Difference between two instructions

Dear friends i have another question please

What is the difference between these two instructions:

Note that :IN1 declared as a pointer of the address of DB18.DBX128.0 (16#84000400)

FIRST INSTRUCTION :

L   P##IN1 //NOW THE VALUE IN ACC1=16#85000010,AND THE VALUE IN ACC2=16#84000400

SECOND INSTRUCTION:

 LAR1   P##IN1 //NOW THE VALUE IN ACC1=16#84000400,AND THE VALUE IN ACC2=16#84000400

Why this happen, both instruction should give same result in both accumulators,But in the first instruction ACC1=16#8500001 which is the address of DI2.0(instance data block 2.0),and ACC2=16#84000400 which is the address of DB18.DBX128.0

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Split from indirect addressing.

Hi.

First instruction loads the address of #in into accu1

Second loads the value of #in to ar1 addres register

Hope it helps.

IZ ALJABRE-JORDAN

FIRST INSTRUCTION :

L   P##IN1 //NOW THE VALUE IN ACC1=16#85000010,AND THE VALUE IN ACC2=16#84000400

SECOND INSTRUCTION:

 LAR1   P##IN1 //NOW THE VALUE IN ACC1=16#84000400,AND THE VALUE IN ACC2=16#84000400

Why this happen, both instruction should give same result in both accumulators,But in the first instruction ACC1=16#8500001 which is the address of DI2.0(instance data block 2.0),and ACC2=16#84000400 which is the address of DB18.DBX128.0

 Hi!

Interesting question!

LAR1 should not change the contents of the accumulators, so I don't understand how did it happen in your case.

With L you get hexa85 because the #IN pointer is stored in the instance DB of the FB, so you get an address from this IDB.

What I don't know and is asking the STL experts,  why gets the address of the pointer P#DB18.DBX128.0 into ACCU1 at the beginning (see attachment)?

P.S.:This means that I could start the first line with LAR1 and get the address of the first,  "original" pointer form the FB interface loaded into AR1. Nice.

Kahl

 LAR1 should not change the contents of the accumulators, so I don't understand how did it happen in your case.

I don't think that the orginal poster used both commands in sucession in the code. It seems to me that he used in the same place the first one and then the second one instead. Mistery revealed?

Dear Friends,

Please try these two codes in the the PLCSIM

Where IN1=DB18.DBX128.0

First Code:

L P##IN1 

Now after performing this instruction the PLCSIM shows the following values:

Acc1=16#85000010 This is OK for me.

Acc2=16#84000400  Why? How this can be?

Second Code:

LAR1 P##IN1 

Now after performing this instruction the PLCSIM shows the following values:

Acc1=16#84000400 Why? How this can be?

Acc2=16#8  Why? How this can be?

Address Register 1=16#85000010 Why? This should be 16#84000400 and not 16#85000010 

IZ ALJABRE-JORDAN

Dear Friends,

Please try these two codes in the the PLCSIM

Where IN1=DB18.DBX128.0

First Code:

L P##IN1 

Now after performing this instruction the PLCSIM shows the following values:

Acc1=16#85000010 This is OK for me.

Acc2=16#84000400  Why? How this can be?

Second Code:

LAR1 P##IN1 

Now after performing this instruction the PLCSIM shows the following values:

Acc1=16#84000400 Why? How this can be?

Acc2=16#8  Why? How this can be?

Address Register 1=16#85000010 Why? This should be 16#84000400 and not 16#85000010 

hi, 

as far as I see, everything correct

1st Code

L P##IN1 

Now after performing this instruction the PLCSIM shows the following values:

Acc1=16#85000010 This is OK for me.

Acc2=16#84000400  Why? How this can be?

 L P##IN1

16#85000010 is loaded to ACCU1, and what was before in ACCU1 is now in ACCU2. I guess this was 16#84000400. So everything is fine

2nd code

LAR1 P##IN1 

Now after performing this instruction the PLCSIM shows the following values:

Acc1=16#84000400 Why? How this can be?

Acc2=16#8  Why? How this can be?

Address Register 1=16#85000010 Why? This should be 16#84000400 and not 16#85000010

 LAR1 P##IN1

This instruction does not change ACCU1 and ACCU2. It is loading P##IN1 in AR1. So what you loaded in the 1st code in ACCU1(16#85000010) is now in AR1. 

hope this helps,

 regards,

Dear mop S04,

You are completely right 

After i replaced the code with :

L 0

L 0

LAR1 P##IN1  OR L P##IN1 

Now every thing is running fine.

But i have another question please :

Now i am writing this code in function block FB9 and i am calling this FB9 in FB 364 which has instance DB364.

So why the pointer points to 16#85000010 which is DI2.0 ,Means it points to DI2 not DI364.

I am not clear in that can you pleas explain how does it work.

IZ ALJABRE-JORDAN

But i have another question please :

Now i am writing this code in function block FB9 and i am calling this FB9 in FB 364 which has instance DB364.

So why the pointer points to 16#85000010 which is DI2.0 ,Means it points to DI2 not DI364.

I am not clear in that can you pleas explain how does it work.

hi IZ ALJABRE JORDAN,

I´m not sure what you mean... 

DI means the Instance-DB, right?

Maybe you could provide a Screenshot with online vision on the block and with showing all posible Registers and ACCU´s.

regards,

Yes Exactly DI is Instance DB

please see the attached shots.

In FB364(which has DI364)  i'm calling FB9.

Now in FB9,IN15 points to DI2 not to DI364