On Oct 3, 2007, at 5:17 PM, Hartmut Kaiser wrote: > Chris, > >>> I'm currently trying to remove some of the warnings spit >> out on me by >>> the >>> VC++ compiler (mostly warnings about possible loss of precision) and >>> recognized, that the bitwith (size of a type in bits) is >> represented >>> using different types all over the place: sometimes unsigned, >>> sometimes uint32_t and sometimes it's even uint64_t. I really would >>> like to change that to a more consistent scheme, but need >> some advice >>> which type to choose. >>> Since I >>> think we can't expect to get anything terribly large as the >> bitwidth >>> of a type I'ld say let's go for uint32_t. >> >> Hi Hartmut, >> >> We should go with uint64_t in general. There can be large >> arrays (for example) on 64-bit targets that need this. For >> "known little things" like integer types, uint32_t would be >> sufficient, but it is probably better to be consistent and >> use uint64_t for everything. > > IIUC, the bitwidth is most of the time the number of bits needed to > represent the size, i.e. 32 for uint32_t. In this case 64 is the > largest > number these kind of variables will ever hold, no?
Yes, except that what is the size of "uint32_t[100000000000LL]" ? -Chris _______________________________________________ cfe-dev mailing list [email protected] http://lists.cs.uiuc.edu/mailman/listinfo/cfe-dev
