Re: Invalid operator name?

["David C. Hoos, Sr." <david.c.hoos.sr@ada95.com>]

As with many of the questions posed in this newsgroup, the
answer is given explicitly in the Ada Reference Manual.

Allow me to quote the relevant excerpt:

4.5 Operators and Expression Evaluation

1 The language defines the following six categories of operators (given in order of increasing 
precedence). The corresponding operator_symbols, and only those, can be used as designators in 
declarations of functions for user-defined operators. See 6.6, "Overloading of Operators".

Syntax

2 logical_operator ::=  and  |  or  |  xor
3 relational_operator ::=  =  |  /=  |  <  |  <=  |  >  |  >=
4 binary_adding_operator ::=  +  |  -  |  &
5 unary_adding_operator ::=  +  |  -
6 multiplying_operator ::=  *  |  /  |  mod  |  rem

7 highest_precedence_operator ::=  **  |  abs  |  not

Please note the words "and only those" near the middle of the second line of 4.5(1)

----- Original Message ----- 
From: "Sandro Magi" <smagi@naasking.homeip.net>
Newsgroups: comp.lang.ada
To: <comp.lang.ada@ada-france.org>
Sent: November 27, 2004 10:23 AM
Subject: Invalid operator name?


> When I try to define my own operator, I receive an "Invalid operator name"
> error during compilation. I've found references to operator overloading in
> reference materials, but no word whether new operators can be defined which
> leads me to believe that it's not possible.
>
> Is it not possible to define new operators for a custom type? Can we only
> override the base operators?
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