（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; } } } }