Thread: a pointer points itself

Hi

I wrote this code for fun, just wanted to see what will happen! I declared o pinter that holds its own address. But there is something wrong!

Code:

#include <stdio.h>

int main()
{
	char foo='f';	
	char *pc;
	char **ppc;

	pc=&foo;

	ppc=&pc;

	printf("pc  -> %p\n",pc);
	printf("&pc -> %p\n",&pc);
	printf("ppc -> %p\n",ppc);
	pc=&pc;
	//ppc=&pc;
	//pc=*ppc;
	printf("pc  -> %p\n",pc);

	return 0;
}

The code is compiled with "incompatible assignment" warning. But it works!

But, i think the right way of doing this is to declare a **char and hold the address of *char in **char. But this does not work! When I uncomment the commented lines and leave out the pc = &pc; line, the code does not work!

Code:

char foo='f';	
char *pc;
char **ppc;

pc=&foo;
ppc=&pc;

printf("pc  -> %p\n",pc);
printf("&pc -> %p\n",&pc);
printf("ppc -> %p\n",ppc);

/* pc=&pc; */
ppc=&pc; 

/* REPLACE 
    pc=*ppc; WITH : */
pc = ppc;
printf("pc  -> %p\n",pc);

Well, pc holds an address information. ppc is also holds an address information. It is logical to equate them but one holds an address of a pointer and the other holds address of an ordinary variable.

I think there should be a difference.If not so, what is the point of using a **char (just to create pointer arrays?)