Is it possible to initialize a C pointer to NULL?

Question

I had been writing things like

char *x=NULL; 

on the assumption that

 char *x=2; 

would create a char pointer to address 2.

But, in The GNU C Programming Tutorial it says that int *my_int_ptr = 2; stores the integer value 2 to whatever random address is in my_int_ptr when it is allocated.

This would seem to imply that my own char *x=NULL is assigning whatever the value of NULL cast to a char is to some random address in memory.

While

#include <stdlib.h> #include <stdio.h>  int main() {     char *x=NULL;      if (x==NULL)         printf("is NULL\n");      return EXIT_SUCCESS; } 

does, in fact, print

is NULL

when I compile and run it, I am concerned that I am relying on undefined behavior, or at least under-specified behavior, and that I should write

char *x; x=NULL; 

instead.

Answer 1

Is it possible to initialize a C pointer to NULL?

TL;DR Yes, very much.

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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;     

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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. The initial value of the object is that of the expression (after conversion); the same type constraints and conversions as for simple assignment apply, taking the type of the scalar to be the unqualified version of its declared type.