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From: "Alex Mentis" <foo@invalid.invalid>
Subject: Re: Abstract operator does not hide predefined operator
Date: Fri, 12 Nov 2010 17:39:21 +0000 (UTC)
Date: 2010-11-12T17:39:21+00:00	[thread overview]
Message-ID: <ibju48$35l$1@news.eternal-september.org> (raw)
In-Reply-To: 76885893-2ff1-4175-91c0-fb45b3dfbe15@o15g2000prh.googlegroups.com

Adam Beneschan wrote:

> On Nov 12, 7:04�am, Stefan.Lu...@uni-weimar.de wrote:
> > Hi all,
> > 
> > I've implemented a small generic package for modular arithmetic.
> > Note that Ada's "mod N" types define addition subtraction and
> > multiplication right, but use the integer division for "/" instead
> > of the proper modular division, where A/B is A * Inverse(B). I also
> > tried to get rid of the unary "not" operator, who's purpose I don't
> > understand for general modular arithmetic. (I understand that it is
> > useful if the modulus is a power of two, just like the binary xor
> > operator.)
> > 
> > This is the spec of my generic package:
> > 
> > generic
> > � �type Mod_T is mod <>;
> > package Mod_Arith is
> > � �type Modular is new Mod_T;
> > � �function "/"(Left, Right: Modular) return Modular;
> > � �function Inverse(Value: Modular) return Modular;
> > � �function "not"(Value: Modular) return Modular is abstract;
> > end Mod_Arith;
> > 
> > For testing that package, I first instantiated it:
> > 
> > � �type M3343 is mod 3343;
> > � �package M is new Mod_Arith(Mod_T => M3343);
> > � �use type M.Modular;
> > � �S,T,U: M.Modular;
> > 
> > (Sidenote: 3343 is a prime, so addition and multiplication over
> > M.Modular are field operations, mathematically.)
> > 
> > Now, my freshly defined division works as expected. The
> > multiplication (and addition and subtraction) are inherited from
> > the type M3343:
> > 
> > � �S := 3;
> > � �T := S/(2*S); -- T becomes the multiplicative Inverse of 2 mod
> > 3343; � � � � � � � � �-- the same expression's result in M3343
> > would be zero.  � �Ada.Text_IO.Put_Line(M.Modular'Image(T) &
> > � � � � � � � � � � � � � M.Modular'Image(M.Inverse(2)));
> > � �-- The output is "1672 1672", as expected.
> > 
> > But, unfortunately, the compiler (gnat) also accepts the following:
> > 
> > � �U := not S; -- this should not be possible, because the "not"
> > operator � � � � � � � �-- has been defined abstract ... but
> > instead, the � � � � � � � �-- predefined "not" from M3343 seems to
> > be used � �Ada.Text_IO.Put_Line(M.Modular'Image(U));
> > � �-- The output is "3339", which actually is not(M3343(S)) =
> > 3342-S.
> > 
> > What can I do to get rid of the predefined "not"? If I change the
> > "not" from an abstract function to a function always raising an
> > exception, the statement
> > 
> > � �U := not S;
> > 
> > raises the desired exception. But I would prefer to be told at
> > compile time that I must not use "not", rather than at run time.
> > Also, I would prefer not to define the type Mod_Arith.Modular as a
> > private type.
> > 
> > Any ideas?
> 
> This appears to be a GNAT bug; the inherited "not" subprogram of
> Modular, which is inherited from the predefined function of M3343,
> exists but should be hidden from all visibility, and therefore cannot
> be a possible meaning of "not" in the expression "not S".  So what
> you're doing should work, I think.  (P.S. I've confirmed that the
> version of GNAT I'm using also accepts the program if I write out M
> instead of making it a generic instantiation, and if I use "use"
> instead of "use type".  So neither of those appears to be an issue.)
> 
>                               -- Adam

Agree that it smells buggy.  It seems that the problem only affects
making unary operators abstract ("-" and "not"). Testing with "+", "-",
"/", "*", "and", "or", and "xor" operators as abstract refused to
compile as expected for me.



      reply	other threads:[~2010-11-12 17:39 UTC|newest]

Thread overview: 3+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2010-11-12 15:04 Abstract operator does not hide predefined operator Stefan.Lucks
2010-11-12 16:56 ` Adam Beneschan
2010-11-12 17:39   ` Alex Mentis [this message]
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