I often see (in Linux kernel for example) that unsigned long is used to hold the pointers. I wonder what is the reason for this given that size of a pointer may be larger than integer type (including long).
Is it portable to keep a pointer in unsigned long rather than in uintptr_t in Linux user space applications? (Although I know that uintptr_t guarantees to convert from void * to uintptr integer and back without loss of information)
Thanks.
Is it portable to keep a pointer in
unsigned longrather than inuintptr_tin Linux user space applications? (Although I know thatuintptr_tguarantees to convert fromvoid *touintptrinteger and back without loss of information)
"Yes" in the sense that it will work on any current port of Linux and quite likely in the future. But why? There is a perfectly good typedef that specifies the intent too: uintptr_t - and it makes your code portable to Win64 too.
uintptr_t was a C99 invention and Linux predates C99 by years. Back then there was no convention for specifying an integer wide enough to hold a pointer - but then, there was no unsigned long long either, except by compiler extension, so unsigned long was all that you could reasonably expect to hold a pointer, if anything did, and so it was. By now, it is rather that for any new architecture that runs Linux need to pick type for long so that it is wide enough for pointer size_t etc.
When 64-bit Windows came, too many things were relying on some representation for unsigned long, and it remained as 32 bit instead of the type required to hold pointer. To my knowledge of all relevant platforms today only on Win64 unsigned long is not wide enough to hold a pointer.
