Indirect Addressing

hello everyone

in the below program ID is 109  and DBW 0 is 32

D_adrstart will be 3488(109*32)

ID_adrlenght will be 3488+P#2.0=3490

when i simulate its 436...could you please explain the below programm...attached is simulation screenshot with values

OPN   "DB_ENCS_SKID_NR_ADR"
L     #ID
L     DBW    0                    // Bits of one entry, normally 32
*D
T     #ID_AdrStart                // 32 * ID = Pointerstartadr
L     P#2.0
+D
T     #ID_AdrLength
L     DBW [#ID_AdrLength]

Hello Dongerpure,

Just like with the normal data types or just like in the real world is the case with the money.

IN a case you have 10$ and 10€, and you want to do a SUM, well you convert $ -> € or vice versa, after that you do SUM.

In your example: 3488+P#2.0=3490

Let's convert all to the same currency :   binary for an example:

//----------------------------------------------------------------------
3488 -> 0000 0000 0000 0000 0000 1101 1010 0000

focus on this part: 0000 1101 1010 0000

Now separate this even more:

Byte address:0000 1101 1010 0   (dec 436)

Bit address: 000 (dec 0 )

//-------------------------------------------------------------------------

//----------------------------------------------------------------------
P#2.0 -> 0000 0000 0000 0000 0000 0000 0001 0000

focus on this part: 0000 0000 0001 0000

Now separate this even more:

Byte address:0000 0000 0001 0   (dec 2)

Bit address: 000 (dec 0 )

//-------------------------------------------------------------------------

 Now = 3488+P#2.0  =  0000 1101 1010 0000  + 0000 0000 0001 0000  = 3504 = 0000 1101 1011 0000.  Now let's separate

Byte address:1101 1011 0   (dec 438)

Bit address: 000 (dec 0 )

So you moved  2 bytes inside opened Data blocks.

Always when you work with Pointers in STL pay attention on this:

Format of the Parameter Type POINTER

Best Regards Soble

 hello soble

Thank you very much for such a brief explaination

Could you please clarify how we get this

Now separate this even more:

Byte address:0000 1101 1010 0   (dec 436) ??

Bit address: 000 (dec 0 )

SOble

Hello Dongerpure,

Just like with the normal data types or just like in the real world is the case with the money.

IN a case you have 10$ and 10€, and you want to do a SUM, well you convert $ -> € or vice versa, after that you do SUM.

In your example: 3488+P#2.0=3490

Let's convert all to the same currency :   binary for an example:

//----------------------------------------------------------------------
3488 -> 0000 0000 0000 0000 0000 1101 1010 0000

focus on this part: 0000 1101 1010 0000

Now separate this even more:

Byte address:0000 1101 1010 0   (dec 436)

Bit address: 000 (dec 0 )

//-------------------------------------------------------------------------

//----------------------------------------------------------------------
P#2.0 -> 0000 0000 0000 0000 0000 0000 0001 0000

focus on this part: 0000 0000 0001 0000

Now separate this even more:

Byte address:0000 0000 0001 0   (dec 2)

Bit address: 000 (dec 0 )

//-------------------------------------------------------------------------

 Now = 3488+P#2.0  =  0000 1101 1010 0000  + 0000 0000 0001 0000  = 3504 = 0000 1101 1011 0000.  Now let's separate

Byte address:1101 1011 0   (dec 438)

Bit address: 000 (dec 0 )

So you moved  2 bytes inside opened Data blocks.

Always when you work with Pointers in STL pay attention on this:

Format of the Parameter Type POINTER

Best Regards Soble

Hello Dongerpure,

Well take a look picture:

Point is that  when it comes to the pointers for an example number : 3488 can not be read as an 1 information. Inside it contains two information. Number 436 and number 0.

How is that possible? Well separate it to the bit format, first 3 bits are 1 information ( number 0), rest of the bits represent second information (number 436).

Best Regards Soble

Dongerpure

Could you please clarify how we get this

Now separate this even more:

Byte address:0000 1101 1010 0   (dec 436) ??

Bit address: 000 (dec 0 ) 

 

Have a look at the given link by SOble. Explanation of bytes 2, 3, 4, 5 are identical as contents of the address registers.

Here is your case in Address register you have:

0000 0000 0000 0000 0000 1101 1010 0000 (bin) = 3488 (dec) = P#436.0 (address)

this is because 3488 / 8 = 436 (byte address) and reminder 0 (bit address)

On the other hand:

P#2.0 (address) = 16 (dec) = 0000 0000 0001 0000 (bin)

and this is because 2 * 8 + 0 = 16