求修改程序，大致上就是將八位數(shù)碼管改成這樣

ID:797839 · 發(fā)表于 2020-7-5 16:10

#include<reg51.h>
#define uchar unsigned char

sbit S1=P3^4;
sbit dula=P2^6;
sbit wela=P2^7;
uchar k;

void delay_ms(uchar z)
{
uchar i,j;
for(i=z;i>0;i--)
for(j=110;j>0;j--);
}

void keyscan()
{

S1=1;
if(S1==0)
{
delay_ms(10);
if(S1==0)
{
k++;
if(k==5)
{
k=1;
}
while(!S1);
}

}
}

void display()
{
P0=0x36;
dula=1;
dula=0;
if(k==1)
{

P0=0xe7;
wela=1;
wela=0;
delay_ms(10);
}
if(k==2)
{

P0=0xc3;
wela=1;
wela=0;
delay_ms(10);
}
if(k==3)
{

P0=0x81;
wela=1;
wela=0;
delay_ms(10);
}
if(k==4)
{

P0=0x00;
wela=1;
wela=0;
delay_ms(10);
}
}

void main()
{
uchar i,count;
P0=0x00;
dula=1;
dula=0;
P0=0xff;
wela=1;
wela=0;
while(1)
{
keyscan();
display();
}
}

ID:332444 · 發(fā)表于 2020-7-5 17:29

也就是一個數(shù)碼管顯示||這樣的符號，送給數(shù)碼管的信號是固定的，只要控制點亮的個數(shù)就可以了？還是要顯示對稱的，若顯示對稱的則一個數(shù)碼管顯示右|一個顯示左|這樣的話送給數(shù)碼管的值就有變化了，不過也就是2個值而已。

ID:568565 · 發(fā)表于 2020-7-5 20:09

這樣的效果?

ID:213173 · 發(fā)表于 2020-7-5 21:45

#include<reg51.h>
#define uchar unsigned char
sbit S1=P3^4;
sbit dula=P2^6;
sbit wela=P2^7;
uchar code table[9][8]={
0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x06,0x30,0x00,0x00,0x00,
0x00,0x00,0x00,0x36,0x36,0x00,0x00,0x00,
0x00,0x00,0x06,0x36,0x36,0x30,0x00,0x00,
0x00,0x00,0x36,0x36,0x36,0x36,0x00,0x00,
0x00,0x06,0x36,0x36,0x36,0x36,0x30,0x00,
0x00,0x36,0x36,0x36,0x36,0x36,0x36,0x00,
0x06,0x36,0x36,0x36,0x36,0x36,0x36,0x30,
0x36,0x36,0x36,0x36,0x36,0x36,0x36,0x36};
uchar code weia[8]={0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f};
uchar i,j,k;
void delay_ms(uchar z)
{
uchar i,j;
for(i=z;i>0;i--)
for(j=110;j>0;j--);
}
void keyscan()
{
if(S1==0)
{
if(++j>10)
j=11;
if(j==10)
{
k++;
if(k>8)
k=0;
}
}
else j=0;
}
void display()
{
P0=0x00;dula=1;dula=0;
P0=weia[i];wela=1;wela=0;
P0=table[k][i];dula=1;dula=0;
i++;i%=8;
}
void main()
{
while(1)
{
keyscan();
display();
delay_ms(1);
}
}

復制代碼

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