[Axiom-developer] RE: [Aldor-l] exports and constants

[Bill Page]

On July 21, 2006 2:38 AM Christian Aistleitner wrote:
> ...
> Ralf wrote:
> >
> > X: CatX == if true then (A add) else (B add); -- (*)
> > ...
> > That is clear, but why does the compiler reject the program 
> > without the "add" in line (*)?
> 
> I'd consider that a bug in comparison of exports. Replacing 
> your (*) line by
> 
>    X: CatX == if true then (address@hidden) else (address@hidden);
> 
> gives a working program with output
> 
>    X has CatA: T
>    X has CatB: F
>    X has CatX: T
> 

Wow! Nice. And

   X: CatX == if false then (address@hidden) else (address@hidden);
 
gives

   X has CatA: F
   X has CatB: T
   X has CatX: T

That's excellent.

> So the problem (wild guess) is that the compiler Has problems
> with seeing that the if statement gives CatX in both branches
> of the if statement. Mainly because the types of A and B aro
> not equal. However, you can hint the compiler. My code is telling
> him "The if part gives CatX and the else part gives CatX". Then
> the compiler can infer, that the whole  "if" statement gives
> CatX. And it is at least the type of X (which is CatX). So it
> matches.
>

I accept your argument and withdraw (most) of my comments about
static types in my previous message.
 
> ...
> > But what about...
> >
> > ---BEGIN aaa3.as
> > #include "aldor"
> > import from String, TextWriter, Character, MachineInteger;
> >
> > define CatA: Category == with;
> > define CatB: Category == with;
> > define CatX: Category == with;
> >
> > A: Join(CatX, CatA) == add;
> > B: Join(CatX, CatB) == add;
> >
> > MyPkg(X: CatX): with {isA?: () -> Boolean} == add {
> >     isA?(): Boolean == X has CatA;
> > }
> > main(): () == {
> >     import from CommandLine, Array String;
> >     X: CatX == if zero? #arguments then (A add) else (B add);
> >     stdout << isA?()$MyPkg(X) << zero? #arguments << newline;
> > }
> > main();
> > ---END aaa3.as
> >
> >  >aldor -fx -laldor aaa3.as
> >  >aaa3
> > FT
> >  >aaa3 1
> > FF
> >
> > It is clear with the "add" in the definition of X that we 
> > will always see a F in the first column of the output. Can
> > someone find a "nice" construction where X is defined by a
> > similar "if" construction and the first F would turn into T?
> 
> Of course. Again, be nice to the compiler. He is old. He is 
> buggy. He is a bit rusty on the edges. And he is locked in,
> in a jail of non-free source code.

Hear, hear! Please, let's find a way to free him! :-)

> But help him. Do not force him to do extra work. Tell 
> him what you know, if he needs to be told.
> 
> X: CatX == if zero? #arguments then (address@hidden) else (address@hidden);
> 
> does the trick for me:
> 
> LC_ALL=C /opt/aldor/bin/aldor -M no-abbrev -C args=-Wopts=-m32
> -Fx -lalgebra -laldor aaa3.as && ./aaa3 && ./aaa3 xx
> cc1: note: -fwritable-strings is deprecated; see 
> documentation for details
> cc1: note: -fwritable-strings is deprecated; see 
> documentation for details
> TT
> FF
>

That is quite fantastic to me. It means that the value of the
"constant" X is not determined at compile time. I did not expect
this to be possible. This variant also works is even better:

  main4(n:Integer): Boolean == {
        X: CatX == if zero? n then (address@hidden) else (address@hidden);
        isA?()$MyPkg(X);
  }

  main4(0)
  main4(1)

But then I must realize that X is local to main4, so in the
dynamic invocation context of main4 X is still "constant".

http://wiki.axiom-developer.org/SandBoxAldorSemantics

> ... 
> I hope we both deem my solution to be "nice" :)
>

Very Nice, indeed!
 
Regards,
Bill Page.