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CONTINUATION OF THE COURSE ON INSTRUCTIONS from the INSTRUC.DOC file----------------------------------------------------------
  
CONTINUATION OF THE COURSE ON INSTRUCTIONS from the INSTRUC.DOC file----------------------------------------------------------
−  
−  
−  
 
LSL #BBB,dn (.B),[.W],(.L) (Logical Shift Left)
  
LSL #BBB,dn (.B),[.W],(.L) (Logical Shift Left)
 
--------------
  
--------------
Line 14: Line 11:
 
------------------
  
------------------
   
−
Allows a shift by #BBB or dm (according to syntax) BITS
 +
Performs a shift of #BBB or dm (depending on the syntax) BITS
 
of the data register dn to the left.
  
of the data register dn to the left.
−
Replacement bits are ZERO.
 +
The replacement bits are ZEROS.
   
−
However, it must be that 1<= #BBB <= 8 for the 1st form.
 +
It is necessary that 1<= #BBB <= 8 for the 1st form.
−
-------------
+
-------------
   
−
For the 3rd form, the addressing modes allowed for the operand
 +
For the 3rd form, the allowed addressing modes for the
−
destination are:
 +
destination operand are:
−
-----------
 +
-----------
 
BBBB
  
BBBB
 
BB
  
BB
Line 31: Line 28:
 
d(an)
  
d(an)
 
d(an,rn)
  
d(an,rn)
  +
⚫
 
 
But in this case, the operation can only shift
  
But in this case, the operation can only shift
 
1 SINGLE BIT.
  
1 SINGLE BIT.
   
  ⚫
For example, if d0=%10000101110001101111011100000110 (word)
⚫
 
⚫
That is to say that for example if d0=%10000101110001101111011100000110 (word)
  
 
------
  
------
−
And I write LSL.W #1,dn
+
And if I write LSL.W #1,dn
   
−
or MOVE.W #1,d1
 +
or MOVE.W #1,d1
−
LSL.W d1,d0 , we obtain:
 +
LSL.W d1,d0, we get:
   
−
d0=%00001011100011011110111000001100
+
d0=%00001011100011011110111000001100
−
<-------- <--:0
+
<-------- <--:0
   
−
All Bits of the LOW word of the data register dn (.W)
 +
All the bits of the lower WORD of the data register dn (.W)
 
have been shifted by 1 bit to the left:
  
have been shifted by 1 bit to the left:
   
−
The incoming least-significant bit is always replaced with a
+
The least significant bit entering is always replaced by a
−
ZERO bit.
 +
NULL bit.
 
 
−
That is to say: That we move #BBB or dm BITS out of dn from the left and that the
 +
Meaning: #BBB or dm BITS are shifted out of dn to the left and the
−
replacement bits entering on the right of dn are ZERO BITS.
 +
replacement bits entering on the right of dn are NULL BITS.
   
−
The content of dn is therefore changed, the MSB also has a chance to be
 +
The contents of dn are therefore changed, the MSB is also likely to change,
−
changed, which is why the SIGN of dn will not necessarily be preserved.
 +
that's why the SIGN of dn may not necessarily be preserved.
   
−
The CCR will be changed according to the new value of dn.
 +
The CCR will change according to the new value of dn.
 
---
  
---
−
N=1 if the MSB is 1, otherwise it is equal to 0
 +
N=1 if the MSB is 1, otherwise it is 0
−
Z=1 if all bits of dn are zero, otherwise it is equal to 0
 +
Z=1 if all bits of dn are null, otherwise it is 0
 
V is always set to 0
  
V is always set to 0
 
C and X are set to the value of the LAST BIT EXITED from dn.
  
C and X are set to the value of the LAST BIT EXITED from dn.
Line 71: Line 67:
 
LSL.B #5,d0
  
LSL.B #5,d0
   
−
We operate a shift of 5 bits of the LOW BYTE of d0
 +
A shift of 5 bits of the lower BYTE of d0 is operated
 
(only bits 0 to 7 of d0 are therefore concerned by lsl.B)
  
(only bits 0 to 7 of d0 are therefore concerned by lsl.B)
   
Line 80: Line 76:
 
X=C=0 <- 1111111110100000 in d0
 
X=C=0 <- 1111111110100000 in d0
   
−
5 zero bits entered to the right of the low byte of d0, the
 +
5 null bits have entered the right of the lower byte of d0, the
−
last bit to exit from d0 was zero: X=C=0 in the CCR
 +
last bit to exit from d0 was null: bit X=C=0 in the CCR
   
   
  +
⚫
...
  
  +
  +
LSR #BBB,dn (.B),[.W],(.L) (Logical Shift Right)
  +
--------------
  ⚫
or
  +
  +
LSR dm,dn (.B),[.W],(.L)
  +
--------------
  +
or
  +
  +
LSR.W destination (.W)
  +
------------------
  +
  +
Performs a shift of #BBB or dm (depending on the syntax) BITS
  +
of the data register dn to the right.
  +
The replacement bits are ZEROS.
  +
  +
It is necessary that 1<= #BBB <= 8 for the 1st form.
  +
-------------
  +
For the 3rd form, the allowed addressing modes for the
  +
destination operand are:
  +
-----------
  +
BBBB
  +
BB
  +
(an)
  +
-(an)
  +
(an)+
  +
d(an)
  +
d(an,rn)
  ⚫
  +
But in this case, the operation can only shift
  +
1 SINGLE BIT.
  +
  +
For example, if d0=%10000101110001101111011100000110 (word)
  +
------
  +
And if I write LSR.W #1,dn
  +
  +
or MOVE.W #1,d1
  +
LSR.W d1,d0, we get:
  +
  +
d0=%01000010111000110111101110000011
  +
0:--> ------->
  +
  +
All the bits of the lower WORD of the data register dn (.W)
  +
have been shifted by 1 bit to the right:
  +
  +
The most significant bit entering is always replaced by a
  +
NULL bit.
  ⚫
  +
Meaning: #BBB or dm BITS are shifted out of dn to the right and the
  +
replacement bits entering on the left of dn are NULL BITS.
  +
  +
The contents of dn are therefore changed, the MSB is also likely to change,
  +
that's why the SIGN of dn may not necessarily be preserved.
  +
  +
The CCR will change according to the new value of dn.
  +
---
  +
N=1 if the MSB is 1, otherwise it is 0
  +
Z=1 if all bits of dn are null, otherwise it is 0
  +
V is always set to 0
  +
C and X are set to the value of the LAST BIT EXITED from dn.
  +
  +
  +
Example of use:
  +
----------------------
  +
MOVE #%1111111101010101,d0
  +
LSR.B #5,d0
  +
  +
A shift of 5 bits of the lower BYTE of d0 is operated.
  +
(only bits 0 to 7 of d0 are therefore concerned by lsr.B)
  +
  +
We get:
  +
  +
------0:>------> CCR
  +
  +
d0 :1111111100000010 -> X=C=1
  +
  +
5 null bits have entered the left of the lower byte of d0, the
  +
last bit to exit from d0 was 1: bit X=C=1 in the CCR
  +
   
 
Back to [[ASM_Tutorial]]
  
Back to [[ASM_Tutorial]]
 
[[Category: ASSEMBLER 68000 on ATARI ST Part 1 ]]
  
[[Category: ASSEMBLER 68000 on ATARI ST Part 1 ]]
−  
−
(Note: Due to the length of the document, only a fragment of the translation is produced. The translation task can continue with subsequent requests if the remaining parts of the document require translation.)
  

Revision as of 12:24, 17 December 2023

CONTINUATION OF THE COURSE ON INSTRUCTIONS from the INSTRUC.DOC file---------------------------------------------------------- 

  LSL    #BBB,dn   (.B),[.W],(.L)             (Logical Shift Left)
  --------------
  or

  LSL      dm,dn   (.B),[.W],(.L)
  --------------
  or 

  LSL.W  destination  (.W)
  ------------------

  Performs a shift of #BBB or dm (depending on the syntax) BITS
  of the data register dn to the left.
  The replacement bits are ZEROS.

  It is necessary that 1<= #BBB <= 8 for the 1st form.
                        -------------

  For the 3rd form, the allowed addressing modes for the
  destination operand are:
  -----------
  BBBB
  BB
  (an)
  -(an)
  (an)+
  d(an)
  d(an,rn)

  But in this case, the operation can only shift
  1 SINGLE BIT.

  For example, if d0=%10000101110001101111011100000110 (word)
  ------
  And if I write LSL.W #1,dn 

              or MOVE.W #1,d1
                 LSL.W d1,d0, we get: 

                           d0=%00001011100011011110111000001100 
                               <--------                       <--:0    

  All the bits of the lower WORD of the data register dn (.W)
  have been shifted by 1 bit to the left:

  The least significant bit entering is always replaced by a
  NULL bit.
  
  Meaning: #BBB or dm BITS are shifted out of dn to the left and the
  replacement bits entering on the right of dn are NULL BITS.

  The contents of dn are therefore changed, the MSB is also likely to change,
  that's why the SIGN of dn may not necessarily be preserved.

  The CCR will change according to the new value of dn.
     ---
  N=1 if the MSB is 1, otherwise it is 0
  Z=1 if all bits of dn are null, otherwise it is 0
  V is always set to 0
  C and X are set to the value of the LAST BIT EXITED from dn.
   

  Example of use:
  ----------------------
  MOVE     #%1111111101010101,d0  
  LSL.B    #5,d0

  A shift of 5 bits of the lower BYTE of d0 is operated
  (only bits 0 to 7 of d0 are therefore concerned by lsl.B)

  We get: 

  CCR        --------<------<:0

  X=C=0  <-  1111111110100000 in d0   

  5 null bits have entered the right of the lower byte of d0, the
  last bit to exit from d0 was null: bit X=C=0 in the CCR




  LSR    #BBB,dn   (.B),[.W],(.L)             (Logical Shift Right)
  --------------
  or

  LSR      dm,dn   (.B),[.W],(.L)
  --------------
  or

  LSR.W  destination (.W)
  ------------------

  Performs a shift of #BBB or dm (depending on the syntax) BITS
  of the data register dn to the right.
  The replacement bits are ZEROS.

  It is necessary that 1<= #BBB <= 8 for the 1st form.
                        -------------   
  For the 3rd form, the allowed addressing modes for the
  destination operand are:
  -----------
  BBBB
  BB
  (an)
  -(an)
  (an)+
  d(an)
  d(an,rn)
  
  But in this case, the operation can only shift
  1 SINGLE BIT.

  For example, if d0=%10000101110001101111011100000110 (word)
  ------
  And if I write LSR.W #1,dn 

              or MOVE.W #1,d1
                 LSR.W d1,d0, we get: 

                           d0=%01000010111000110111101110000011 
                          0:-->                        ------->   

  All the bits of the lower WORD of the data register dn (.W)
  have been shifted by 1 bit to the right:

  The most significant bit entering is always replaced by a 
  NULL bit.
  
  Meaning: #BBB or dm BITS are shifted out of dn to the right and the
  replacement bits entering on the left of dn are NULL BITS.

  The contents of dn are therefore changed, the MSB is also likely to change,
  that's why the SIGN of dn may not necessarily be preserved.

  The CCR will change according to the new value of dn.
     ---
  N=1 if the MSB is 1, otherwise it is 0
  Z=1 if all bits of dn are null, otherwise it is 0
  V is always set to 0
  C and X are set to the value of the LAST BIT EXITED from dn.
   

  Example of use:
  ----------------------  
  MOVE     #%1111111101010101,d0  
  LSR.B    #5,d0

  A shift of 5 bits of the lower BYTE of d0 is operated.
  (only bits 0 to 7 of d0 are therefore concerned by lsr.B)

  We get: 

      ------0:>------>     CCR

  d0 :1111111100000010  -> X=C=1         

  5 null bits have entered the left of the lower byte of d0, the
  last bit to exit from d0 was 1: bit X=C=1 in the CCR


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