2 ways of accessing value of an array of pointers in C -

the first block

#include <stdio.h>  const int max = 3;  int main() {      int  var[] = { 10, 100, 200 };     int i, *ptr[max];      (i = 0; < max; i++) {         ptr[i] = &var[i]; /* assign address of integer. */     }      (i = 0; < max; i++) {         printf("value of var[%d] = %d\n", i, *ptr[i]);     }      return 0; } 

easy understand, since ptr array of int pointers. when need access i-th element, need dereference value *ptr[i].

now second block, same, points array of char pointer:

#include <stdio.h>  const int max = 4;  int main() {      char *names[] = {         "zara ali",         "hina ali",         "nuha ali",         "sara ali",     };      int = 0;      (i = 0; < max; i++) {         printf("value of names[%d] = %s\n", i, names[i]);     }      return 0; } 

this time, when need access element, why don't add * first?

i tried form correct statement print value, seems if dereference, single char. why?

printf("%c", *names[1]) // z 

i know there no strings in c, , char array. , know pointers, still don't under syntax here.

for printf() %s format specifier, quoting c11, chapter §7.21.6.1

s
if no l length modifier present, argument shall pointer initial element of array of character type. [...]

in case,

 printf("value of names[%d] = %s\n", i, names[i] ); 

names[i] is pointer, required %s. why, don't dereference pointer.

fwiw,

• %c expects int type argument (converted unsigned char,), need dereference pointer value.
• %d expects int argument, have go ahead , dereference pointer, mentioned in question.