The logic operation of the accumulator A for the microcontroller:
CLR A; 0. The value of the value is 0, the single-cycle single-byte instruction is the same as the MOV A, and the effect.
CPL A; reflect the value in A
RL A; logic to the value of A
RLC A; Logic left by plusing the value in A
RR A; logically right of the value in A
RRC A; Logic movement to the value of the value plus the value
Swap A; high value, low value, low 4 digits.
Example: (a) = 73h, execute CPL A, this:
73H is chemical as 01110011,
It is a 10001100 in opposing position, which is 8CH.
RL A is sent to the value of the value in (a) to 0th, and the 0th is 1 bit, and push it according to the class.
Example: The value in A is 68h, executes RL A. 68H is chemically binary 01101000, moving by the above figure. 01101000 is 11010000, that is, D0H.
RLC A is shifted by the inlet bits (c) on (a).
Example: The value of the value is 68h, and the value is 1, then execute RLC A
After 1 01101000, the result is 0 11010001, that is, the value of the C-entertel bit becomes 0, and (a) becomes D1H.
RR A and RRC A don't talk more, please refer to the two routines above.
SWAP A is exchanged for high, low 4 digits in A.
Example: (a) = 39h, after SWAP A, the value in A is 93h. How is it just right? Because this is a 16-based number, each 16-in number represents 4 bits. Note that if this is: (a) = 39, there is no H, after SWAP A, may not be (a) = 93. To make it a binary and then calculated: 39 is a binary 10111, that is, 0001, 0111 high 4 bits are 0001, low 4 bits are 0111, and the exchange is 0110001 after exchange, that is, 71h, ie 113.
Exercise, known (a) = 39h, perform the results of each step after performing the following single-chip command
CPL a
RL A
CLR C
RRC A
Setb C
RLC A
SWAP A
Through the previous study, we have mastered a considerable single-chip degree, everyone may be a little tired of these boring microcontroller instructions, let us easily make an experiment.
Experiment 5:
ORG 0000H
LJMP START
ORG 30H
Start:
Mov sp, # 5fh
Mov A, # 80h
LOOP:
MOV P1, A
RL A
LCALL DELAY
Ljmp loop
DELAY:
Mov r7, # 255
D1: MOV R6, # 255
D2: NOP
NOP
NOP
NOP
Djnz R6, D2
Djnz R7, D1
RET
End
Let us write the program into the film, put it into the experiment board, look at the phenomenon.
I saw a phenomenon of dark point flow, let's analyze it.
Instead ORG 0000h, LJMP Start, ORG 30H, etc. We will later analyze. Starting from START, Mov SP, # 5FH, this is the initialization stack, there is no tight in this program, but we slowly start contacting regular programming, I will slowly give you habits.
Mov A, # 80h, send 80h this number to a. What do you do? I don't know, look down.
MOV P1, A. Send the value in A to the P1 port. At this time, the value of A is 80h, so it is 80h, so the value of the P1 port is 80h, that is, 10000000B, through the previous analysis, we should know that the LED connected in P1.7 is not bright. And other LEDs are bright, so a "dark point" is formed. Continue to see, RL A, RL A is left to move the value in A, count, what is the result of the shift? Yes, it is 01h, that is, 00000001B, which should be connected in the LED on P1.0, and other bright, from the phenomenon, "dark point" flows back. Then call the delay program, this is very familiar, let this "dark point" "dark" for a while. Then be turned to the LOOP (LJMP loop). Please calculate it, which light should be not lit. . . . . By the way, it should be connected to the light on P1.1. In this way, it is cycled to form a "dark point flow".
problem:
How to achieve highlights?
How to change the direction of the flow?
Answer:
1. Change the initial value in A to 7FH.
2, change RL A to RR a. Read more
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