Thread: Doubt regarding pointer

Yep. Addresses are typically given in hex, so the location is actually 0x0000000C.

Originally Posted by karthik537

In the code below

Code:

int main()
{
     char *p;
     p = 12;
     *p = 20;
}

I want a clarification regarding pointer p that in the second stmt, p is getting an address 12 and in 3rd stmt we are assigning the value 20 at the address 12...right?

If not please give the justifiction in detail....

Correct.

The posted code, however, will give warnings in most compilers, as the compiler thinks you are doing something "strange" if you assign an integer constant to a pointer - a cast will remove the warning.

It will also, almost certainly, cause something to go wrong in your system - either it will change memory that you shouldn't change, or it will crash, because address 12 is not at all available to you.

--
Mats

Originally Posted by matsp

It will also, almost certainly, cause something to go wrong in your system - either it will change memory that you shouldn't change, or it will crash, because address 12 is not at all available to you.

It's my computer, I can do what I want with it!!

can u please explain

BEN10: you do not have to (and usually should never try to) assign pointer addresses unless it's an address that you know has already been assigned to a pointer.

Code:

char *ptr, *p2;
p2=ptr;  //okay
ptr=12; //not okay

If you want to see the address assigned by the compiler (or I guess actually the linker?):

Code:

printf("%p",ptr);

Notice, as matsp sayeth, this is the hexadecimal address of a location in the physical memory of the computer.

Originally Posted by MK27

Notice, as matsp sayeth, this is the hexadecimal address of a location in the physical memory of the computer.

Pedantically, it's the virtual address - if you do not deal with kernel code, you never see the actual physical location for anything [assuming we're talking 32-bit OS's on x86, that is].

--
Mats

Strictly speaking, the %p format specifier is only used to print a pointer to void, so it should be:

Code:

printf("%p", (void*)ptr);

Also, it is not guaranteed that the representation of the address will be in hexadecimal, though it is typically the case.