用定时器在数码管上依次扫描出1,2,3,4

2019-04-14 20:07发布

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#include //#include #define uchar unsigned char #define uint unsigned int sbit dula=P2^6; sbit wela=P2^7; uchar code tabdu[]={ 0xc0,0xf9,0xa4,0xb0, 0x99,0x92,0x82,0xf8, 0x80,0x90,0x88,0x83, 0xc6,0xa1,0x86,0x8e}; uchar code tabwe[]={ 0xfe,0xfd,0xfb,0xf7}; uchar tt,numdu,numwe; void init(); void main() { init(); //居然加上了VOID 结果我白白调试了将近一个小时气死人了 while(1) { } } void init() { numdu=1; numwe=0; TMOD=0x01; EA=1; TH0=(65536-50000)/256; TL0=(65536-50000)%256; ET0=1; TR0=1; } void time1() interrupt 1 { TH0=(65536-50000)/256; TL0=(65536-50000)%256; tt++; if(tt==10) { tt=0; wela=1; P2=tabwe[numwe]; wela=0; numwe++; if(numwe==4) numwe=0; dula=1; P0=tabdu[numdu]; dula=0; numdu++; if(numdu==16) numdu=1; } }啥也不说了，细节的问题好久才看出来。越是细节越是最最简单的错误越不容易发现

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