I don't know of a portable representation of a byte other than a char. And while no one promises that a char is really a byte, I expect that too much code would break and so it would be many many years before this is changed in practice.
Likewise, no one promises that sizeof(int) == 4, but vendors haven't changed this to 8 on 64 bit platforms either, opting instead to use "long long" or "long int". AFAIK there's nothing wrong with using void *, and most of the C runtime functions like malloc use this. C++, however, explicitly disallows pointer arithmetic on a void *, so if you're programming in C++, you may end up casting to a char * anyway. On GNU/Linux I might include /usr/include/stdint.h, which typedefs char to "int8_t" but, once again, AFAIK, there's no guarantee that stdint.h is going to exist on every platform. Wilson On 12/8/05, Allan Wind <[EMAIL PROTECTED]> wrote: > On 2005-12-08T10:37:36-0800, Wilson Yeung wrote: > > You're expected to cast your structure into an unsigned char *, > > because in C/C++, the only portable way to represent a byte is with a > > char/unsigned char. > > Off-topic, I suppose, but what is a portable representation of a byte? > What does unsigned char pointer give you that a void pointer does not > (other than support for legacy compilers)? > > > /Allan >
