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From: =?utf-8?Q?Yannick_Duch=C3=AAne_=28Hibou57?=
 =?utf-8?Q?=29?=
 <yannick_duchene@yahoo.fr>
Newsgroups: comp.lang.ada
Subject: Re: Noob question: universal_integer type
Date: Sat, 22 May 2010 01:33:02 +0200
Organization: Ada At Home
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Date: 2010-05-22T01:33:02+02:00
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Le Sat, 22 May 2010 01:15:19 +0200, Duke Normandin <dukeofperl@ml1.net> =
a  =

=C3=A9crit:

> Coronado's Ada tutorial - Chapt3
>
> [quote]
> The constant 17 is of a very special type defined by Ada as type
> "universal_integer" which can be combined with any of the integer type=
s
> without specific conversion.
> [/quote]
>
> Let me get this right... if I use an undeclared integer in an expressi=
on,
> Ada will "deem it" to be a "universal_integer" and not choke at
> compile-time?
On this subject, you may like to read a previous thread named =E2=80=9CI=
nteger  =

questia=E2=80=9D.
Here is a Google Group link for conveniance:
http://groups.google.com/group/comp.lang.ada/browse_thread/thread/d9d2bc=
cce5d4fc93#

The key, is that universal integer, is a type like other types you may  =

defined. Every type integer-like type can be converted to and from  =

universal integer.

So, yes, the compiler will not complain, because if you have

    My_Constant : constant :=3D 1; -- Named number, that is here, univer=
sal  =

integer.
    type My_Integer_Type is range 1 .. 9;
    My_Entity : My_Integer_Type :=3D My_Constant;

the compiler will not see a contradiction, it will not see the declarati=
on  =

of a given type which is initialized with a value of a different type, i=
t  =

will see a literal, which it will automatically convert to the target ty=
pe.

-- =

There is even better than a pragma Assert: a SPARK --# check.