Isn't the line 1 and line 4 means the same that is in *ptr and hello both contain manash so why hello runs but *ptr doesn't?

Question

#include <stdio.h>
#include <string.h>
int main(){
char hello[] = "manash";

char * ptr = &hello;
printf("%s\n",hello);

printf("%s\n",ptr);
printf("%s\n",&hello);

printf("%s\n",*ptr);

return 0;}

Answer 1

#include <stdio.h>
#include <string.h>
int main(){
char hello[] = "manash";

char * ptr = hello; //no need of & since it is charecter array, already points to starting addrress
printf("%s\n",hello); //%s prints total array

printf("%s\n",ptr);
printf("%p\n",&hello); // &hello returns address of hello array..

printf("%c\n",*ptr); //*ptr points to the value of first charecter array , not total array. and ptr returns total array.

return 0;
}

//hope it helps...

Answer 2

There *ptr returns a single first charecter so need %c ..

Answer 3

Okay I got your point so does that mean
Printf("%c\n,*ptr); gives output m

My coding c app give me error but it's okay if It atleast give me m as an output

Answer 4

Ooo the small details
You Gotcha me
Well thanks for the help!