Re: EXTERNAL: Re: [vhdl-200x] Modular types

> On 08/07/2014 09:24, tgingold@free.fr wrote:
> >> I agree that the resolution proposal inherently feels wrong to use
> >> as
> >> a solution.
> >>
> >> I'm not sure I understand what the problems  with (1) are (apart
> >> from
> >> overflowing) in performing all the arithmetic in normal integer
> >> mode
> >> and implicitly converting to the modular type on assignment.
> > Well, the arithmetic is not the same.  Eg: (3 + 1) mod 3
> >
> > For a signed type, the result is 1.
> > For a modular type with a modulus of 4, the result is 0.
> >
> > Regards,
> > Tristan.
> >
> That's why the proposal is carry out the 'mod' function implicitly on
> the final assignment to the actual.

Let's me clarify:

  type T is mod 4;
  variable A : T;

  A := 3; -- 3
  A := A + 1; -- 0
  A := A mod 3; -- 0

  So using modular arithmetic for each operator, you get for type T: (3 + 1) mod 3 = 0 

  No, using only 'mod' on the final assignment:
    A := (3 + 1) mod 3;
<=> A := 4 mod 3;
<=> A := 1;

  Conclusion: doing reduction on the final value is not equivalent to modular arithmetic.

Regards,
Tristan.

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