>> > >> > 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. regards, -- Cédric _______________________________________________ cfe-dev mailing list [email protected] http://lists.cs.uiuc.edu/mailman/listinfo/cfe-dev
