这是一个51hei网友测试通过的程序、电路很简单就不画了的段选是P0口位选是P1口
#include <reg52.h> typedef unsigned char uint8; typedef unsigned int uint16; typedef unsigned long uint32; code uint8 number[]={0xc0,0xf9,0xa4,0xb0, 0x99,0x92,0x82,0xf8,0x80,0x90,0x88,0x83， 0xa7,0xa1,0x86,0x8e}; //数字0到F uint16 counter = 0; //定义初始化 uint8 a[6]; //六个数码管 sbit ENLED = P1^4; sbit A0 = P1^0; sbit ADDR1 = P1^1; sbit ADDR2 = P1^2; sbit ADDR3 = P1^3; void timer1_led() { TMOD |= 0x10; //寄存器状态M0位置1，或运算 TMOD &= 0xDF; TH1 = 0xFC; //12*(65535-x)/11059200 =1s = 0x67; TR1 = 1; //定时器置1打开，计数器开始计时 } void int_init() { ET1 = 1; // 打开定时器1溢出中断使能置1 EA = 1; //总阀门打开 } refresh_led() { static uint8 j = 0; //静态变量刷新 switch(j) { case 0:ADDR0 = 0;ADDR1 = 0;ADDR2 = 0;j++;P0 = number[a[0]];break; case 1:ADDR0 = 1;ADDR1 = 0;ADDR2 = 0;j++;P0 = number[a[1]];break; case 2:ADDR0 = 0;ADDR1 = 1;ADDR2 = 0;j++;P0 = number[a[2]];break; case 3:ADDR0 = 1;ADDR1 = 1;ADDR2 = 0;j++;P0 = number[a[3]];break; case 4:ADDR0 = 0;ADDR1 = 0;ADDR2 = 1;j++;P0 = number[a[4]];break; case 5:ADDR0 = 1;ADDR1 = 0;ADDR2 = 1;j = 0;P0 = number[a[5]];break; default :break; } } main() { ENLED = 0;ADDR3 = 1; //138译码器使能 timer1_led(); //定时初始化 int_init(); while(1); } void interrupt_timer1() interrupt 3 //中断函数 { static uint32 sec=0; //定义变量 TH1 = 0xFC; TL1 = 0x67; counter++; if(1000==counter) { sec++; counter=0; a[0] = sec%10; //个位数取余 a[1] = sec/10%10; //十位取余 a[2] = sec/100%10; //百位取余 a[3] = sec/1000%10; //............... a[4] = sec/10000%10; a[5] = sec/100000%10; } refresh_led(); }
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