When printf a string we are not using *. Why?

Question

In c programming when we print a string. We are not using * . But when print a number using printf we are using *. So how it is understanding, i am printing a string or int. Is understanding using %s operator?

Attaching an example code

#include<stdio.h>

int main(int argc,char* argv){

char data[]="This is an example of pointer";
char *pointerstringdata =data;
printf("print the string data is >> %s\n",pointerstringdata);   /* Here we are not using * why?  case -1*/


int numberdata =100;
int *pointerintdata=&numberdata;
printf("print the int data is >> %d\n",*pointerintdata);   /* Here we are  using * why?  case -2*/
return 0;
}

Answer 1

Because pointers hold reference to the object. * dereference this object. So if the pointer holds the reference to the char object when we dereference it we get this object. So dereference char pointer is just the single char not the address of the first char in the string.