為什么數(shù)碼管顯示數(shù)字時(shí)會(huì)跳動(dòng)，只想知道原因

ID:105333 · 發(fā)表于 2016-2-14 11:34

#include<reg52.h>

sbit ADDR0 = P1^0;
sbit ADDR1 = P1^1;
sbit ADDR2 = P1^2;
sbit ADDR3 = P1^3;
sbit ENLED = P1^4;

unsigned char code ledchar[] ={0xC0, 0xF9, 0xA4, 0xB0, 0x99, 0x92, 0x82, 0xF8,
                           0x80, 0x90, 0x88, 0x83, 0xC6, 0xA1, 0x86, 0x8E};

unsigned char ledbuff[] = {0xff,0xff,0xff,0xff,0xff,0xff};

void main()
{
unsigned int cnt = 0;
unsigned long sec = 0;
unsigned char i = 0;

ADDR3 = 1;
ENLED = 0;
TMOD = 0x01;
TH0 = 0xfc;
TL0 = 0x67;
TR0 = 1;
while(1)
{
if(TF0 == 1)
{
TF0 = 0;
TH0 = 0xfc;
TL0 = 0x00;
cnt++;
if(cnt == 1000)
{
cnt = 0;
sec++;
ledbuff[0] = ledchar[sec%10];
ledbuff[1] = ledchar[sec/10%10];
ledbuff[2] = ledchar[sec/100%10];
ledbuff[3] = ledchar[sec/1000%10];
ledbuff[4] = ledchar[sec/10000%10];
ledbuff[5] = ledchar[sec/100000%10];
}
      P0 = 0xff;
   switch(i)
   {
      case 0: ADDR2=0; ADDR1=0; ADDR0=0; i++; P0=ledbuff[0];break;
case 1: ADDR2=0; ADDR1=0; ADDR0=1; i++; P0=ledbuff[1];break;
case 2: ADDR2=0; ADDR1=1; ADDR0=0; i++; P0=ledbuff[2];break;
case 3: ADDR2=0; ADDR1=1; ADDR0=1; i++; P0=ledbuff[3];break;
case 4: ADDR2=1; ADDR1=0; ADDR0=0; i++; P0=ledbuff[4];break;
case 5: ADDR2=1; ADDR1=0; ADDR0=1; i=0; P0=ledbuff[5];break;
default:break;    //會(huì)有跳動(dòng)這種情況
   }
}
}
}

ID:72781 · 發(fā)表于 2016-2-14 19:24

加了break語(yǔ)句，每次case選擇后，i++，相應(yīng)位數(shù)碼管顯示就退出了，退出后需要在下一次定時(shí)時(shí)間到，即TF0==0才再次case選擇下一個(gè)數(shù)碼管顯示，這個(gè)定時(shí)是1ms，然后總體一個(gè)數(shù)碼管就會(huì)是亮1ms，滅5ms（這5ms是其他數(shù)碼管亮的時(shí)間），可能是這點(diǎn)導(dǎo)致閃爍。