C Pointer question

[girish1026]

Hi to all.

I have a small question regarding pointers concept is C. The code is below

#include<stdio.h>
#include<conio.h>

int main(void);      //Prototype for main() function. Ofcourse not required

int main(void)
{

   int i=2,*j;
   j=&i;
   clrscr();

     printf("Value of i is %d\n",i);
     printf("Address of i is %u", j);

   getch();
   return 0;

}

Output:

Value of i is 2
Address of is is 65524

Here my question is, we declared *j as int. The int data type range is -32768 to 0 to +32767. It is in 2 Bytes size. But how it able to print the address value which is beyond the range 32767. Actually we need to declare *j as

unsigned int *j;

Because unsigned int range is 0 to 65535. But instead of this, if we declare *j as

int *j;

it is able to print the value beyond specified range. How this is possible. Please let me know.

Thanks to all.

[laserlight]

Here my question is, we declared *j as int. The int data type range is -32768 to 0 to +32767. It is in 2 Bytes size. But how it able to print the address value which is beyond the range 32767.

In the second printf, you did not print *j. You printed j. Normally, we would use %p instead of %u and cast j to void*.

[noobiept]

'j' is not an int, its a pointer to int.
So that range doesn't apply to pointers.

The value of 'j' is an address of memory, in your case, the address of the variable 'i'.

[girish1026]

In the second printf, you did not print *j. You printed j. Normally, we would use %p instead of %u and cast j to void*.

I am sorry laserlight. I did not understand your answer in clear way. Please let me know it in detail. Thanks.

[laserlight]

Actually, I think noobiept got your question better: you are confusing the pointer with what it points to. That is why you asked about *j when you printed j.

[girish1026]

Actually, I think noobiept got your question better: you are confusing the pointer with what it points to. That is why you asked about *j when you printed j.

Ya now i understood. Actually j points to the address of i and not actually it stores. So that we can print the address directly. Thanks for your reply.

[Elysia]

What? A pointer stores the address of whatever it points to, so of course you can print the address directly.
Also note that you declare j to be an int*. There is no variable named *j if that's what you believe.

[nonoob]

To answer your question, girish1026, which everyone is dancing around...

Declaring a variable int vs unsigned int makes no difference in how its values are represented internally. For your compiler/machine it may be a 2 byte integer for example. Therefore formatting for printf is not affected by the storage type either. It's because you said "%u" that the value is interpreted to be unsigned. Or you could specify "%d" if you want to see it as a signed number. "%p" is available for pointers which are usually interpreted as unsigned.

The only difference signed vs unsigned makes is during calculations, conversions during assignment, etc. where overflow conditions may be handled differently for either type.

[laserlight]

To answer your question, girish1026, which everyone is dancing around...

Declaring a variable int vs unsigned int makes no difference in how its values are represented internally. For your compiler/machine it may be a 2 byte integer for example. Therefore formatting for printf is not affected by the storage type either. It's because you said "%u" that the value is interpreted to be unsigned. Or you could specify "%d" or "%p" if you want to see it as a signed number.

That does not actually answer the question correctly. The problem, as noobiept, Elysia and I have stated is that this:

printf("Address of i is %u", j);

Prints the pointer j, whose value is the address of i. Therefore, that i is of type int or unsigned int does not matter. What matters is that j is of a pointer type, and it turns out that printing the value of this pointer prints 65524. However, girish1026 expected that the type of j must have the same range as the type of i (i.e., *j), but this is not the case.

[Elysia]

Do not print a non-pointer type with %p. You will easily invoke undefined behavior.