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求一个51单片机矩阵加法方式,目前矩阵可以弄出来,但加减法不会

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ID:541708 发表于 2019-5-18 21:52 | 显示全部楼层 |阅读模式
1黑币
#include<reg51.h>
#define anjian P1
unsigned int keyshu;
unsigned int a;
unsigned int shumaguan[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};
void delay(unsigned int i)
{
   while(i--);
}
void xianshi(unsigned int m)
{
unsigned int sw,gw;
sw=m/10;
gw=m%10;
P2=0X04;
P0=shumaguan[sw];
delay(1000);
P2=0X00;
P0=shumaguan[gw];
delay(1000);
}
void KEY()
{
P1=0X0F;
if(P1!=0x0f)
{
  delay(1000);
  if(P1!=0x0f)   
  {
   switch(anjian)
   {
    case(0x07): keyshu=0;break;
    case(0x0b): keyshu=1;break;   
    case(0x0d): keyshu=2;break;
    case(0x0e): keyshu=3;break;
   }
   P1=0XF0;
   switch(anjian)
   {
    case(0x70): keyshu=keyshu;break;
    case(0xB0): keyshu=keyshu+4;break;
    case(0xD0): keyshu=keyshu+8;break;
    case(0xE0): keyshu=keyshu+12;break;
   }
   while(a<50&&anjian!=0x0f)
   {
    delay(1000);
    a++;
   }
  }
}
}
void main()
{ while(1)
  {
  KEY();
  xianshi(keyshu);
  }
}#include<reg51.h>
#define anjian P1
unsigned int keyshu;
unsigned int a;
unsigned int shumaguan[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};
void delay(unsigned int i)
{
   while(i--);
}
void xianshi(unsigned int m)
{
unsigned int sw,gw;
sw=m/10;
gw=m%10;
P2=0X04;
P0=shumaguan[sw];
delay(1000);
P2=0X00;
P0=shumaguan[gw];
delay(1000);
}
void KEY()
{
P1=0X0F;
if(P1!=0x0f)
{
  delay(1000);
  if(P1!=0x0f)   
  {
   switch(anjian)
   {
    case(0x07): keyshu=0;break;
    case(0x0b): keyshu=1;break;   
    case(0x0d): keyshu=2;break;
    case(0x0e): keyshu=3;break;
   }
   P1=0XF0;
   switch(anjian)
   {
    case(0x70): keyshu=keyshu;break;
    case(0xB0): keyshu=keyshu+4;break;
    case(0xD0): keyshu=keyshu+8;break;
    case(0xE0): keyshu=keyshu+12;break;
   }
   while(a<50&&anjian!=0x0f)
   {
    delay(1000);
    a++;
   }
  }
}
}
void main()
{ while(1)
  {
  KEY();
  xianshi(keyshu);
  }
}


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ID:541708 发表于 2019-5-18 21:53 | 显示全部楼层
矩阵可以弄出来,但加减法不会弄
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