点阵式LED“0－9”数字显示程序+仿真（C语言源码系统仿真实例）.rar

#include <reg51.h> unsigned char code tab[]={0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f}; unsigned char code digittab[10][8]={{0x00,0x00,0x3e,0x41,0x41,0x41,0x3e,0x00}, //0 {0x00,0x00,0x00,0x00,0x21,0x7f,0x01,0x00}, //1 {0x00,0x00,0x27,0x45,0x45,0x45,0x39,0x00}, //2 {0x00,0x00,0x22,0x49,0x49,0x49,0x36,0x00}, //3 {0x00,0x00,0x0c,0x14,0x24,0x7f,0x04,0x00}, //4 {0x00,0x00,0x72,0x51,0x51,0x51,0x4e,0x00}, //5 {0x00,0x00,0x3e,0x49,0x49,0x49,0x26,0x00}, //6 {0x00,0x00,0x40,0x40,0x40,0x4f,0x70,0x00}, //7 {0x00,0x00,0x36,0x49,0x49,0x49,0x36,0x00}, //8 {0x00,0x00,0x32,0x49,0x49,0x49,0x3e,0x00} //9 }; unsigned int timecount; unsigned char cnta; unsigned char cntb; void main(void) { TMOD=0x01; TH0=(65536-3000)/256; TL0=(65536-3000)%6; TR0=1; ET0=1; EA=1; while(1) {; } } void t0(void) interrupt 1 using 0 { TH0=(65536-3000)/256; TL0=(65536-3000)%6; P3=tab[cnta]; P1=digittab[cntb][cnta]; cnta++; if(cnta==8) { cnta=0; } timecount++; if(timecount==333) { timecount=0; cntb++; if(cntb==10) { cntb=0; } } }