

       printf ("the value of %nc :  \n"
   "", &c);

In the call above the function printf does not output the value of the variable c itself. You need an additional call of the function printf to output the value of the variable c .

If you want to do this in one line then you can write as it is seen in the demonstrative program below.

#include <stdio.h>

int main(void) 
{
   int c;

   printf( "%d\n",  ( printf ("the value of %nc :  ",  &c ), c ) );

    return 0;
}

The program output is

the value of c :  13

Or if you want to include a new line character in the outputted string literal you can rewrite the call pf printf the following way

printf( "%d\n",  ( printf ("the value of %nc :  \n   ",  &c ), c ) );

In this case the program output will look like

the value of c :  
   13