>> The first 16 bytes of the DTF is a CCB. One of the 16-bit fields in the
CCB is the residual count.
Then I am confused, because it looks like the CCB, in my case, is as
follows after reading the first card. The compile listing also seems to
say that the first two bytes are the residual count. If that is correct,
did I read 80? ...or did I read only 48? This seems reversed. However,
the 7th half-word would be right.
00500300 00500340 00066150 00308200
CARDIN DTFCD DEVADDR=SYSRDR,DEVICE=3505,IOAREA1=CARD_IA,BLKSIZE=128, X
ERROPT=IGNORE,EOFADDR=DATAIN
+* IOCS AND DEVICE INDEPENDANT I/O - DTFCD -5686-007-02-C440 @U31TUJS
+* 5686-007 (C) COPYRIGHT IBM CORP. 1981,1989 @U31TUJS
+* LICENSED MATERIAL - PROGRAM PROPERTY OF IBM @D149DBF
+ DC 0D'0'
+CARDIN DC X'000082000000' RES. COUNT, COM. BYTES STATUJJ
+ DC AL1(0) LOGICAL UNIT CLASS
+ DC AL1(0) LOGICAL UNIT
+ DC A(IJCX0017) CCW ADDRESS @D33FDVS
+ DC 4X'00' CCB-ST BYTE,CSW CCW ADDR.
Sincerely,
Dave Clark
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On Mon, Dec 15, 2025 at 12:12 PM Charles Mills <[email protected]> wrote:
> 50-year-old memory so take this with a grain of salt: The first 16 bytes
> of the DTF is a CCB. One of the 16-bit fields in the CCB is the residual
> count. So if the DTF is expecting 128 bytes and reading 80, the field will
> contain 48 = x'0030'.
>
> But I would think that scanning for binary zeroes or translating them to
> blanks would be a perfectly acceptable approach.
>
> Charles
>
>
