Re: defining functions

Re: defining functions

[Matthew Heaney]

In article <81gthb$kej$1@bunyip.cc.uq.edu.au> , "Riyaz Mansoor" 
<s800032@student.uq.edu.au> wrote:

> when i try to define a function as below my compiler tells me the name is
> invalid. when i change the name to "=" it works fine. is this a compiler
> specific problem? or ada restriction?

This is not an Ada "restriction."  This is an Ada *feature*.


>    function  "=="  (Stack1, Stack2 : in Stack) return Boolean;

Wrong.  Do this instead:

  function "=" (L, R : Stack) return Boolean;


> if this is a compiler problem, i'm using AdaGIDE (GNAT 3.12), is there
> anyway i can tell the compiler to accept "==" as a valid name?

No.  Look up the list of valid operators in the RM.


> (i find the single equal sign really irritating!!!)

I'm sorry to hear that.



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Re: defining functions

[Lutz Donnerhacke]

* Riyaz Mansoor wrote:
>anyway i can tell the compiler to accept "==" as a valid name?
>(i find the single equal sign really irritating!!!)

May be, but "==" is not a valid binary Operator in Ada. So you can not
define it.

Re: defining functions

[Ted Dennison]

In article <81gthb$kej$1@bunyip.cc.uq.edu.au>,
  "Riyaz Mansoor" <s800032@student.uq.edu.au> wrote:

> when i try to define a function as below my compiler tells me the name
> is invalid. when i change the name to "=" it works fine. is this a
> compiler specific problem? or ada restriction?
>
>    function  "=="  (Stack1, Stack2 : in Stack) return Boolean;

No. You do not get to create your own binary operators with any series
of random characters you choose. You are restricted to the ones the
language provides. I was a smidge diappointed when I first figured that
out myself.

> anyway i can tell the compiler to accept "==" as a valid name?
No. You'll have to stick to Ada/Pascal/Modula/Oberon syntax for equality
and assignment. Sorry.

But on the bright side, you'll make less compilable typos this way.

--
T.E.D.


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Re: defining functions

[Gautier]

> (i find the single equal sign really irritating!!!)

Ah.

The fact that "=" means "equal" does worry you ?!

How strange...

Unless you had too much violent images in your childhood,
like "==" meaning "equal", 
     "="  meaning "let... be..."
     "a=-b" meaning either "substract b to a", or "let a be -b"
etc.

;-)

-- 
Gautier

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defining functions

[Riyaz Mansoor]

hi there

when i try to define a function as below my compiler tells me the name is
invalid. when i change the name to "=" it works fine. is this a compiler
specific problem? or ada restriction?

   function  "=="  (Stack1, Stack2 : in Stack) return Boolean;

if this is a compiler problem, i'm using AdaGIDE (GNAT 3.12), is there
anyway i can tell the compiler to accept "==" as a valid name?

(i find the single equal sign really irritating!!!)

thanx

----------------------------------------------------

Re: defining functions

[Robert Dewar]

In article <81gthb$kej$1@bunyip.cc.uq.edu.au>,
  "Riyaz Mansoor" <s800032@student.uq.edu.au> wrote:
>    function  "=="  (Stack1, Stack2 : in Stack) return Boolean;

> (i find the single equal sign really irritating!!!)


One of the things that is important when you come to a new
language is to understand the style and conform to it. Nothing
would be more horrible (*) than seeing == used in an Ada
program, and it is in this case an excellent thing that the
compiler prevents this abuse.

(*) well I guess a comparable thing would be to see an Ada
programmer writing C, and starting off with

#define begin {
#define end   }

etc.

Unfortunately the C preprocessor WOULD allow this appalling
abuse of the macro processor.


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Re: defining functions

[Riyaz Mansoor]

> No. You do not get to create your own binary operators with any series
> of random characters you choose. You are restricted to the ones the
> language provides. I was a smidge diappointed when I first figured that
> out myself.

> No. You'll have to stick to Ada/Pascal/Modula/Oberon syntax for equality
> and assignment. Sorry.
>
> But on the bright side, you'll make less compilable typos this way.

thanx TED. u really answered my question.