Why i can init char pointer but not int pinter?

Question

I'm recollecting my C programming skills after 1 year. So I decided to start from scratch. I got stuck with poineters

why i can do :

char *string="Hello";
printf ("%s",string);

But i can't do :

int *pt=23;
printf("%d",*pt);

Doesn't the pointer need to be an address ? but why the first example works?

Answer 1

Because using string literals to initialize a character pointer or array is a special allowed syntax, with specific rules. In your case you set a pointer to point at a string literal's address, the string literal itself having type char[] and existing in read-only memory.

For the integer case, yes it needs to be an address, or more specifically another integer pointer. You can't initialize a pointer with an integer. See:
"Pointer from integer/integer from pointer without a cast" issues

Answer 2

The string literal "Hello" is stored in memory as a character array like

char unnamed_literal[] = { 'H', 'e', 'l', 'l', 'o', '\0' };

So in this declaration

char *string="Hello";

the pointer string is assigned with the address of the first character of the already existent array. It can be imaging like

char unnamed_literal[] = { 'H', 'e', 'l', 'l', 'o', '\0' };
char *string = unnamed_literal;;

As for this declaration

int *pt=23;

then the value 23 is not a valid address that points to a valid object defined in your program. The compiler should issue a message that you are trying to assign an integer to a pointer. Thus this call

printf("%d",*pt);

invokes undefined behavior.

To make an analogy with the initialization of a pointer by string literal you could write for example

int a[] = { 1, 2, 3 };
int *pt = a;

Answer 3

This is syntaxic sugar from standard C:

What is actually stored in your char pointer is the address at which the string is loaded in memory when starting the program.

For an integer however, it will try to set the address of the pointer to 23 which is (and should be) invalid.

Answer 4

For example an integer, like:

int VAL = 10;

int *ptr = &VAL; /* ptr now points to val */

*ptr = DESIRED_VALUE; /* *(asterisk) here means dereference, meaning that the value of VAL will change */

or

int *&VAL = DESIRED_VALUE;

Answer 5

    int variable_1 = 23;
    int *pointer_to_int = &variable_1;

The first line is an integer variable, initialized to an integer value.

The second line is a pointer to integer variable, initialized to an address of an integer variable (the address of the variable variable_1, defined in the previous declaration.

    char *p = "Charles Dickens";

This is a char pointer variable, initialized to the address of a string literal, which is a constant array of characters, composed of the characters between the quotes, plus an aditional \0 character, that represents the end of the string literal.

    char c = 'a'; /* char variable initialized to value 'a' */
    char *pointer_to_c = &c;

Probably you will find this last declaration very similar to the one above. The different thing is that in C you can use string literals, but a single character is similar to an integer, and this is the way pointer variables initialize: with variables addresses.