On 19 August 2016 at 07:53, Jeffrey Walton wrote: >>> Forgive my ignorance... 'unsigned long' is 64 bits, which is what that >>> interface needs (if I am reading the docs correctly). >> >> The functions takes a pointer to an unsigned 64-bit type, not >> necessarily unsigned long. > > Thanks again Jonathon. I guess this is what I don't quite > understand... From the preprocessor, GCC tells me the UINT64 type is > the following on x86_64: > > #define __UINT64_TYPE__ long unsigned int > > If 'long unsigned int' is the 64-bit integer and it meets alignment It's not *the* 64-bit integer, it's *a* 64-bit integer. On LP64 platforms > and size requirements (as a 'minimum' of sorts), then why can't it be > used where the 64-bit type is required (i..e, the pointer in the call > to _rdrand64_step). Because that's how the C++ type system works. __UINT64_TYPE__ is the underlying type of uint64_t. The signature for the function you're calling is: ../gcc/config/i386/immintrin.h:_rdrand64_step (unsigned long long *__P) That's not uint64_t, so don't try to use uint64_t. Use the type the function is declared to take. The rules of C++ don't say you can freely mix pointers to distinct types as long as they are a bit samey. You can't mix int* and long* when sizeof(long) == sizeof(int) and you can't mix unsigned long long* and unsigned long* when sizeof(unsigned long long) == sizeof(unsigned long). You've basically gone chasing after some unrelated definition (that of uint64_t) which has nothing to do with _rdrand64(), looking at unrelated implementation details of that different type, and the simple solution is to just use the type in the function declaration.
