HCS12 Assembly Programming - The HCS12/9S12: An Introduction to Software and Hardware Interfacing - page 77

The result is shown in Figure 2.15b.

Original value New value

[A] = 10111110
C = 1

[A] = 11011111
C = 1

Figure 2.15b ■ Execution result of rora

▲

Example 2.23
▼ Write a program to count the number of 0s contained in memory locations $1000,$1001
and save the result at memory location $1005.
Solution: The logical shift-right instruction is available for double accumulator D. We can load
this 16-bit value into D and shift it to the right 16 times or until it becomes 0. The algorithm of
this program is as follows:
Step 1
Initialize the loop count to 16 and the zero count to 0.
Step 2
Place the 16-bit value in D.
Step 3
Shift D to the right one place.
Step 4
If the C flag is 0, increment the zero count by 1.
Step 5
Decrement the loop count by 1.
Step 6
If the loop count is 0, then stop. Otherwise, go to step 3.

The program is as follows:

org $1000

dc.w $2355

zero_cnt rmb 1

lp_cnt rmb 1

org $1500

clr zero_cnt ; initialize the zero count to 0

movb #16,lp_cnt ; initialize the loop count to 16

ldd $1000 ; place the 16-bit number in D

again lsrd

bcs chk_end ; branch if the lsb is 1

inc zero_cnt