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In the code below
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?Code:int main()
{
char *p;
p = 12;
*p = 20;
}
If not please give the justifiction in detail....
Seems right, I think..
Yep. Addresses are typically given in hex, so the location is actually 0x0000000C.
Quote:
Originally Posted by karthik537
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.
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Mats
can u please explainQuote:
Addresses are typically given in hex
It's my computer, I can do what I want with it!!Quote:
Originally Posted by matsp
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.Quote:
can u please explain
If you want to see the address assigned by the compiler (or I guess actually the linker?):Code:char *ptr, *p2;
p2=ptr; //okay
ptr=12; //not okay
Notice, as matsp sayeth, this is the hexadecimal address of a location in the physical memory of the computer.Code:printf("%p",ptr);
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].Quote:
Originally Posted by MK27
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Mats
Strictly speaking, the %p format specifier is only used to print a pointer to void, so it should be:
Also, it is not guaranteed that the representation of the address will be in hexadecimal, though it is typically the case.Code:printf("%p", (void*)ptr);