hope someone can help. I try to understand how a SMF RDW works. Basically I have no problems to understand ?the whole thing? as long as the record is not ?spanned?. To make it clearer: For example a RDW (first 4 Bytes) of ?33.14.01.00?. OK, x?3314? tells me that the record is 13.076 Bytes long, but how to deal with the remaining part the ?0100?? The only thing I know is that the record is spanned actually. Does this mean that the next 256 Bytes ?after? the first 13.076 Bytes belong together or how is this calculated?!
RDWs are 4 bytes. The first two are always the lenght of the record. The fourth byte is reserved and is always x'00'. The third byte, the one you're talking about, is always x'00' unless the record format is VBS, .i.e. spanned records. With non spanned records, a logical record must always fit into a single block. For spanned format, the logical records may be split into multiple parts, each one being written to a separat block. So, there is a first part, there are zero or more intermiediate parts, and there is a final part. The rightmost 2 bits of the third byte in the RDW indicat just this: x'01' -> first part, .i.e. a new spanned record starts here. x'11' -> intermediate part. Note that there is no sequence number. The sequence is given by the sequence of the blocks. x'10' -> The final part, .i.e. the end of the spannded record. Note that even with spanned format, non-spanned records are allowed besides spanned records. I.e. there may be records that do fit in a block, and they will have the third byte as x'00'. -- Peter Hunkeler ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@listserv.ua.edu with the message: INFO IBM-MAIN
