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From: "Dmitry A. Kazakov" <mailbox@dmitry-kazakov.de>
Newsgroups: comp.lang.ada
Subject: Re: Ada needs some modernization
Date: Tue, 31 May 2022 21:55:05 +0200
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On 2022-05-31 19:54, Matt Borchers wrote:
> Throughout my career, I often find myself writing code similar to:
> 
> if (A and B) or else (not A and C) then...
> 
> and I always wished there was a better and clearer way to write this in Ada.  Then along came if expressions.  But, if expressions don't help that much with readablity although it is arguably simpler:
> 
> if (if A then B else C) then...

Not same. In the former A may be computed twice.

> What amendment can we suggest to the Ada syntax so the if expression be better written when used in an if statement?

I newer felt it necessary. To me much more aggravating is code that 
combines test/allocator with renaming, i.e.

    if P /= null then
       declare
          X : T renames P.all;
       begin
          ...
       end;
    end if;
------------
    if X in T'Class then
       declare
          XX : T'Class renames T'Class (X);
       begin
          ...
       end;
    end if;
-----------
    P : access T'Class := new S;
    X : S renames S (P.all);

If one could come up with some syntax for if-then-declare and 
new-then-declare that would cover a lot of cases.

> I know other languages support this and it often looks like  A ? B : C  or something similar.  That's certainly not Ada-like IMO, but I can't think of something better. These same languages often also have a null check operator  A ?? B  (where A and B are access types of the the same Type) such that if A is not null then A is returned otherwise B is returned.  So useful and helpful!

Not in a strongly typed language IMO.

[...]

> Maybe something like:
> 
> for i in arr'Range loop
>      if arr(i) = match then
>          --do something A
>          exit;
>      end if;
> then
>      --do something else B
> end loop;

I usually use a nested function, e.g. search with a fallback:

    function Get_Me_Something return Element is
    begin
       for I in arr'Range loop
          if Arr (I) = match then
             return Arr (I);
          end if;
       end loop;
       return Defaul;
    end Get_Me_Something

-- 
Regards,
Dmitry A. Kazakov
http://www.dmitry-kazakov.de