[LMH] Re: [LMH]bored? read pages 7,8,9 and explain

[Steve Krueger]

Brad Parker wrote:

> Hi,
>
> If anyone (1) bored, (2) interested in CADR architecture and (3)
> familiar with ldb and dpb, would you mind reading "A. I. memo 528" page
> 7,8 & 9
>
> and then talking with me about how the ldb and dpb parts of the byte
> instruction should work?

Hi Brad,

I can probably answer any question you might have.

I know how these instructions work and so find the description in AI 
Memo 528 to be completely clear (well mostly clear).  I suggest you pose 
a question and I'll try to explain it.  Maybe this programmers' summary 
will help:

You use LDB to extract a bit field from a word, aligning it to the lsb 
end of the word.  If the A-SRC is all zero, this produces a result that 
is zero extended.  This is a very frequent operation.  The rotation 
count is the bit number that the field begins at, and the length is the 
field width.

             |<- len ->|<--- rot ---->|
     +--------------------------------+
     |       |  field  |              |  M-SRC
     +--------------------------------+


                            |<- len ->|
     +--------------------------------+
     |AAAAAAAAAAAAAAAAAAAAAA|  field  |  RESULT
     +--------------------------------+

"A" are bits from the A-SRC that are not modified.  Only the len bits at 
the right of the word are modified.

DPB is the logical opposite of LDB, it takes an lsb-justified bit field 
and places it at the bit field described by the rotation and length.

                             |<- len ->|
      +--------------------------------+
      |                      |  field  |  M-SRC
      +--------------------------------+

              |<- len ->|<--- rot ---->|
      +--------------------------------+
      |AAAAAAA|  field  |AAAAAAAAAAAAAA|  RESULT
      +--------------------------------+

"A" represents bits from the A-SRC that are not modified.

Selective deposite replaces the corresponding bits of the A-SRC with a 
bit field from the M-SRC described by its position and length.  It does 
a DPB without actually rotating the M-SRC first.


             |<- len ->|<--- rot ---->|
     +--------------------------------+
     |       |  field  |              |  M-SRC
     +--------------------------------+



             |<- len ->|<--- rot ---->|
     +--------------------------------+
     |AAAAAAA|  field  |AAAAAAAAAAAAAA|  RESULT
     +--------------------------------+

"A" represents bits of the A-SRC that are not modified.

Note that the length is encoded in the microinstruction as length-1.

You could read the Q-DATATYPE field of a Lisp word into an untyped 
(pure) number with:
   ((M-1) LDB M-T (BYTE 5. 25.) A-ZERO)

This encodes ROT as 25 decimal (31 octal) and length-1 as 4.

Note too that fields can be counted from the left instead of the right 
by doing a bit of arithmetic on the position and width. 

Does this help or are there more questions?

BTW, I don't think I found the left mask and right mask description 
helpful when I read AI Memo 528 almost 25 years ago either.

    -Steve