Thread: Pointer pointing to itself

Code:

include<conio.h>
#include<stdio.h>
void main()
{
int number = 100;
int *numPtr =(&numPtr);
clrscr();  //sorry for the wrong positioning earlier :(
if(numPtr == *numPtr)
{
printf("wow!!!");
}
printf("hello!!! mr. %d %d %d",++*numPtr,numPtr,numPtr+1);
getch();
}

OUTPUT :
hello!!! mr. -13 -14 -12


Query 1 : Why is "wow" not printed?...The pointer 'numPtr' contains its own address(since it points to itself), and since the dereference happens on this 'self pointer' (*numPtr), it refers to the contents of 'numPtr' . Then why does it not print "wow"?
When i replace the above 'if' statement with 'if(numPtr != *numPtr)', amazingly the "wow" is again not printed?

Query 2:
The output too is confusing.
My numPtr somehow gets the value -14(ok...fine).
' ++*numPtr ' does -14+1 = -13(fine..again)
Now printing ' numPtr ' should give me -13(the address it contains, but the output shows -14)
Lets assume ' numPtr ' contains -14. Now ' numPtr+1 ' should give -13, but the output gives me -12.

Please help! . Am i making some basic conceptual mistake?

Thanks in advance

What is the type of numPtr?
What is the type of &numPtr?

1) Is fundamentally broken. A pointer to a pointer is not a pointer. So if you have an int, and take the address of that, you get an int*. Now, take the address of that and you get an int**. Type mismatch and therefore you are getting undefined behavior. Don't bother trying to figure it out. It just. won't. work. The end.

2) Where is the code?

3) Use int main.

Code:

int number = 100;
int *numPtr =(&numPtr);
if(numPtr == *numPtr)
{
printf("wow!!!");
}
clrscr();

Does clrscr(); have anything to do with not seeing "wow!!!", I don't that function available to me.

>Now printing ' numPtr ' should give me -13(the address it contains, but the output shows -14)
No, the address of numPtr is still -14 ( or the hexadecimal equivalent ). You only changed the value pointed to by numPtr in the last operation. This goes for the last question as well.

[EDIT : REASON: In reply to other posts]
Am I wrong in saying that &numPtr will be assumed by the compiler to be the address of an integer variable through implicit type coercion and hence when dereferenced it will retrieve the value of the address of numPtr in integer form?

clrscr is a non-standard function that clears the console screen.
The reason for not seeing the wow is undefined behavior. That statement is simply broken is so many horrible ways. Just give it some thought.

Originally Posted by CSaw

Code:

int number = 100;
int *numPtr =(&numPtr);
if(numPtr == *numPtr)
{
printf("wow!!!");
}
clrscr();

Does clrscr(); have anything to do with not seeing "wow!!!", I don't that function available to me.

>Now printing ' numPtr ' should give me -13(the address it contains, but the output shows -14)
No, the address of numPtr is still -14 ( or the hexadecimal equivalent ). You only changed the value pointed to by numPtr in the last operation. This goes for the last question as well.

[EDIT : REASON: In reply to other posts]
Am I wrong in saying that &numPtr will be assumed by the compiler to be the address of an integer variable through implicit type coercion and hence it will receive the value of the address of numPtr in integer form?

Firstly, you are wrong in not following the forum rules and posting ON THE SAME THREAD from multiple accounts just to confuse the hell out of us.

Secondly, the compiler doesn't assume anything, because unlike us it can't ponder things. All the compiler sees is <token> = <token>.

Thirdly, numPtr will not equal *numPtr(unless by some crazy chance) because numPtr is an address of an int and *numPtr is the int contained at that address. Also, numPtr is not even pointing to some VALID memory address therefore *numPtr could be anything. That is why WOW will probably NEVER get printed. However, as the posts above have indicated, welcome to the realm of undefined behavior.

EDIT: Disregard the first paragraph. I thought you were the same person.

Originally Posted by CSaw

Code:

//....
printf("hello!!! mr. %d %d %d",++*numPtr,numPtr,numPtr+1);
//....

From original post.

Ah, that. Well, once again undefined behavior. You cannot write and read from the same variable in the same statement because the order of evaluation is implementation defined.

Originally Posted by CSaw

Actually I tried it out on my compiler and wow was printed...here is my code:

Code:

int main()
{
	int *numPtr = &numPtr;
	if(numPtr == *numPtr)
	{
		puts("woow");
	}
	return 0;
}
/* OUTPUT VIA DEBUGGER */
// numPtr = 0x0012ff54
// *numPtr = 1245012
// 0x0012ff54 is the same as 1245012 in decimal

However I hear what you both ( claudiu and Elysia ) are saying about undefined behaviour but this makes sense to me, because it dereferences the value of numPtr and converts to an int, both of them are the same because it is pointing to itself. Am I getting this all wrong?

Whatever compiler you are using to compile that is retarded. Any decent compiler will warn you not to assign incompatible pointer types to one another.

Here is what GCC has to say about that piece of code:

"test.c: In function `main':
test.c:5: warning: initialization from incompatible pointer type
test.c:6: warning: comparison between pointer and integer
"

As a rule of thumb, in 99% of cases, don't try to outsmart the compiler. You won't.

The fact that you get those warnings means it's undefined behavior.

Btw, not all warnings mean 'undefined behaviour'.
This is what OP is doing exactly(idea).
Except it has(should has?) defined behaviour?

Code:

void *p = NULL;
p = &p;
if( p == *(void**)p ) {
  printf("WOW\n");
}