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    Re: iSCSI: keys/parameter dependence



    --- Julian Satran <Julian_Satran@il.ibm.com> wrote:
    
    > Almost everything is correct - except the mechanism
    > for 
    > Very-Long-responses - (TTT).
    > This is different than the spanning mechanism (C-
    > bit) in the sense that 
    > has in theory no bounds (it was devised mainly for
    > SendTargets but can be 
    > used by any mechanism that has to send a lot of data
    > - not negotiations).
    
    Hmm, let me verify that I'm understanding this...
    
    Let's suppose the following scenario:
    
    A target is planning to send a very long response
    (exceeding the mandatory buffering capabilities
    of the other side) to a TextRequest that contained
    the SendTargets key. This is legal, 
    because it is not a negotiation, right? 
    
    Let's suppose the text it is sending
    will get broken in m TextResponse PDUs. 
    For any of these PDUs, if the key=value pairs fit
    completely (i.e., including the terminating '\0'),
    and if the target doesn't mind the possibility
    that the initiator may send some data back
    (for example, do a little negotiation in parallel
    with hearing what SendTargets will return...),
    then it need not set the C-bit. Correct?
    
    But it is also permissible to set the C-bit, 
    since the target may not consider itself as 
    "being done with this set of keys". Right?
    
    If, however, in some PDU the last key=value
    pair is split (i.e., will continue in the next
    PDU), then it MUST set the C-bit, because
    such a PDU cannot possibly "end a set of keys".
    
    Can anybody comment on this understanding, please?
    
    Applying this understanding to the example in
    section 9.10.4, I can conjecture that the C
    bit may or may not be set; it depends on whether
    <part x> contains only complete (with '\0' at the 
    end) pairs or not, and possibly on target's
    preferences. The PDUs travelling in the other
    direction need not necessarily be empty, unless
    the C-bit in the preceding PDU forces them to be.
    Is this true?
    
    Many thanks in advance,
    
    Martins Krikis, Intel Corp.
    
    Disclaimer: these are my own opinions and are
                not necessarily Intel's opinions.
    
    
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Last updated: Wed Jun 05 16:18:33 2002
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