questions from a newbie

questions from a newbie

[zork]

Hi, I just started a course in ada. I just have 2 questions at present.

-------------
q1) can:

c : character;

if c in 'A'..'Z' or c in 'a'..'z' or c in '0'..'9' then
   ....
end if;

be written as something like:

c : character;

if c in ('A'..'Z', 'a'..'z', '0'..'9') then
   ....
end if;
-------------

Also, I know you can do the following:

type new_type is array(1..20) of string(1..50);
words : new_type;
index : integer := 20;
words (15) (index ..index) := "K";

however I find that I cannot instead say:

words(15)(index):="K";

why is this so? I get a "Type mismatch in assignment statement, continuing"
error. It does however work when I use words(15)(index):='K'. The rational
behind this is that (index..index) represents a range - hence a string -
whereas (index) represents a single character?

Any insight most helpful.

Cheers,
zork

Re: questions from a newbie

[Marius Amado Alves]

> words(15)(index):="K";
> 
> why is this so? I get a "Type mismatch in assignment statement, continuing"
> error. It does however work when I use words(15)(index):='K'. The rational
> behind this is that (index..index) represents a range - hence a string -
> whereas (index) represents a single character?

Yes. 'K'. Single quotes.

Re: questions from a newbie

[Steve]

"zork" <zork@nospam.com> wrote in message news:40f684a8@dnews.tpgi.com.au...
> Hi, I just started a course in ada. I just have 2 questions at present.
>
> -------------
> q1) can:
>
> c : character;
>
> if c in 'A'..'Z' or c in 'a'..'z' or c in '0'..'9' then
>    ....
> end if;
>
> be written as something like:
>
> c : character;
>
> if c in ('A'..'Z', 'a'..'z', '0'..'9') then
>    ....
> end if;
> -------------

Not directly, but you can make use the of standard Ada libraries:
  Ada.Strings.Maps
  Ada.Strings.Maps.Constants
and do something like:

  if Is_In( c, Alphanumeric_Set ) then
    ...
  end if;

Since you're a newbie, I recommend you peruse annex A of the Ada 95
reference manual; it describes the predefined language enviroment, which is
basically a list of all of the standard libraries.

You can download a copy from:
    http://www.adaic.org/standards/ada95.html

>
> Also, I know you can do the following:
>
> type new_type is array(1..20) of string(1..50);
> words : new_type;
> index : integer := 20;
> words (15) (index ..index) := "K";
>
> however I find that I cannot instead say:
>
> words(15)(index):="K";
>
> why is this so? I get a "Type mismatch in assignment statement,
continuing"
> error. It does however work when I use words(15)(index):='K'. The rational
> behind this is that (index..index) represents a range - hence a string -
> whereas (index) represents a single character?
>

You answered your own question here.

If you're coming from a different programming language, you may find that
Ada is kind of a "hard ass" about the syntax it will accept.  When you get
over the frustration and find that more programs work correctly after you
get past the compiler, you may never want to go back.

Steve
(The Duck)

> Any insight most helpful.
>
> Cheers,
> zork
>
>

Re: questions from a newbie

[Georg Bauhaus]

zork <zork@nospam.com> wrote:
: 
: if c in 'A'..'Z' or c in 'a'..'z' or c in '0'..'9' then
: 
: 
: if c in ('A'..'Z', 'a'..'z', '0'..'9') then

This is the purpose of Ada.Strings.Maps etc.
In Ada.Strings.Maps.Constants you will find predefined
Character_Set constant that match yours.

: Also, I know you can do the following:
: 
: type new_type is array(1..20) of string(1..50);
: words : new_type;
: index : integer := 20;
: words (15) (index ..index) := "K";
: 
: however I find that I cannot instead say:
: 
: words(15)(index):="K";
: 
: why is this so?

...(index) denotes a Character, one component of an array,
...(index .. index) denotes an array slice.
The index values can in general be results of computations at
runtime, i.e., ...(n .. m).  What kind of thing other than
an array slice should (n .. m) denote? OTOH, ...(n) where n
is an index value cannot but denote one array component.

: I get a "Type mismatch in assignment statement, continuing"
: error. It does however work when I use words(15)(index):='K'. The rational
: behind this is that (index..index) represents a range - hence a string -
: whereas (index) represents a single character?

Re: questions from a newbie

[Jacob Sparre Andersen]

Zork wrote:

> q1) can:
> 
> c : character;
> 
> if c in 'A'..'Z' or c in 'a'..'z' or c in '0'..'9' then
>    ....
> end if;
> 
> be written as something like:
> 
> c : character;
> 
> if c in ('A'..'Z', 'a'..'z', '0'..'9') then
>    ....
> end if;

Sort of:

   case C in
      when 'A' .. 'Z' | 'a' .. 'z' | '0' .. '9' =>
         ....
      when others =>
         null;
   end case;

Greetings,

Jacob
-- 
"Banning open source would have immediate, broad, and
 strongly negative impacts on the ability of many sensitive
 and security-focused DOD groups to protect themselves
 against cyberattacks"                        -- Mitre Corp.

Re: questions from a newbie

[zork]

Thanks everyone for the insight. Much appreciated. Im liking ada. Previously
programmed in c++ and java.

cheers,
zork


"Jacob Sparre Andersen" <sparre@nbi.dk> wrote in message
news:pl3c3tcnny.fsf@sparre.crs4.it...
> Zork wrote:
>
> > q1) can:
> >
> > c : character;
> >
> > if c in 'A'..'Z' or c in 'a'..'z' or c in '0'..'9' then
> >    ....
> > end if;
> >
> > be written as something like:
> >
> > c : character;
> >
> > if c in ('A'..'Z', 'a'..'z', '0'..'9') then
> >    ....
> > end if;
>
> Sort of:
>
>    case C in
>       when 'A' .. 'Z' | 'a' .. 'z' | '0' .. '9' =>
>          ....
>       when others =>
>          null;
>    end case;
>
> Greetings,
>
> Jacob
> -- 
> "Banning open source would have immediate, broad, and
>  strongly negative impacts on the ability of many sensitive
>  and security-focused DOD groups to protect themselves
>  against cyberattacks"                        -- Mitre Corp.