Re: udigits_to_I

[Bruno H. <br...@cl...>]

Sam wrote:
> digits points to a sequence of bytes, so it should probably be uint8*.

Yes. uint8* or uintB*, it's the same. (I use uint8 only when the
number of bits matters, whereas uintB is more the elementary unit of
addressable memory.)

> >   is len > 0 or can it be 0 ?
>
> dunno.  it depends on the behavior of UDS_to_I()

UDS_to_I() is specified to take an UDS as input; this term is defined
in arilev1.d.

> BTW, what does "there must be room for 1 digit below of MSDptr" mean?

It means that MSDptr[-1] may be written to. That's also the reason why
the argument is not a 'const' pointer.

> > Is the input in big-endian or little-endian order?)
>
> I don't know what "big-endian" and "little-endian" are.
> In my kindergarten they taught only one way to record numbers with
> digits: 123 = 1*100 + 2*10 + 3*1
> So, byte sequence "123" represents the integer 1*256^2 + 2*256 + 3*1.

This is big-endian. Little-endian would be the opposite: when the
byte sequence 1 2 3, ordered from low to high addresses, represents the
integer 1*1 + 2*256 + 3*256^2.

> > If not, then why is the 'digits' pointer not 'const'?
>
> because I don't know what "const" means

If a function takes a    'const uintB *' argument, it promises to not
write to memory through that pointer. So that means a :IN parameter,
whereas 'uintB *' means a :IN-OUT or :OUT parameter (in our FFI's notation).

> Maybe you could fix the comments of UDS_to_I() first?

I'm updating the definitions in arilev1.d.

For the implementation of udigits_to_I:
Note that you cannot always interpret an array of type uint8[] as an
array of type uintD[]: The number of elements may not be a multiple
of sizeof(uintD)/sizeof(uint8) (what do you want to do with the last few
bytes in that case?), and the alignment of the memory may not be a multiple
of alignof(uintD).

Bruno