pointer

Dear Friends,

I have written a function with two pointer parameters - one for input and other for output. Say for example #input_pointer of type pointer and #ouput_pointer of type pointer.

When I call that function from OB1 with the parameter values p#i0.0 and p#DB1.DBX0.0 respectively, it is taking the parameter values without any problem.

But in my function where I have put the codes:

L p##input_pointer
LAR1
.
.
.
.
L p##output_pointer
LAR2
.
.
.
.

I am getting the values in Acc1 16#870000A8 for input pointer
and 16#870000D8 for output pointer. But I am expecting the values - 16#81000000 and 16#84000000 respectively. Where is the actual problem? Please help to identify the problem. Hope there must be somebody who can clarify the matter.

With best regards,

Shyamal

In this case you are loading a pointer to a pointer.

Before calling your function, code in OB1 generates the pointers and stores them in the L area of OB1. The parameters to your FC are then assigned pointers to the L area, but are changed to the V area (local data of the calling block).

Your function needs the following to access the pointers:

Here's the code presented by the editor

Here's the actual instructions behind the code

Daer L D,

It is very much justified that you wants to be known to all with the name L D[AR2, P#0.0]. You are really great. I had always got very helpful informations of this type of questions from you.

I will try to understand the mechanism you described. Would you kindly help me to have any documents from which I can learn more about this mechanism.

Best Regards,

Shyamal

Read Berger. Experiment with STL.

would you please give me the website address.

Here's a link to a book store:

http://www.allbookstores.com/author/Hans_Berger.html

Hi there LD,

In the above post you broke this:

CALL FC 4
pInput :=P#I 0.0
pOutput:=P#DB1.DBX 0.0

into this:

BLD 1
= L 20.0
L 0
T LW 21
L P#I 0.0
T LD 23
L 1
T LW 27
L P#DBX 0.0
T LD 29
UC FC 4
P#L 21.0
P#L 27.0
BLD 2

Q1)How??
Q2)Why L 0 and and L 1 is happening??


Burger is very expensive ;-)

manmeetvirdi said:

Burger is very expensive ;-)

Click to expand...

Compared to what - Macdonalds, or a Step 7 License ?

See the following thread on how to reveal the "broken" version of the code.

The 0 and 1 are the DB numbers for the pointers. P#I0.0 has a DB number of 0

http://www.plctalk.net/qanda/showthread.php?t=31273&highlight=hidden

Wow that's great. Delighted.
Felt like roaming around in Egypt in search of treasure beneath Pyramids! And found it !
Now thinking how to utilize the treasure

All thanks to you for giving detailed maps