Re: "Must instantiate controlled types at library level." Why?

[Martin Krischik <krischik@users.sourceforge.net>]

Peter C. Chapin wrote:

> Jeffrey Carter <spam@spam.com> wrote in news:NJeoc.16758$V97.5496
> @newsread1.news.pas.earthlink.net:
> 
>> The call to Library_Level.Op would attempt to dispatch to
>> Outer.Inner.Op, which is out of scope. The references to I in
>> Outer.Inner.Op refer to a variable that no longer exists.
> 
> Thank you for your detailed post. It was very helpful. I haven't yet
> gotten around to studying Ada's object oriented features in detail but I
> can see that the issue is related to the fact that Ada allows procedures
> to be defined inside other procedures.
> 
> Just for fun I attached an equivalent C++ program below. Interestingly, it
> seems to compile and work using three different C++ compilers. I wonder
> its behavior is technically "undefined". It seems like it probably should
> be. Perhaps I'll ask on the C++ group.

It is not quite complete:

> -----> cut <-----
> #include <iostream>
> 
> class X {
> public:
>   virtual void f();
> };
> 
> void X::f()
> {
>   std::cout << "I'm in X::f\n";
> }
> 
> X *ptr;
> 
> void helper()
> {

auto int Z := 5;

>   class Y : public X {
>   public:
>     virtual void f() { 
>     std::cout << "I'm in Y::f\n";

std::cout << "and Z is " << Z << std:endl;

>     }  
>   };
> 
>   ptr = new Y;
> }
> 
> int main()
> {
>   helper();
>   ptr->f();
>   return 0;
> }

But you can reduce the probem to a 5 liner:

int&
Very_Silly ()
  {
  auto int Short_Lived = 5;
  return Short_Lived;
  }

A good compiler will warn you. But it does compile.

With Regards

Martin

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