Confused with syntax of String pointer to print values

I just learned pointers and I am quite confused with the syntax.

In my code, I’ve created a pointer variable ptr_int that points to int x[].
To print the value of the first element in the array, I would *ptr_int
To print the address of the first element in the array, I would use ptr_int

However, now I create a pointer variable str_ptr that points to an array of characters (string).

My question is,

1. Why can’t I use *str_ptr to print the value of the first element in the array?

2. Why can’t I use str_ptr to print the address of the first element in the array?

3. Why does str_ptr prints out the entire String? Doesn’t str_ptr points to the address of the first element only?

    int x[5] = {1, 2, 3, 4, 5};
    int *ptr_int;
    ptr_int = x;
   
    printf("\n\n The address of ptr_int is: %u", ptr_int);
    printf("\n The value of ptr_int is: %d", *ptr_int);
   
    char *str_ptr = "Character string to be printed";
    // Confused??
    printf("\n %s", str_ptr);
   

My expected output for str_ptr

1. *str_ptr = C
2. str_ptr = <address of first element, C>

Actual output

>Solution :

0. The format string %p is used to print void pointers. So it should be:

printf("\n\n The address of ptr_int is: %p", (void *) ptr_int);

1. If you want to print the first character you need to do either:

printf("%c\n", *str_ptr); // or
printf("%.1s\n", str_ptr);

2. In order to print the address of the first element in the array you would do:

printf("%p\n", (void *) str_ptr); // or
printf("%p\n", (void *) &str_ptr[0]);

3. A pointer to array points to the whole array. Pointer to first element of array points to a single value. Both pointers point to same memory address.