How to determine if a variable is a pointer?

Question

I was reading the Wolfenstein 3D code, and I encountered ISPOINTER macro:

#define ISPOINTER(x) ((((uintptr_t)(x)) & ~0xffff) != 0)

I know we have std::is_pointer, but how does this macro work? I tried and failed with strange behavior which I couldn't explained why it's happend:

#define ISPOINTER(x) ((((uintptr_t)(x)) & ~0xffff) != 0)
int main()
{
        int* ptr;
        int val;

        if (ISPOINTER(ptr)) {
                std::cout << "`ptr`: Is Pointer" << std::endl;
        }
        if (ISPOINTER(val)) {
                std::cout << "`val`: Is Pointer" << std::endl;
        }
}

I don't have any output, but if I add another pointer:

#define ISPOINTER(x) ((((uintptr_t)(x)) & ~0xffff) != 0)
int main()
{
        int* ptr;
        int val;
        int* ptr2;

        if (ISPOINTER(ptr)) {
                std::cout << "`ptr`: Is Pointer" << std::endl;
        }
        if (ISPOINTER(val)) {
                std::cout << "`val`: Is Pointer" << std::endl;
        }
        if (ISPOINTER(ptr2)) {
                std::cout << "`ptr2`: Is Pointer" << std::endl;
        }
}

The output will be:

`ptr`: Is Pointer

What does ISPOINTER doing? It's undefined behavior?

FWIW, std::is_pointer exits and will do the job, plus it's standard approved ;) It's also a compile time thing, so no run time overhead. — NathanOliver
– NathanOliver, Commented Aug 26, 2021 at 13:35
@NathanOliver I just want to know what is that pointer trick in the game code? — Ghasem Ramezani
– Ghasem Ramezani, Commented Aug 26, 2021 at 13:36
@GhasemRamezani if this came from game code then what's likely happening is the game is making (unsafe) assumptions about its runtime situation and using bits of a pointer for other information. Then differentiating the two by seeing if those bits are/are not there. — Mgetz
– Mgetz, Commented Aug 26, 2021 at 13:40
"Safe" is a matter of perspective. If you know your target machine and compiler's behavior, you can operate under added assumptions. Granted, some code-bases don't always do that and are simply written naively, but still. Seeing as this comes from code written in the early 90's (and by id software no less, who are known for such things), this isn't a surprising macro. — StoryTeller - Unslander Monica
– StoryTeller - Unslander Monica, Commented Aug 26, 2021 at 13:43
@GhasemRamezani so let's take x86-64 for a second. The implementation requires (currently) that all pointers have 48bits of valid info, everything else is garbage. If you know the format required by the platform you can use the unused bits for other information insofar as you revert those changes prior to actually using the pointer as a pointer. — Mgetz
– Mgetz, Commented Aug 26, 2021 at 13:44

perivesta · Accepted Answer · 2021-08-26 13:49:05Z

Let's do this in steps:

((uintptr_t)(x)) is simply a cast from whatever x is into a uintptr_t (an unsigned integer type capable of storing pointer values)

~0xffff is a bit-wise complement of 0xffff (which is 16 bits of all 1s). The result of that is a number that is all 1s except the last 16 bits.

((uintptr_t)(x)) & ~0xffff is a bit-wise AND of the pointer value with the above number. This will effectively zero-out the 16 lowest bits of whatever the pointer value is.

The full expression now just checks if the result is zero or not. So the whole expression basically checks if any bits except the least-significant 16 are set and if so it considers it a pointer.

Since this came from Wolfenstein 3D, they probably made the assumption that all dynamically allocated memory lives in high memory addresses (higher than 2^16). So this is NOT a check if a type is a pointer or not using the type system like std::is_pointer does. This is an assumption based on the target architecture Wolfenstein 3D will likely run on.

Keep in mind that this is not a safe assumption, since "normal" values above 2^16 would also be considered pointers and the memory layout of your process can be very different depending on a lof of factors (e.g. ASLR)

JDługosz · Accepted Answer · 2021-08-26 14:06:30Z

2

It might be similar to certain parameters in Windows, which can be an ordinal value or a pointer. An ordinal value will be <64K. On that operating system, any legal pointer will be >64K. (On older 16-bit versions of Windows, only 4K was reserved.)

This code may be doing the same thing. A "resource" may be a built-in or registered value referred to by a small number, or a pointer to an ad-hoc object. This macro is used to decide whether to use it as-is or look it up in the table instead.

edited Aug 26, 2021 at 14:06

answered Aug 26, 2021 at 14:00

JDługosz

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