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Isn't that the number of records for sequential "get" operations? Not a
block size?  Year ago, I remember Dick Bains mentioning that using the
NBRRCDS() parm with the number of records set to the record count of the
file was a way to cache the file into storage.



On 12/23/05 11:49 AM, "Tony Carolla" <carolla@xxxxxxxxx> wrote:

> Okay, first off,
> 
> MERRY CHRISTMAS!!!  May the holiday spirit hit you like a model B50 falling
> off a loading dock ;-)
> 
> I have a question.  I am trying to calculate the optimum block size to use
> with OVRDBF SEQONLY(YES XXXX), for an input file from which I will read all
> records.  The file contains the following layout:
> 
> FMACRO  3A
> FMNUM  9P 0
> FMUPDAT  L
> 
> DSPFD indicates that the record length is 18.  Now, I know once I READ, the
> field that will be created to represent FMUPDAT is a length 10 (ISO) field.
> But on disk, it is a four-byte field.  So my question is this:  My
> understanding is that a size of I/O from disk is 128K.  If this is to be
> read directly from disk, into the buffer when the READ op executes, will
> there be a need for ten bytes to store this field?  I know that the single
> variable inside the program will never see that four-byte value directly,
> but why would a disk buffer need to know anything else other than the four
> bytes?
> 
> --
> "Enter any 11-digit prime number to continue..."
> "In Hebrew SQL, how do you use right() and left()?..." - Random Thought
> "If all you have is a hammer, all your problems begin to look like nails"
> --
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