> NoSize ::= NumericString
> emptyNoSize NoSize ::= ""
>
> SmallSize ::= NumericString SIZE(0..6)
> emptySmallSize SmallSize ::= ""
>
> Both of these require a length determninate with value 0. There is of course
> nothing encoded for the string itself, but will the value for the next type
> be encoded on an octet boundary due to to the alignment requirements of this
> string type? I think not, and Bancroft point out NOTE 2 for 10.9.3.3/X.691
> which makes it clear. I just wanted to make sure that that note applies to
> my situation at hand.
In the case of NoSize, where the length (unconstrained) is itself one
octet long, the length is aligned, so one octet of 0 will be used, and it
ends there since the value needs no value.
In the case of SmallSize, where the length (constrained) is 3 bits long,
the length is not aligned, so 3 unaligned bits will be used. It also ends
there since the value needs no value.
Thus...
SEQUENCE {a NoSize, b NoSize, c NoSize}
where all three are empty will be encoded (in octets) 00 00 00.
SEQUENCE {a SmallSize, b SmallSize, c SmallSize}
where all three are empty will be encoded (in bits) 000 000 000.
Care to try encoding
SEQUENCE {a SmallSize, b NoSize, c SmallSize, d NoSize} again where all
strings are empty?
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