# Symbol

The # indicates indirection. I will have to be careful not to explain it wrong, so I'll stop on that subject, but

Also note the size mismatch between the data types. In a COP instruction, the destination determines the size of each element, so if the destination is a Float, thats 32 bits per chunk and five chunks. This means that you are copying 160 bits (10 integers or 5 floats) from the Source to the Destination. In your case the source is a float and the destination is an INT so with a length of 5, you are copying 80 bits, or 2.5 floats!

The COP is a bit for bit image...the MOV will do data type conversion.
 
Even simpler.....

It indicates that the address you see is the starting point. #F8:61 with a length of 5 indicates that we will start copying at F8:61 and grab five consecutive values. That means F8:61, F8:62, F8:63, F8:64, F8:65 are your source data.

If your destination is #N80:43 then the destination will start with N80:43 and continue through N80:47.

So,

F8:61 is copied to N80:43
F8:62 is copied to N80:44
F8:63 is copied to N80:45
F8:64 is copied to N80:46
F8:65 is copied to N80:47

Hope that helps.

OG
 
Thanks.
So a copy instruction with the source being #I:1.0, Destination being #N23.0 with a Length of 4.
What would this mean?

Thanks
 
...
So a copy instruction with the source being #I:1.0, Destination being #N23.0 with a Length of 4.
What would this mean?

Assuming that you meant N23:0 for the destination, 4 elements (16 bit integers in this case) will be copied beginning at I:1.0. Exactly which input words get copied would depend on the I/O configuration of the PLC.

It might be:
I:1.0 -> N23:0
I:1.1 -> N23:1
I:1.2 -> N23:2
I:1.3 -> N23:3

Or:
I:1.0 -> N23:0
I:2.0 -> N23:1
I:3.0 -> N23:2
I:4.0 -> N23:3

Or something else entirely. It depends on what cards are in those slots.
 

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