On Oct 4, 2007, at 7:44 AM, Cédric Venet wrote: >>>> 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]" ? >> >> sizeof(uint32_t[100000000000LL]) < 2^64, >> >> so the size is representable with 64 bits, which means the >> corresponding >> bitwidth is not even '64', which conveniently fits into a int8_t. >> >> But probably I'm getting something really wrong here. >> Regards Hartmut > > I think we lost the correct definition of bitwith along the way. > quoting earlier post from this thread: > >>>>> recognized, that the bitwith (size of a type in bits) is > > for me, bitwith( X ) == 8*sizeof( X ) > > If we speak of the same things, this was done in order to thread the > size of bitfield member in an uniform manner with other 'normal' > variable.
Yep, exactly! -Chris _______________________________________________ cfe-dev mailing list [email protected] http://lists.cs.uiuc.edu/mailman/listinfo/cfe-dev
