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From: Howard Brazee on 22 Feb 2007 15:04
On Thu, 22 Feb 2007 12:51:05 -0700, Louis Krupp
<lkrupp(a)pssw.nospam.com.invalid> wrote:

>> The record matches the layout perfectly!
>> Micro Focus S9(4) COMP is two bytes binary
>> and 9V9(4) COMP is three bytes binary.
>>
>>
>>
>
>How do you fit a sign and four digits into two bytes? (Or am I being
>unusually dense today?)

A byte holds 16 bits - 15 plus the sign. +9999d = 10011100001111b

From: Richard on 22 Feb 2007 15:06
On Feb 23, 8:51 am, Louis Krupp <lkr...(a)pssw.nospam.com.invalid>
wrote:


> > The record matches the layout perfectly!
> > Micro Focus S9(4) COMP is two bytes binary
> > and 9V9(4) COMP is three bytes binary.
>
> How do you fit a sign and four digits into two bytes? (Or am I being
> unusually dense today?)

In binary (as stated). The value in 16 bit can be -32368 to 32367
which is a larger range than -9999 to 9999.

COMP (in this case) however is byte reversed compared to intel format
being big-endian.



From: Rick Smith on 22 Feb 2007 15:35

"Richard" <riplin(a)Azonic.co.nz> wrote in message
news:1172174815.695241.198480(a)l53g2000cwa.googlegroups.com...
[snip]
> In binary (as stated). The value in 16 bit can be -32368 to 32367
> which is a larger range than -9999 to 9999.

Pedantically, the range is -32768 to +32767, which
is the range that applies when the NOTRUNC
compiler directive is used, as is true in the case
of the record layout in question. [Important for any
who may wish to write the requested program.]



From: William M. Klein on 22 Feb 2007 15:57
Also important that the NOIBMCOMP directive is in effect. This is important for
why you can have 3 (or 1) byte binary fields.

--
Bill Klein
wmklein <at> ix.netcom.com
"Rick Smith" <ricksmith(a)mfi.net> wrote in message
news:12trvl1q5rqfp29(a)corp.supernews.com...
>
> "Richard" <riplin(a)Azonic.co.nz> wrote in message
> news:1172174815.695241.198480(a)l53g2000cwa.googlegroups.com...
> [snip]
>> In binary (as stated). The value in 16 bit can be -32368 to 32367
>> which is a larger range than -9999 to 9999.
>
> Pedantically, the range is -32768 to +32767, which
> is the range that applies when the NOTRUNC
> compiler directive is used, as is true in the case
> of the record layout in question. [Important for any
> who may wish to write the requested program.]
>
>
>


From: jacodeguy on 22 Feb 2007 16:00
On Feb 22, 2:06 pm, "Richard" <rip...(a)Azonic.co.nz> wrote:
> On Feb 23, 8:51 am, Louis Krupp <lkr...(a)pssw.nospam.com.invalid>
> wrote:
>
> > > The record matches the layout perfectly!
> > > Micro Focus S9(4) COMP is two bytes binary
> > > and 9V9(4) COMP is three bytes binary.
>
> > How do you fit a sign and four digits into two bytes? (Or am I being
> > unusually dense today?)
>
> In binary (as stated). The value in 16 bit can be -32368 to 32367
> which is a larger range than -9999 to 9999.
>
> COMP (in this case) however is byte reversed compared to intel format
> being big-endian.

I have some printouts of what the sketch should look like, I cannot
find the first record (pin 01-000-001-00), but I do have the second
(pin 01-000-006-00). Based on looking at the printout, I think the
first part should be like this:

01-000-006-00, A, -40, 16, 1, 11, 13, 33, 13, 49, 31, 49, 31, 53, 47,
53, 47, 27, 23, 27, 23, 33, 13, 33, 0, -2,

C I can do, and would love too, but a little rusty. The bytes and
strings I think I can get out, but I'm not sure about the S9(4) COMP
(two bytes reversed) thing.

if I have:

byte *buffer;
fread(buffer, 2, f);
// lets pretened I just read a comp, you previously said:
// "Just byte switch the COMP fields."
// How would I do that?
int x = (buffer[1] << 16) && buffer[0]; // something like this?

-Andy




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