Question

In the following code:

#include<stdio.h>

int main(int argc,char *argv[]){
  int index;
  for(index = 0; index < argc; index++) {
    printf("The %d is %s\n",index,argv[index]);
  }
  return 0;
}

from the declaration char *argv[],cdecl says it means
declare argv as array of pointer to char, so I think the code should write like this :
printf("The %d is %s\n",index,*argv[index]);

as the following code:

void give_me_ptr(int *ptr){
  printf("the value of ptr is %p\n", ptr);
  printf("the value of ptr is %d\n", *ptr);
}

int main(void){
  int a = 10,index;
  give_me_ptr(&a);
  return 0;
}

it turns out

the value of ptr is 0x7fff76010b48
the value of ptr is 10

to me it means I should add * before pointer variable.

Answer 1

In printf function, format specifier %s requires a pointer to the first character of the string as the corresponding argument. This is why you pass argv[index] to it. argv[index] is a pointer.

If you passed *argv[index], you would be passing only the first character of the argv[index] string, which is incorrect. %s requires a pointer to the first character, not the character itself.