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Section
04 Part 07 – The ROL & ROR Instructions |
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“I’ve learned over the years that it doesn’t matter where you pitch in the rotation. For me, preparation is everything.” ~Cory Lidle |
Introduction
Now
that we’ve looked into bit shifting, we can now look at bit rotation. We’ll start by using the byte C4 as an example,
and here it is in binary:
|
1100 0100 |
When
rotating left, all bits are shifted to the left. Any bits that go outside the byte on the left
come back in on the right. As shown by
this graphical example:
So if
we rotate the byte C4 a bit to the left, we’ll get:
|
v 1000 1001 < > > > > >
> ^ |
Which
is 89 in hex.
As you can see, all of the bits have shifted to the left, and the MSB
was taken out and put on the far right as the LSB. If we rotate it an extra 2 times to the left,
we’ll get:
|
v 0010 0110 < > > > > >
> ^ |
Where
the bits 10 on the far left have shifted out of the byte, and back round to the
right side.
For
rotating right, the same rules apply.
Any bits that go outside the byte on the right come back in on the
left. As shown by this graphical
example:
So
we’ll use the byte C4 once again:
|
1100 0100 |
And
we’ll rotate it to the right by 2 bits:
|
> 0011 0001 v ^ < < < < <
< |
All
of the bits have shifted to the right, and the LSB bits were taken out and put
on the far left as the MSB bits.
And
that is the basic idea of bit rotation.
The ROL
Instruction
ROL – ROtate Left (without
extended)
This
instruction will rotate the bits of the destination operand to the left. The number of
times that the bits are rotated is decided by the source operand.
Examples
|
rol.b #$02,d0 |
This
will rotate the byte of d0, 02 bits to the left. We’ll pretend that d0 contains 009F00B2 in
this example, and since it’s a byte,
only the B2 is change. B2 in binary is:
|
1011 0010 |
After
rotating the bits left by 02,
we get:
|
v 1100 1010 < > > > > >
> ^ |
1100
1010 in hex is CA, which is saved back into d0.
Now d0 contains 009F00CA.
Here’s
another example using long-word:
|
rol.l #$04,d2 |
This
will rotate the long-word of d2, 04 bits to the left. We’ll pretend that d2 contains 76543210, in
binary that’s:
|
0111 0110 0101 0100 0011
0010 0001 0000 |
Rotating
left 04 bits results in:
|
v 0110 0101 0100 0011 0010
0001 0000 0111 < > > > > >
> > > > > > > > > > > > > > >
> ^ |
Now
d2 contains 65432107.
There
is a small different between the shift and rotation instructions when it comes
to using a data register, for example:
|
rol.l d0,d1 |
Now
as I mentioned before; for shifting, the maximum is the number of bits that can
be shifted over depends on the size you’ve chosen.
- .b for byte gives a maximum shift of
$08 bits.
- .w for word gives a maximum shift of
$10 bits.
- .l for long-word gives a maximum
shift of $1F bits.
For
rotation, there is no maximum (technically speaking):
- Going any higher than $07 bits for
byte will wrap around to $00
- Going any higher than $0F bits for
word will wrap around to $00
- Going any higher than $1F bits for
long-word will wrap around to $00
So,
rotating a byte 09 bits is the same as rotating a byte 01 bit. 0A is the same as 02. And so on...
The
rules for the shift instructions used on memory apply here for the rotation
instructions too; see the previous part for details.
The ROR
Instruction
ROR – ROtate Right (without
extended)
This
instruction will rotate the bits of the destination operand to the right. The number
of times that the bits are rotated is decided by the source operand.
Examples
|
ror.b #$02,d0 |
This
will rotate the byte of d0, 02 bits to the right. We’ll pretend that d0 contains 009F00B2 in
this example, and since it’s a byte,
only the B2 is change. B2 in binary is:
|
1011 0010 |
After
rotating the bits right by 02,
we get:
|
> 1010 1100 v ^ < < < < <
< |
1010
1100 in hex is AC, which is saved back into d0.
Now d0 contains 009F00AC.
Here’s
another example using word:
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ror.w #$04,d2 |
This
will rotate the word of d2, 04 bits to the right. We’ll pretend that d0 contains 0000BA98. Only a word is changed, so BA98 in binary is:
|
1011 1010 1001 1000 |
Rotating
right 04 bits results in:
|
> 1000 1011 1010 1001 v ^ < < < < <
< < < < < < |
Now
d2 contains 00008BA9.
The
rules for the shift instructions used on memory apply here for the rotation
instructions too; see the previous part for details.