Setting the high bit of an integer type of unknown length

▲
1
▼

I'm looking for a portable and elegant way to set the high bit of a pointer type in a situation where we don't know if the pointer will have 32, or 64 bits.

I'm working on some legacy code that we inherited from a defunct company. It used to work in a 32 bit environment and now we need a working 64 bit version too. That means much larger pointers than before, and larger magic numbers to represent invalid values. Currently the best proposal will look broadly like this:

#if ENVIRONMENT_IS_32_BIT
#define INVALID_PTR_CODE 0x80000000
#else
#define INVALID_PTR_CODE 0x8000000000000000
#endif

As I say, I am hoping for something more elegant, and something that will work across different compilers (MSVC, gcc, clang at least). A perfect solution would work for any environment / size of pointer.

c++
asked on Stack Overflow Oct 19, 2018 by Tim Randall

3 Answers

▲
4
▼
static_cast<uintptr_t>(1) << (CHAR_BIT * sizeof(void*) - 1)

would do it.

(Note that the current crop of Intel chips has 48 bit pointers under the hood.)

answered on Stack Overflow Oct 19, 2018 by Bathsheba • edited Oct 19, 2018 by Bathsheba
▲
3
▼

To get the size of a pointer in bytes you can use sizeof(void *), in bits that's sizeof(void *) * CHAR_BIT. Now you need to get a 1 in that position (keep in mind it's 0 based) so you would do 1ULL << (sizeof(void *) * CHAR_BIT - 1). Then just OR that with your number.

answered on Stack Overflow Oct 19, 2018 by ODYN-Kon • edited Oct 19, 2018 by ODYN-Kon
▲
0
▼

Consider the following code:

int *x = (int*)-1;                           //You set all the bits of `x`
int *y = (int*)((unsigned long long)x >> 1); //y has all bits set besides the msb
x = (int*)(x - y);                           //subtract y from x to have only th msb set
answered on Stack Overflow Oct 19, 2018 by Lorand • edited Oct 20, 2018 by Lorand

User contributions licensed under CC BY-SA 3.0