Hi,
ENUMERATED { a,b(100),c} is equivalent to ENUMERATED{a(0), b(100), c(1)}
So in your exemple :
a ENUMERATED { a,b(100),c} ::= c
affect the value c(1) to a.
If you used a mix of NamedNumbers and identifier in an EnumeratedType, you
affect an integer value to each identifier of your Enumerated, beginnning to
0, increasing 1, except the values already defined in your NamedNumbers.
As an exemple :
T ::= ENUMERATED{a(3) ,b ,c(1), d, e} is equivalent to ENUMERATED{a(3),
b(0), c(1), d(2), e(4)}
Best regards,
HETAULT Pierre-Marie
ATOS ORGIN
www.marben-products.com
-----Original Message-----
From: Bujji krishna R B <[EMAIL PROTECTED]>
To: [EMAIL PROTECTED] <[EMAIL PROTECTED]>
Date: vendredi 21 septembre 2001 14:19
Subject: [ASN.1] Enumerated types in ASN.1
>Hi ,
>
> I have a doubt regarding the representation of Enumerated types in ASN.1.
> Consider an enumerated type given below
> A ::= ENUMERATED { a,b(100),c} ::= c
>
> Now what is the Value of A , is it either 2 or 101.
>
> I would also like to know the significance of having a mix of NamedNumbers
>and
> identifier in the Enumerated types.
>
>Thanks
>krishna
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