Question

{
    int *p=12;
    printf("%p",p);
    printf("\n%d",p);
}

OUTPUT:

0000000C

12

Question: So is p assigned the address 0x0000000C?

{ 
    int *p=12;
    *p=22;
}    

But this one doesn’t run. So what’s actually happening?

Answer 1

int *p=12;

This declares a pointer and sets the address to which it points to 12.

*p=22;

This de-references the pointer and writes 22 to the int at that memory address 12. Since you did not allocate any memory and just set the pointer to point at a random address, it results in a runtime error.

What is confusing you is that both pieces of code contain *p=.... However, the first assignment is to the pointer, and the second assignment is to the pointee. This is just one of those notational overloadings that you have to get used to when programming in C.