在lcd1602上显示出按键的按值，就是显示出手机号码，下面是我在keil5上自己敲的代码，现在问题是只能显示出一位数字，有没有人教一下?
#include <reg51.h>
#include <intrins.h>
#include <string.h>
#define uchar unsigned char
#define uint unsigned int
sbit RS=P1^7;
sbit RW=P1^6;
sbit E=P1^5;
uchar code str1[]={0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39};
uchar code str2[]={0x4e,0x65,0x77,0xfe,0x54,0x65,0x6c,0x4e,0x75,0x6d,0x2e,0x2e,0x2e,0xfe,0xfe,0xfe};
uchar code str3[]={0x49,0x6e,0x70,0x75,0x74,0xfe,0x54,0x65,0x6c,0x4e,0x75,0x6d,0x2e,0x2e,0x2e,0xfe};
uchar keyval=11;
void delayms(uint n)
{
uchar i;
while(n--)
for(i=0;i<113;i++);
}
void check_busy(void)
{
uchar dt;
do
{
dt=0xff;
E=0;
RS=0;
RW=1;
E=1;
dt=P0;
}while(dt&0x80);
E=0;
}
void write_command(uchar com)
{
check_busy();
E=0;
RS=0;
RW=0;
P0=com;
E=1;
_nop_( );
E=0;
delayms(1);
}
void write_data(uchar dat)
{
check_busy();
E=0;
RS=1;
RW=0;
P0=dat;
E=1;
_nop_();
E=0;
delayms(1);
}
void LCD_initial(void)
{
write_command(0x38);
_nop_( );
write_command(0x0C);
_nop_( );
write_command(0x06);
_nop_( );
write_command(0x01);
delayms(2);
}
void write_string(uchar addr,uchar str[])
{
uchar i;
write_command(0x80+addr);
for(i=0;i<strlen(str);i++)
{
write_data(str[i]);
}
}
uchar keyscan()
{
uint row,rol;
P2=0xf0;
if((P2&0xf0)!=0xf0)
{
delayms(20);
if((P2&0xf0)!=0xf0)
{
switch(P2&0xf0)
{
case 0x70:rol=0;break;
case 0xb0:rol=1;break;
case 0xd0:rol=2;break;
case 0xe0:rol=3;break;
}
P2=0x0f;
if((P2&0x0f)!=0x0f)
{
switch(P2&0x0f)
{
case 0x07:row=0;break;
case 0x0b:row=1;break;
case 0x0d:row=2;break;
case 0x0e:row=3;break;
}
while((P2&0x0f)!=0x0f);
keyval=rol*4+row;
}
}
}
return keyval;
}
void main()
{
LCD_initial();
write_string(0x00,str3);
while(1)
{
keyscan();
if(keyval<=9)
{
write_command(0xc0);
write_data(str1[keyval]);
}
if(keyval==10)
{
write_string(0x00,str2);
write_command(0xc0);
write_data(0xfe);
}
}
}
参考GPT和自己的思路：你可以尝试在keyscan函数中添加一个判断，如果当前已经按下了一个数字键，再次按下数字键时，可以将当前的数字与新的数字拼接起来，组成一个手机号码。例如：
void main()
{
LCD_initial();
write_string(0x00,str3);
uchar phone[12]=""; // 定义一个字符串，存储手机号码
while(1)
{
keyscan();
if(keyval<=9)
{
write_command(0xc0);
write_data(str1[keyval]);
if(strlen(phone)<11) // 手机号码最多只能11位
{
phone[strlen(phone)] = str1[keyval]; // 将当前按下的数字加入手机号码中
}
}
if(keyval==10)
{
write_string(0x00,str2);
write_command(0xc0);
write_data(0xfe);
memset(phone, 0, sizeof(phone)); // 清空手机号码
}
if(strlen(phone)==11) // 如果手机号码已经11位，显示手机号码
{
write_string(0x00, phone);
}
}
}
参考GPT和自己的思路：你的代码中只显示一位数字是因为在keyscan()函数中，当按键按下后keyval的值被记录，但没有重置，所以当下一次循环中keyval的值还是上一次按下的值，导致只显示一位数字。解决方法是在处理完keyval后要将其重置为一个不可能的值，比如keyval=11，即可避免此问题。另外，你在显示手机号码时，只有10个数字键和1个特殊键（#键），如果按下的键值超过这个范围，代码中并没有对keyval进行处理，这也需要注意。
有些错误不一定会显示，如内存溢出这种就不会显示，这些还需要自己去了解，通过上网搜索等方式寻找解决方法
#include <reg51.h>
#include <math.h>
sbit RS=P2^4;
sbit RW=P2^5;
sbit E=P2^6;
sbit BUZ=P1^0;
sbit IN=P2^0;
sbit DE=P2^1;
sbit BA=P2^2;
unsigned char code mod1[]=" workmod1:clock ";
unsigned char code mod2[]=" workmod2:setup ";
unsigned char code mod3[]=" workmod3:alarm ";
unsigned char code mod4[]=" workmod4:timer ";
unsigned int code timenumber[]={0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39};//0,1,2,3,4,5,6,7,8,9
unsigned int o,p;
unsigned char bs,i;
unsigned int c1,c2,c3,c4,c5,c6,z,i3;//mod1
unsigned int s;//mod2
long int a;
long int kk,k2;//mod4
long int t,k,i1,i2;//timer
void delay(int t)//50us
{
unsigned int x;
for(;t>0;t--)
for(x=19;x>0;x--);
}
void writecom(unsigned char com)
{
E=0;
RS=0;
RW=0;
P0=com;
delay(10);
E=1;
delay(20);
E=0;
}
void writedate(unsigned char dat)
{
E=0;
RS=1;
RW=0;
P0=dat;
delay(10);
E=1;
delay(20);
E=0;
}
void innit()
{
delay(30);
writecom(0x38);
delay(10);
writecom(0x38);
delay(10);
writecom(0x38);
writecom(0x38);
writecom(0x08);
writecom(0x01);
writecom(0x06);
writecom(0x0c);
}
void workmod1()
{
innit();
s=0;
while(bs==0)
{
writecom(0x80);
for(i=0;i<16;i++)
{
writedate(mod1[i]);
}
writecom(0x80+0x40);
if(s==0)writedate('O'),writedate('F'),writedate('F');
if(s==1)writedate('O'),writedate('N'),writedate(' ');
c1=i1%10,c2=i1/10,c3=i2%60%10,c4=i2%60/10,c5=i2/60%24%10,c6=i2/60%24/10;
writecom(0x80+0x45);
writedate(timenumber[c6]);writedate(timenumber[c5]);
writedate(':');
writedate(timenumber[c4]);writedate(timenumber[c3]);
writedate(':');
writedate(timenumber[c2]);writedate(timenumber[c1]);
i3=i2/60;
if(i3!=z)//strike
{
o=1;
z=i3;
}
if(s==1&&a==i2)o=1;//alarm
}
}
void workmod2()
{
innit();
s=0;
writecom(0x80);
for(i=0;i<16;i++)
{
writedate(mod2[i]);
}
while(bs==1)
{
i1=0;
writecom(0x80+0x4a);
writedate(':');writedate('0');writedate('0');
writecom(0x80+0x40);
switch(s)
{
case 0:
{
writedate('M');
if(IN==0)
{
delay(500);
i2++;
delay(500);
}
else if(DE==0)
{
delay(500);
i2--;
delay(500);
}
}break;
case 1:
{
writedate('H');
if(IN==0)
{
delay(500);
i2=i2+60;
delay(500);
}
else if(DE==0)
{
delay(500);
i2=i2-60;
delay(500);
}
}break;
}
c1=i1%10,c2=i1/10,c3=i2%60%10,c4=i2%60/10,c5=i2/60%24%10,c6=i2/60%24/10;
writecom(0x80+0x45);
writedate(timenumber[c6]);writedate(timenumber[c5]);
writedate(':');
writedate(timenumber[c4]);writedate(timenumber[c3]);
}
}
void workmod3()
{
innit();
s=0;
writecom(0x80);
for(i=0;i<16;i++)
{
writedate(mod3[i]);
}
while(bs==2)
{
if(a==1440)a=0;
writecom(0x80+0x40);
switch(s)
{
case 0:
{
writedate('M');
if(IN==0)
{
delay(500);
a++;
delay(500);
}
else if(DE==0)
{
delay(500);
a--;
delay(500);
}
}break;
case 1:
{
writedate('H');
if(IN==0)
{
delay(500);
a=a+60;
delay(500);
}
else if(DE==0)
{
delay(500);
a=a+60;
delay(500);
}
}break;
}
writecom(0x80+0x45);
c3=a%60%10,c4=a%60/10,c5=a/60%10,c6=a/60/10;
writedate(timenumber[c6]);writedate(timenumber[c5]);
writedate(':');
writedate(timenumber[c4]);writedate(timenumber[c3]);
}
}
void workmod4()
{
innit();
kk=0;
writecom(0x80);
for(i=0;i<16;i++)
{
writedate(mod4[i]);
}
while(bs==3)
{
k2=0;
if(s==0)
{
writecom(0x80+0x45);
writedate('R');
writedate('E');
writedate('A');
writedate('D');
writedate('Y');
}
else if(s==1)
{
writecom(0x80+0x45);
for(i=0;i<5;i++)
writedate(' ');
while(kk!=60&&bs==3&&s==1)
{
kk=k2/20;
writecom(0x80+0x47);
writedate(timenumber[(60-kk)/10]);writedate(timenumber[(60-kk)%10]);
}
if(kk==60)o=1,s=0,kk=0;
}
}
}
void main()
{
innit();
TMOD=0x01;TH0=(65536-50000)/256;TL0=(65536-50000)%256;
EA=1;EX0=1;EX1=1;
IT0=1;IT1=1;
ET0=1;TR0=1;
k=0;bs=0;s=0;z=0;k2=0;i1=0;i2=0;kk=0;a=0;o=0;p=0;
while(1)
{
switch(bs)
{
case 0:workmod1();break;
case 1:workmod2();break;
case 2:workmod3();break;
case 3:workmod4();break;
}
}
}
void int0 () interrupt 0//buttion:affirm
{
s++;s=s%2;
BUZ=1;
}
void timer0 () interrupt 1
{
for(t=0;t<20;t++)
{
TH0=(65536-50000)/256;
TL0=(65536-50000)%256;
}
k++;
if(k==20) i1++,k=0;
if(i1==60) i2++,i1=0;
if(i2==1440) i2=0;
k2++;//only mod4
if(o)BUZ=0,p++;
if(p==40)BUZ=1,o=0,p=0;
}
void int2 () interrupt 2//buttion:set up
{
kk=0;BUZ=1;
bs++;
bs=bs%4;
}