89c52，不用锁存器的数码管，如何编出动态显示的程序

ID:334421 · 发表于 2018-5-20 20:08

#include<reg52.h>
#define uchar unsigned char
#define uint unsigned int
sbit wela1=P2^0;
sbit wela2=P2^1;
sbit wela3=P2^2;
sbit wela4=P2^3;
sbit wela5=P2^4;
sbit wela6=P2^5;
uchar code table[]={0xc0,0xf9,0xa4,0xb0,
                  0x99,0x92,0x82,0xf8,
0x80,0x90,0x88,0x83,
0xc6,0xa1,0x86,0x8e};
void delay(uint);
void main()
{
while(1)
{
wela1=0;
     P0=table[0];
     delay(5);
      wela1=1;

wela2=0;
     P0=table[1];
     delay(5);
     wela2=1;

wela3=0;
     P0=table[2];
     delay(5);
     wela3=1;

wela4=0;
     P0=table[3];
     delay(5);
     wela4=1;

wela5=0;
     P0=table[4];
     delay(5);
     wela5=1;

wela6=0;
     P0=table[5];
     delay(5);
     wela6=1;
}
}
void delay(uint xms)
{
uint i,j;
for(i=xms;i>0;i--)
for(j=110;j>0;j--);
}

ID:330198 · 发表于 2018-5-22 09:45

你现在的程序不是常量嘛，用一个变量tab[n]，n变化就能动态显示了

ID:296634 · 发表于 2018-5-22 10:23

#include<reg51.h> #define SEG P0 unsigned char SMGD[]={0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x83,0xf8,0x80,0x98}; //共阳极段显0~9 unsigned char SMGW[]={0x01,0x02,0x04,0x08}; //位选1~4 void delay(unsigned int i) //延时函数 { while(i--); } void main(void) //主函数 { unsigned char Temp; while(1) { switch (Temp%4) { case 0:P0 = SMGD[Temp%10];break; case 1:P0 = SMGD[Temp%10];break; case 2:P0 = SMGD[Temp%10];break; case 3:P0 = SMGD[Temp%10];break; } P2 =SMGW[Temp%4]; Temp++; delay(50000); } }

ID:137653 · 发表于 2018-5-22 11:53

显示第一个字，然后清除，然后显示第二个，如此循环就行

ID:213173 · 发表于 2018-5-22 20:45

使用锁存主要是为了达到I/O口复用的目的，节省I/O口。如果I/O口够用当然可以不用锁存器。示例如下：

#include <AT89X51.H>
#define uint unsigned int
#define uchar unsigned char
uchar code table[]={
0xc0,0xf9,0xa4,0xb0,
0x99,0x92,0x82,0xf8,
0x80,0x90,0x88,0x83,
0xc6,0xa1,0x86,0x8e};
uchar code table1[]={
0x01,0x02,0x04,0x08,
0x10,0x20,0x40,0x80};
void delay(uint k)
{
uint i,j;
for(i=k;i>0;i--)
for(j=114;j>0;j--);
}
void main()
{
uchar i=0;
while(1)
{
for(i=0;i<8;i++)
{
P0=0xff;
P2=table1;
P0=table;
delay(1);
}
}
}

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