Is it possible to initialize a C pointer to NULL?

TL;DR Yes, very much.

The actual claim made on the guide reads like

On the other hand, if you use just the single initial assignment, int *my_int_ptr = 2;, the program will try to fill the contents of the memory location pointed to by my_int_ptr with the value 2. Since my_int_ptr is filled with garbage, it can be any address. [...]

Well, they are wrong, you are right.

For the statement, (ignoring, for now, the fact that pointer to integer conversion is an implementation-defined behaviour)

int * my_int_ptr = 2;

my_int_ptr is a variable (of type pointer to int), it has an address of its own (type: address of pointer to integer), you are storing a value of 2 into that address.

Now, my_int_ptr, being a pointer type, we can say, it points to the value of "type" at the memory location pointed by the value held in my_int_ptr. So, you are essentially assigning the value of the pointer variable, not the value of the memory location pointed to by the pointer.

So, for conclusion

 char *x=NULL;

initializes the pointer variable x to NULL, not the value at the memory address pointed to by the pointer.

This is the same as

 char *x; x = NULL;    

Expansion:

Now, being strictly conforming, a statement like

 int * my_int_ptr = 2;

is illegal, as it involves constraint violation. To be clear,

• my_int_ptr is a pointer variable, type int *
• an integer constant, 2 has type int, by definition.

and they are not "compatible" types, so this initialization is invalid because it's violating the rules of simple assignment, mentioned in chapter §6.5.16.1/P1, described in Lundin's answer.

In case anyone's interested how initialization is linked to simple assignment constraints, quoting C11, chapter §6.7.9, P11

The initializer for a scalar shall be a single expression, optionally enclosed in braces. Theinitial value of the object is that of the expression (after conversion); the same typeconstraints and conversions as for simple assignment apply, taking the type of the scalarto be the unqualified version of its declared type.

The tutorial is wrong. In ISO C, int *my_int_ptr = 2; is an error. In GNU C, it means the same as int *my_int_ptr = (int *)2; . This converts the integer 2 to a memory address, in some fashion as determined by the compiler.

It does not attempt to store anything in the location addressed by that address (if any). If you went on to write *my_int_ptr = 5;, then it would try to store the number 5 in the location addressed by that address.

To clarify why the tutorial is wrong, int *my_int_ptr = 2; is a "constraint violation", it is code which is not allowed to compile and the compiler must give you a diagnostic upon encountering it.

As per 6.5.16.1 Simple assignment:

Constraints

One of the following shall hold:

• the left operand has atomic, qualified, or unqualified arithmetic type, and the right has arithmetic type;
• the left operand has an atomic, qualified, or unqualified version of a structure or union type compatible with the type of the right;
• the left operand has atomic, qualified, or unqualified pointer type, and (considering the type the left operand would have after lvalue conversion) both operands are pointers to qualified or unqualified versions of compatible types, and the type pointed to by the left has all the qualifiers of the type pointed to by the right;
• the left operand has atomic, qualified, or unqualified pointer type, and (considering the type the left operand would have after lvalue conversion) one operand is a pointer to an object type, and the other is a pointer to a qualified or unqualified version of void, and the type pointed to by the left has all the qualifiers of the type pointed to by the right;
• the left operand is an atomic, qualified, or unqualified pointer, and the right is a null pointer constant; or
• the left operand has type atomic, qualified, or unqualified _Bool, and the right is a pointer.

In this case the left operand is an unqualified pointer. Nowhere does it mention that the right operand is allowed to be an integer (arithmetic type). So the code violates the C standard.

GCC is known to behave poorly unless you explicitly tell it to be a standard C compiler. If you compile the code as -std=c11 -pedantic-errors, it will correctly give a diagnostic as it must do.