s5 to s7 - many doubts

DO DW 34

L FY 0

L DBB 69 // L DBW 68 ?

SLW 3

LAR1

L MB [AR1,P#0.0]

--------

:LW =DB-P (BLOCK_DB) load number of db ????

// L #DB_P;

// // *** Error in Line 13881 (FB 80): Invalid operator, can possibly be replaced with the

//statement:\"L P##DB_P\" ***

--------

:DO FW 222

:L DW 0

L MB 223

SLW 4

L 2

* I

LAR1

L DBW [AR1,P#0.0]

I have others doubts...., in attachment , help me....,

Thanks

Alex

DO DW 34

L FY 0

This loads in the accu theflag byte specified in DW34

L DBB 69 // L DBW 68 ?

SLW 3

LAR1

L MB [AR1,P#0.0]


Because in step 7 the pointeruses the 3 last significant bits to specify a bit refernce, you must shift 3 positions to the left if you start from a byte number.
So firts you loas DBB69 that contains the number of the flag byte to load.
Ten you shift 3 positions to the left to compute a local pointer.
YouLoad this computed pointer in the register AR1.
Finally, you load the accumulator with the flagbyte specified in AR1 + a nul offset in this case (P#0.0)

Conclusion, both programs do the same. But S5 starts with MW34, and S7 with the content of DBB69.
If you use DBB69, the byte number can only be from 0..255
if you use DBW68, the byte number can be higher than 255.

--------

:DO FW 222

:L DW 0

Here in S5 you load in the accu the DW specified in FW222
Pay attention that in S5, a DW is copmposed of DL (Data Left)and DR (Data right)

L MB 223

SLW 4

L 2

* I

LAR1

L DBW [AR1,P#0.0]

In S7, however you have a byte structure. Therefor, to reach the equivalent of DW10 in S5, you must get dbb20 in S7.This explains whyyou see why there is a multiplication by 2 in computing the pointer placed in AR1.

I hope this clarifies some points.


In

:LW =DB-P (BLOCK-DB) ??, this is a FB

:L KH 00FF
:AW

// L #DB_P;

// // *** Error in Line 13881 (FB 80): Invalid operator, can possibly be replaced with the

//statement:\"L P##DB_P\" ***

thanks