STATIC types in ADA?

STATIC types in ADA?

[jhuber]

Do STATIC data types (like C++ has) exist in ADA?

By this I mean, is there a way I can define a variable so that
when I call a procedure and set a value (say 5.0) and then
exit the procedure and come back into it at a later time, the
variable would still have the same value (i.e., 5.0)?

I think private is the closest thing I've found that will do it.

Re: STATIC types in ADA?

[Jim Rogers]

In Ada you simply declare the variable in the package body,
but outside any subprogram.

Jim Rogers

jhuber@nlxcorp.ellsworth.com wrote:

> Do STATIC data types (like C++ has) exist in ADA?
> 
> By this I mean, is there a way I can define a variable so that
> when I call a procedure and set a value (say 5.0) and then
> exit the procedure and come back into it at a later time, the
> variable would still have the same value (i.e., 5.0)?
> 
> I think private is the closest thing I've found that will do it.
> 

Re: STATIC types in ADA?

[Matthew Heaney]

<jhuber@nlxcorp.ellsworth.com> wrote in message
news:3c7f8fd6$0$7055$43695faf@reader...
> Do STATIC data types (like C++ has) exist in ADA?
>
> By this I mean, is there a way I can define a variable so that
> when I call a procedure and set a value (say 5.0) and then
> exit the procedure and come back into it at a later time, the
> variable would still have the same value (i.e., 5.0)?

Just declare the object in a static scope (a package):

package P is
   F : Float := 0.0
end;

with P;
procedure Op is
begin
   P.F := 5.0;
end;

Realize that a variable declared in a package is equivalent to an object
declared in a C++ namespace, or as a static member of a class:

//n.hpp
namespace N
{
   extern float f;
}

//n.cpp
float N::f;

or

//c.hpp
class C
{
public:
   static float f;
//...
};

//c.cpp
float C::f;


Often variables are hidden in the body of the package, so you could write it
as:

package P is
   procedure Op;
end;

package body P is
   F : Float := 0.0;

   procedure Op is
   begin
      F := 5.0;
   end;
end P;

That would translate as:

//n.hpp
namespace N
{
   void op();
}

//n.cpp
namespace
{
   float f;
}

void N::op()
{
   f = 5;
}


or

//c.hpp
class C
{
public:
   static void op();
//...
private:
   static float f;
//...
};

//c.cpp
float C::f;

void C::op()
{
   f = 5;
}

Actually, you could declare f in an anonymous namespace in the
implementation file c.cpp, instead of as a static variable of the class.
This lets you hide it completely.  One of my complaints about C++ is that
you often have to put things in the private part of the class declaration,
up in the header file, instead of in the implementation file.  Thus in order
to get visibility, I pay with a compilation dependency.  I haven't found a
satisfactory way around this.

Re: STATIC types in ADA?

[Jeffrey Carter]

jhuber@nlxcorp.ellsworth.com wrote:
> 
> Do STATIC data types (like C++ has) exist in ADA?
> 
> By this I mean, is there a way I can define a variable so that
> when I call a procedure and set a value (say 5.0) and then
> exit the procedure and come back into it at a later time, the
> variable would still have the same value (i.e., 5.0)?

In Ada such a variable (not a type) is considered to contain state
information. State information is typically saved in variables in the
same package body as the subprograms that manipulate the state. In
multitasking situations (which don't arise in C++), state information is
typically saved in a protected object and manipulated by the protected
object's operations.

-- 
Jeffrey Carter

Re: STATIC types in ADA?

[Gautier Write-only-address]

>Do STATIC data types (like C++ has) exist in ADA?
...
>I think private is the closest thing I've found that will do it.

It's much simpler. Ada is a block-structured language.
A variable declared outside of procedure or package P is
"static" relatively to P. Inside of P, it is a local variable
just to be used by P. Where a variable is visible, it is alife
and keep its value.

Ex:

package body P is
  i: Integer;

  procedure Q is
    j: integer;
    function R(...) return ... is
      k: integer;
    begin
      -- here i,j,k are visible
      ...
    end R;
  begin
    j:= ...
    i:= i+j;
  end Q;

begin
  i:= 0;
  Q;
  -- i has changed
  Q;
  -- again...
end P;

________________________________________________________
Gautier  --  http://www.mysunrise.ch/users/gdm/gsoft.htm

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