怎么让这几个函数同时运行，求分析

ID:840694 · 发表于 2020-11-8 19:23

单片机源程序如下:
#include <reg52.h>
#include <stdio.h>
#include<intrins.h>
#define uint unsigned int
#define uchar unsigned char
sbit LEDR=P1^0;//???
sbit LEDY=P1^1;//???
sbit LEDG=P1^2;//???
sbit DU = P2^6;
sbit WE = P2^7;
uchar  code tabel[]= {0x3F, 0x06, 0x5B, 0x4F, 0x66, 0x6D, 0x7D, 0x07, 0x7F, 0x6F,};
void delay (uint i)
{
uint x,y;
for (x=0;x<i;x++)
for (y=0;y<110;y++);
}

void UARTInit()
{
      SM0 = 0;       SM1 = 1;
      TR1 = 1;
      TMOD |= 0x20;
      TH1 = 0xfd;
      TL1 = 0xfd;
}
uchar k;

void display(uchar i)
{
      uchar  shi, ge;
      shi = i % 100 / 10;       //236 % 100 / 10 = 3
      ge  = i % 10;//236 % 10 =6

      P0 = 0XFF;
      WE = 1;
      P0 = 0XFE;
      WE = 0;

      DU = 1;
      P0 = tabel[shi];//
      DU = 0;
      delay(5);

      //???λ?????
      P0 = 0XFF;//???????
      WE = 1;//??λ???????
      P0 = 0XFD; //1111 1101
      WE = 0;//????λ?????

      DU = 1;//???????????
      P0 = tabel[ge];//
      DU = 0;//??????????
      delay(5);

}
//?????0?????
void timer0Init()
{
      TR0 = 1;       //????????0
      TMOD = 0X01; //???????????1??16λ???????????
      TH0 =(65535-46082 ) / 256;
      TL0 =(65535-46082 ) % 256; //???50ms
}

void main()

{

      UARTInit();
      {
            TI = 1;
            puts("?????");
            while(!TI);
            TI = 0;
            delay(15000);
            TI= 1;
            puts("?????");
            while(!TI);
            TI=0;
            delay(3000);
            TI=1;
            puts("?????");
            delay(10000);
            TI=0;
            }
LEDR=0;

{
      uchar mSec, Sec;//????????????
      Sec=15;
      timer0Init();//?????0?????
      while(1)
      {
            if(TF0 == 1)//?ж???????
            {
                     TF0 = 0;//?????????????λ
                  TH0 =(65535-46082 ) / 256;
                     TL0 = (65535-46082) % 256; //???50ms
                     mSec++;//50ms??
                     if(mSec == 20)
                     {
                              mSec = 0;
                              Sec--;//1?????
                     }
            }
            display(Sec); //????????????
      }
}
delay(12000);
LEDR=1;
LEDR=0;
delay(250);
LEDR=1;
delay(250);
LEDR=0;
delay(250);
LEDR=1;
delay(250);
LEDR=0;
delay(250);
LEDR=1;
delay(250);
LEDR=0;
delay(250);
LEDR=1;
delay(250);
LEDR=0;
delay(250);
LEDR=1;
delay(250);
LEDR=0;
delay(250);
LEDR=1;
delay(250);

LEDY=0;
delay(3000);
LEDY=1;
LEDG=0;
delay(10000);
for(k=0;k<3;k++);
{
LEDG=0;
delay(250);
LEDG=1;
delay(250);
LEDG=0;
delay(250);
LEDG=1;
delay(250);
LEDG=0;
delay(250);
LEDG=1;
delay(250);
LEDG=0;
delay(250);
LEDG=1;
delay(250);
LEDG=0;
delay(250);
LEDG=1;
delay(250);
LEDG=0;
delay(250);
LEDG=1;
delay(250);
}
}

ID:472390 · 发表于 2020-11-9 11:27

程序不可能同时执行的，都是挨个执行。即使加入操作系统也是挨个执行，只是从宏观上看好像是在并行执行而已。单片机推荐使用定时器设置不同的时间标志位，进行任务的分配。比如设置10毫秒的定时器时基，然后在定时器中用一个变量进行加加，当这个变量等于10或者20 的时候，设置一个标志位，在主函数中检测这个标志位即可完成定时器对任务的分配。切记分配的函数执行时间不能超过设置的调度时间，否则调度任务将毫无意义。