I can t kneel down and ask the master for help! MCU program explained

The first floor is right, C51 has several control functions for the LCD screen. Help people: the interface of the LCD screen has RS signal (register selection, H data, L instruction) connection, R W read and write signal, LCDEN read and write allowed, D0 D7 data.

void delay(uint z)

A time-delay subroutine used for timing coordination when sending data or instructions to an LCD.

void write com (uchar com) sends instruction byte functions to lcd.

void write date (uchar date) is a function that sends data bytes to the lcd.

void write_nyr(uchar add,uchar date)

Display the 10 digits and single digits of the Date on the first line of the LCD screen.

void write_sfm(uchar add,uchar date)

Display the 10 digits and single digits of the date on the second line of the LCD screen.

You can't afford to kneel for a long time with such a **? You can't afford to kneel for a long time to take the information, and it's already done.

1.It's a cycle, and it's an endless cycle. He is often used in the microcontroller, the function is that as long as the microcontroller is powered on, it makes the program continue to execute the content in the endless loop!

2.This depends on the hardware circuit of your single-chip microcomputer, and you had better understand the hardware circuit before programming. Usually the common end of the LED is gated with a triode!

It seems that your LED is connected to the common anode, that is, the 8 LED positive electrodes are connected to the emitter of the triode, the collector is connected to the VCC, and the base is connected to the P1 port. P1=0, the transistor is forward conducted, VCC is added to the LED, if the other end of the LED is low, so that the LED is positively biased, the LED is on. Think about it

Execute the statement in while(1) non-stop, but when an interrupt occurs, the CPU has to execute the interrupt first, and then return to the while to continue the execution, which is quite a wait for the interrupt! In other words, if there is an interrupt, the CPU should immediately execute the interrupt, and if there is none, it will always execute the one in while(1).

1.SP=40H, A=30H before the CALL command is executed, but in the SUBR, the SP is pressed to the stack for 2009H, and the 2009H is sent back to the PC after the RET command is executed, causing the MCU to jump to the 0x2009 to continue executing the instruction, but this position is not the instruction start address specified in the program, so after running, A=30H, Sp=42H

It's an 8-bit register, right? 0+1+2+..100 gets 16 bits of data, a can't fit it, the program writes like this, the high byte of the result is in r3, and the low byte is in r2:

mov r0,#101 ;0 100 is 101 digits mov r1,

mov a,#

mov b,#

loop:clr c

mov a,r2

add a,r1

mov r2,a

mov r3,a

addc a,#

mov r3,a

inc r1

djnz r0,loop

3.I don't know the order, then I think 50h and 60h are high two digits, and 51h and 61h are low two digits:

clr cmov a,51h

add a,61h

damov 41h,a

mov a,50h

addc a,60h

da mov 40h,amov

(1) After the following procedure is executed, sp=?a=?b=?

org 2000h

mov sp, #40h

mov a , #30h a=30hlcall subr

add a, #10h a=40hmov b, a b=40hl1： sjmp l1

subr: mov dptr,#2009hpush dpl 09h

push dph 20h

RET So, A=40h, B=40h, Sp=2009h