Re: if X in 1..35000 versus Boolean

[Keith Thompson <kst@cts.com>]

Jan Prazak <janp9@gmx.net> writes:
> On Sat, 13 Jul 2002 15:27:03 -0100, Frank J. Lhota wrote:
> > Also, 1 .. 35000 is an integer range, not an enumeration type.
> 
> But the word "in" tests, if given object (variable...) is
> in given enumeration type.
> 
> type num : (1, 2, 3, 4);
> 
> if 3 in num then ...
> 
> -- this should be the same as: if 3 in (1,2,3,4);
> -- and also the same as: if 3 in 1..4;
> 
> So I assume that "if X in 1..35000" has the same rules.

I think you're thinking of Pascal sets, not enumeration types.  Ada
doesn't have built-in set types, though they're easy enough to
implement.

If I recall correctly (it's been a while), in Pascal you can write
something like:

    if X in [1 .. 4] then (* This is Pascal, not Ada *)
    begin
        ...;
    end;

where [1 .. 4] is a set value constructor.  A naive Pascal compiler
would build a set value with bits 1, 2, 3, and 4 set, and all other
bits cleared, and test whether bit X is set.  A more clever Pascal
compiler might generate code equivalent to X >= 1 and X <= 4.

You can also do more complex things like

    if X in [1 .. 4, 7 .. 9, 12] then (* This is Pascal, not Ada *)
    begin
        ...;
    end;

The Ada construct 1 .. 4 is *not* a set value constructor, it's a
range.  Given the condition "X in 1 .. 4", there's no reason for the
compiler to generate anything more complicated than "X >= 1 and X <= 4".
There is no direct Ada equivalent to "X in [1..4, 7..9, 12]".

An enumeration type is something else; for example:
    type Color is (Red, Blue, Green);
Note that the values are identifiers, not numeric literals.

I suggest you find a good Ada textbook or online tutorial.

-- 
Keith Thompson (The_Other_Keith) kst@cts.com  <http://www.ghoti.net/~kst>
San Diego Supercomputer Center           <*>  <http://www.sdsc.edu/~kst>
Schroedinger does Shakespeare: "To be *and* not to be"