%n specifier use in C

[Hetfield]

Can you help me why %n specifier is used for in this programming example that i get from deitels book.

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

int main()
{
  int a[10] = { 0 };            /* random integers from 1 to 1000 */
  int i;                        /* loop counter */
  int count;                    /* number of characters in current value */
  int totalCount = 0;           /* total characters in array */

  srand(time(NULL));

  /* fill the array with random numbers */
  for (i = 0; i <= 9; i++) {
    a = 1 + rand() % 1000;
  }                             /* end for */

  /* print table headers */
  printf("%s\t%s\n", "Value", "Total characters");

  /* loop through 10 elements */
  for (i = 0; i <= 9; i++) {
    printf("%d%n", a, &count);
    totalCount += count;        /* update totalCount */
    printf("\t%d\n", totalCount);
  }
}

[Corrosio]

The percent symbol is a wildcard. The percent symbol tells the compiler where to substitute the value from the variable in the string.
In your case


printf ("% s \ t% s \ n", "Value", "Total characters");


you say you want to print two string values. But I assume the output will be integers, so you should use the following:
pro integer
printf ("% i \ t% i \ n", "Value", "Total characters");


or
or a number of the double data type.
printf ("% d \ t% d \ n", "Value", "Total characters");

[hamster_nz]

From "man fprintf":

n The number of characters written so far is stored into the integer pointed to by the corresponding argument. That
argument shall be an int *, or variant whose size matches the (optionally) supplied integer length modifier. No
argument is converted. (This specifier is not supported by the bionic C library.) The behavior is undefined if
the conversion specification includes any flags, a field width, or a precision.

Never knew that....