Can someone explain to me with an example how implicit tagging reduces overhead of number of bytes transfered in ASN.1 ? Thanks. asked 08 Jan '12, 12:17  mmalik10 |
2 Answers:
When using BER (BASIC ENCODING RULES) or DER (DISTINGUISHED ENCODING RULES), data for types are encoding using a Type-Length-Value format. Each primitive ASN.1 Type such as INTEGER has a UNIVERSAL TAG assigned by the ASN.1 standard. If you have just A ::= INTEGER This has a tag of UNIVERSAL 2, so an encoding of the interger value 5 in BER would be in hex 02 01 05. B ::= [2] IMPLICIT INTEGER For B, with an implicit tag, this says to replace the existing tag on INTEGER with [2], so the encoding in BER of the value 5 would be in hex 82 01 05. C ::= [2] EXPLICIT INTEGER For C, with an explicit tag, this says to add [2] in front of the existing tag, so the encoding in BER of the value 5 would be in hex A2 03 02 01 05. There is a free ASN.1 book you can download from http://www.oss.com/asn1/resources/books-whitepapers-pubs/asn1-books.html which explains tagging in much more detail. Paul answered 18 Jan '12, 09:40  Paul edited 18 Jan '12, 10:46 |
Take this ASN1 description Foo SEQUENCE of { bar [1] Bar, bar2 [2] Bar2 } Bar ::= INTEGER When EXPLICIT the encoding of the INTEGER type for Bar will be included adding at least 2 bytes, for IMPLICIT only the encoding of the tag will be there. answered 09 Jan '12, 22:49  Anders ♦ the encoding is clear but How decoder knows the real type of ASN.1 data? (15 Dec '16, 02:05) IBrahim El-K... |
which one of the books in that page do you recommend?
Either of the first two books on that page are excellent. I have heard comments that the first one (by Olivier Dubuisson) might be easier to understand, while others enjoy the British humor punctuating the thorough explanations in the second one (by John Larmouth). Both are excellent resources.