Step 5 TNB instruction with operand of zero!

Slogging my way through re-writing a Step 5 program in RS Logix, I have come across this segment:

A F 60.7
A T 28
A I 3.4
A I 3.5
AN I 11.2
A I 11.7
= F 70.6
TNB 0
A(
A I 30.6
A I 17.1
O
A I 30.7
A I 17.0
)
A I 37.7
AN I 38.0
L DW 28
SD T 28
NOP 0
NOP 0
NOP 0
NOP 0
***

All fairly simple except that "TNB 0" statement.

As far as I can tell this is a bytewise transfer of zero bytes, i.e. do nothing.

Am I missing something or is there some "programming trick" being employed? I can't see why this is not two separate segments with the second segment starting where the TNB is.

Any thoughts gratefully appreciated.

Strange.
The 'segment end' instruction could have been manipulated in the CPU to make the segment too long to be viewable by the Step5 software - a primitive kind of knowhow prot.

What does the preceeding code look like?

Kalle

The code either side of that segment is very simple bit manipulation, so simple it will display in ladder as would the above if it was two segments without the TNB 0.

Since then I have come accross another one:



Logically segment 5 should start A Q 24.6 if the code for T 55 follows the code for T56 - they are a left / right pair of devices.

This program is something like 20+ years old and has been modified continuously during that time. Could it simply be a corruption - original software was DOS and floppy based?

I'm using IBH S5/S7 for windows to view the code.

Yes it might be corruption or maybe a typo. We'll probably never know.

Kalle