Intermediate expressions and modular types

Intermediate expressions and modular types

[Martin Kristensson]

-- Ada95
type Modular is mod 2**32;
procedure Test(ModA, ModB : in Modular) is
   X : Integer;
begin
   X := Integer(ModA - ModB);
end Test;

The intermediate expression (ModA - ModB) is calculated modulo 2**32 by
compiler A, but not by compiler B.
Which is right?

Surely the correct behavior must be stated in the RM, but where?

Thanks for your help,
/Martin.

Re: Intermediate expressions and modular types

[Robert A Duff]

In article <344898BA.7361@dtek.chalmers.se>,
Martin Kristensson  <d95krima@dtek.chalmers.se> wrote:
>-- Ada95
>type Modular is mod 2**32;
>procedure Test(ModA, ModB : in Modular) is
>   X : Integer;
>begin
>   X := Integer(ModA - ModB);
>end Test;
>
>The intermediate expression (ModA - ModB) is calculated modulo 2**32 by
>compiler A, but not by compiler B.
>Which is right?

Compiler A.

>Surely the correct behavior must be stated in the RM, but where?

Not sure what to quote here.

3.5.4(19) defines the normal "wrap-around" semantics of the "-" operator
on modular types -- in fact it talks about all operators.

So perhaps the issue is that Compiler A thinks it's a Modular "-",
whereas Compiler B thinks it's an Integer "-".  The former is correct,
which follows from all the visibility and overloading rules and so
forth, but 4.6(7) says that the "-" chosen comes from the types of the
operands.

Most likely, it's just a plain old code generation bug, so there's no RM
exegesis necessary to convince the maintainers of Compiler B that they
have a bug.

Why "Compiler A" and "Compiler B"?  Why can't names be named here?  Is
someone afraid of being sued for saying so-and-so compiler has a bug?!
Of course, to indicate a compiler bug, one has to produce a complete
program, which prints out a wrong answer.

- Bob