单数码管循环怎么显示0到9？
答案:1 悬赏:50 手机版
解决时间 2021-01-06 11:23
- 提问者网友：相思似海深
- 2021-01-06 07:07
单数码管循环怎么显示0到9？
最佳答案
- 五星知识达人网友：思契十里
- 2021-01-06 08:29
方法如下:
#include "reg51.h"
#define uchar unsigned char
#define uint unsigned int
uchar code table_numberP[] = {0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f};
uchar code table[] = {0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};
uint n;
uint ge;
void main()
{
ge = 0;
TMOD = 0x01;
TH0 = (65536-45872)/256;
TL0 = (65536-45872)%256;
EA = 1;
ET0 = 1;
TR0 = 1;
while(1);
}
void T0_time()interrupt 1
{
TH0 = (65536-45872)/256;
TL0 = (65536-45872)%256;
n++;
if(n%20 == 0)
{
if(ge<=9){
P1 = table_numberP[0];
P0 = table[ge];
ge++;
if(ge == 10){
ge = 0;
}
}
}
}
主函数中的P2 = LED_CODE[led];led++;可以删去，没有任何的意义
，因为定时和循环显示都在中断函数里面完成了，主函数只是起了一个初始化的作用，led++影响LED_CODE[led]的值，它使LED_CODE[10] = { 0xC0,0xF9,0xA4,0xB0,0x99,0x92,0x82,0xF8,0x80,0x90 }数组中的10个值依次赋给P2，依次显示0--9的数字，即当led = 0时，P2接受0xC0,对应数码显示0，当led = 1时，P2接受LED_CODE[1]=0xF9,对应显示1，依次类推，从而显示0--9。
#include "reg51.h"
#define uchar unsigned char
#define uint unsigned int
uchar code table_numberP[] = {0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f};
uchar code table[] = {0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};
uint n;
uint ge;
void main()
{
ge = 0;
TMOD = 0x01;
TH0 = (65536-45872)/256;
TL0 = (65536-45872)%256;
EA = 1;
ET0 = 1;
TR0 = 1;
while(1);
}
void T0_time()interrupt 1
{
TH0 = (65536-45872)/256;
TL0 = (65536-45872)%256;
n++;
if(n%20 == 0)
{
if(ge<=9){
P1 = table_numberP[0];
P0 = table[ge];
ge++;
if(ge == 10){
ge = 0;
}
}
}
}
主函数中的P2 = LED_CODE[led];led++;可以删去，没有任何的意义
，因为定时和循环显示都在中断函数里面完成了，主函数只是起了一个初始化的作用，led++影响LED_CODE[led]的值，它使LED_CODE[10] = { 0xC0,0xF9,0xA4,0xB0,0x99,0x92,0x82,0xF8,0x80,0x90 }数组中的10个值依次赋给P2，依次显示0--9的数字，即当led = 0时，P2接受0xC0,对应数码显示0，当led = 1时，P2接受LED_CODE[1]=0xF9,对应显示1，依次类推，从而显示0--9。
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