(新手求助!)51单片机
以下这段代码中程序执行到break后是怎么继续进行的?如何保证switch中每一条语句执行1ms。望解答,谢谢了!
#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;
ENLED=0;
ADDR3=1;
TMOD=0x01;
TH0=0xFC;
TL0=0x67;
TR0=1;
while(1)
{
if(TF0==1)
{
TF0=0;
TH0=0xFC;
TL0=0x67;
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];
}
switch(i)
{
case 0: ADDR2=0;ADDR1=0;ADDR0=0;i++;P0=LedBuff[0]; ENLED=0;break;
case 1: ADDR2=0;ADDR1=0;ADDR0=1;i++;P0=LedBuff[1]; ENLED=0;break;
case 2: ADDR2=0;ADDR1=1;ADDR0=0;i++;P0=LedBuff[2]; ENLED=0;break;
case 3: ADDR2=0;ADDR1=1;ADDR0=1;i++;P0=LedBuff[3]; ENLED=0;break;
case 4: ADDR2=1;ADDR1=0;ADDR0=0;i++;P0=LedBuff[4]; ENLED=0;break;
case 5: ADDR2=1;ADDR1=0;ADDR0=1;i=0;P0=LedBuff[5]; ENLED=0;break;
default:break;
}
}
}
}