Why don't I need to dereference a character pointer in C before printing it?

Question

Why does this code work? I would expect that I would need to dereference ptr, printf("%s\n", *ptr); before I could print it out, but I get a Segmentation Fault if I try to do it that way.

#include <stdio.h>

int main(int argc, char *argv[])
{
        char name[] = "Jordan";
        char *ptr = name;
        printf("%s\n", ptr);
}

Hope you guys could give me some insight.

Answer 1

When you print string we need starting address of string.

printf("%s\n", ptr);
                ^ address with %s   

it prints chars till \0 nul encounter.

Whereas to print chat int .. we need value variable:

printf("%c\n", *ptr);
               ^ * with %c print first char

Where as in scanf() a string you always need to give address:

scanf("%s", ptr);
            ^ string address

Also for int scanf() a char

scanf("%c", ptr);
            ^ read at first location char address 

Note: Scanf() need address with %c to store a scanned value in memory.

Be careful your ptr points to a constant string so you can't use in scanf.

Why Segmentation fault with following code ?

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

When you do like this, because of %s printf interprets *ptr as an address, but it's actually not an address and if you treat it as address it points to some location that is read protected for your program(process) So it causes a segmentation fault.

Your ptr via name points to some constant string in memory ("Jordan") as in below diagram:

name 2002
┌─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┐
│ 'J' │ 'o' │ 'r' │ 'd' │ 'a' │ 'n' │'\0' │ ........
└─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┘
  ^
  |
 ptr = name 

 ==> ptr = 2002
     *ptr = 'J'

In printf("%s\n", *ptr); the *ptr = 'J' and ASCII value of char 'J' is 74 but 74 address is not under your process control and you are trying to read from that memory location and its a memory violation and segmentation fault occurs.

If you compile you code containing printf("%s\n", *ptr); then with proper option say -Wall with GCC you will get a warning like below:

warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’

Says %s need (expects ) an address of type char* but you are putting value

notice:

printf("%s\n",   *ptr);
        ^          ^ argument-2
        argument-1