Question

本人很菜。利用单片机写了一个电子时钟C语言程序，但无法通过按键来实现设置时间。请各位高手指教。

#include<reg51.h>
#include<intrins.h>
sbit k1=P1^0;
sbit k2=P1^1;
sbit k3=P1^2;
sbit k4=P1^3;
sbit k5=P1^4;
sbit k6=P1^5;
sbit k7=P1^6;
sbit k8=P1^7;
sbit ko1=P3^0;
sbit ko2=P3^1;
sbit ko3=P3^2;
sbit ko4=P3^3;
unsigned char dis1[11]={ 0x28, 0x7E, 0xA2, 0x62, 0x74, 0x61, 0x21, 0x7A, 0x20, 0x60} ;
unsigned char dis2[11]={ 0x28, 0x7E, 0xA2, 0x62, 0x74, 0x61, 0x21, 0x7A, 0x20, 0x60} ;
unsigned char dis3[11]={ 0x28, 0x7E, 0xA2, 0x62, 0x74, 0x61, 0x21, 0x7A, 0x20, 0x60} ;
unsigned char dis4[7]={ 0x28, 0x7E, 0xA2, 0x62, 0x74, 0x61, 0x21} ;
unsigned char dis5[11]={ 0x28, 0x7E, 0xA2, 0x62, 0x74, 0x61, 0x21, 0x7A, 0x20, 0x60} ;
unsigned char dis66[7]={ 0x28, 0x7E, 0xA2, 0x62, 0x74, 0x61, 0x21} ;
unsigned char dis7[11]={ 0x28, 0x7E, 0xA2, 0x62, 0x74, 0x61, 0x21, 0x7A, 0x20, 0x60} ;
unsigned char dis8[3]={ 0x28, 0x7E, 0xA2};
unsigned char h,g,f,e,d,c,b,a,i,j,k,l,m,w;
void Delay(unsigned int DelayTime ) /* 延时子程序 */
{
//if( DelayTime == 0 )return; /* 如果延时为0 则立即返回 */
while( DelayTime != 0)
{
DelayTime--;
} /* 利用减1 的执行时间达到延时目的 */
}
void kw()
{
ko1=1;
ko2=1;
ko3=1;
ko4=1;
if(ko1==0)
{
Delay(5);
while(1)
{
if(ko1!=0)
{
ko1=0;
break;
}
}
if(w!=0)
{
if(ko2==0)
{
Delay(5);
while(1)
{
if(ko2!=0)
{
ko2=0;
break;
}
}
}
if(ko3==0)
{
Delay(5);
while(1)
{
if(ko3!=0)
{
ko3=0;
break;
}
}
}
if(ko4==0)
{
Delay(5);
while(1)
{
if(ko4!=0)
{
ko4=0;
break;
}
}
}
}
}
}
程序过长贴不上来。自己在想办法了

Answer 1

可按键调节的C语言时钟程序 （郭天祥版）
#include<reg52.h>
#define uchar unsigned char
#define uint unsigned int
uchar count,key1num;
char shi,fen,miao;
uchar code hh[]=" 2010-5-4 Tue";
uchar code yy[]=" 00-00-00";
sbit ledrs=P2^4;
sbit ledrw=P2^5;
sbit leden=P2^6;
sbit key1=P1^0;
sbit key2=P1^1;
sbit key3=P1^2;
sbit key4=P1^4;

void delay(uchar x)
{
int i,j;
for(i=0;i<x;i++)
for(j=0;j<120;j++);
}
void write_com(uchar com)
{
ledrs=0;
P0=com;
leden=1;
delay(5);
leden=0;
delay(5);
}
void write_date(uchar date)
{
ledrs=1;
P0=date;
leden=1;
delay(5);
leden=0;
delay(5);
}
void init()
{
uchar i;
ledrw=0;
write_com(0x38);
write_com(0x0c);
write_com(0x06);
write_com(0x80);
write_com(0x01);

for(i=0;i<15;i++)
{
write_date(hh[i]);
}
write_com(0x80+0x40);
for(i=0;i<12;i++)
{
write_date(yy[i]);
}
TMOD=0x01;
TH0=(65536-50000)/256;
TL0=(65536-50000)%256;
EA=1;
ET0=1;
TR0=1;
}
void write_sfm(uchar add,uchar date)
{
uchar shi,ge;
shi=date/10;
ge=date%10;
write_com(0x80+0x40+add);
write_date(0x30+shi);
write_date(0x30+ge);

}
void keyscan()
{
key4=0;
if(key1==0)
{
delay(5);
if(key1==0)
{
key1num++;
while(key1==0);
if(key1num==1)
{
TR0=0;
write_com(0x80+0x40+10);
write_com(0x0f);
}

if(key1num==2)
{
write_com(0x80+0x40+7);
}
if(key1num==3)
{
write_com(0x80+0x40+4);
}
if(key1num==4)
{
key1num=0;
write_com(0x0c);
TR0=1;
}

}

}
if(key1num!=0)
{
if(key2==0)
{
delay(5);
if(key2==0)
{
while(key2==0);
if(key1num==1)
{
miao++;
if(miao==60)
miao=0;
write_sfm(10,miao);
write_com(0x80+0x40+10);

}
if(key1num==2)
{
fen++;
if(fen==60)
fen=0;
write_sfm(7,fen);
write_com(0x80+0x40+7);
}
if(key1num==3)
{
shi++;
if(shi==24)
shi=0;
write_sfm(4,shi);
write_com(0x80+0x40+4);
}
}
}
if(key3==0)
{
delay(5);
if(key3==0)
{
while(key3==0);
if(key1num==1)
{
miao--;
if(miao==-1)
miao=59;
write_sfm(10,miao);
write_com(0x80+0x40+10);
}
if(key1num==2)
{
fen--;
if(fen==-1)
fen=59;
write_sfm(7,fen);
write_com(0x80+0x40+7);
}
if(key1num==3)
{
shi--;
if(shi==-1)
shi=23;
write_sfm(4,shi);
write_com(0x80+0x40+4);
}
}
}
}
}

void main()
{
init();

while(1)
{
keyscan();
if(count==20)
{
count=0;
miao++;
if(miao==60)
{
miao=0;
fen++;
if(fen==60)
{
fen=0;
shi++;
if(shi==24)
{
shi=0;
}
write_sfm(4,shi);
}
write_sfm(7,fen);
}
write_sfm(10,miao);
}
}

}
void timer0() interrupt 1
{

TH0=(65536-50000)/256;
TL0=(65536-50000)%256;
count++;

}

Answer 2

说实话你编的我确实看不懂，还么学到那么高深的了，
不过用scanf可以实现输入
scanf("%d%d%d%d",&ch1,&ch2,&ch3,&ch4);
不知道能不能对你有些启发，可以试试

追问

谢谢谢你的回答
scanf("%d%d%d%d",&ch1,&ch2,&ch3,&ch4);
是针对键盘输入输出的吧，我是想直接扫描P3口的方式，

追答

scanf只能输入
printf是输出的
大致的格式是
printf（"%d%d:%d%d\n",ch1,ch2,ch3,ch4）；

追问

这问题先放一边了，
很感谢你的帮助

Answer 3

没有注释，好难了解你的意图的，写程序时注释一下，别人也容易明白。

Answer 4

我不知道你在干嘛...
//注释很重要..