我做一个单片机定时器1的中断矩阵按键的实验，但是没有结果显示，求大佬帮忙。

ID:383038 · 发表于 2018-11-9 09:29

#include "reg51.h"

#define GPIO_KEY P1

typedef unsigned int u16;
typedef unsigned char u8;

u8 code smgduan[17]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,
                  0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};

u8 KeyDown(void)
{
u8 KeyValue = 0;

if(GPIO_KEY!=0x0f)
{
      GPIO_KEY=0X0F;
      switch(GPIO_KEY)
      {
         case(0X07): KeyValue=0;break;
         case(0X0b): KeyValue=1;break;
         case(0X0d): KeyValue=2;break;
         case(0X0e): KeyValue=3;break;
      }
      GPIO_KEY = 0xf0;
      switch(GPIO_KEY)
      {
         case(0X70): KeyValue=KeyValue;break;
         case(0Xb0): KeyValue=KeyValue+4;break;
         case(0Xd0): KeyValue=KeyValue+8;break;
         case(0Xe0): KeyValue=KeyValue+12;break;
      }
      while(GPIO_KEY!=0xf0)
      {

      }

}

return KeyValue;
}

void timer1Init(void)
{
TMOD |= 0x10;

TH1 = 0xff;
TL1 = 0xff;

TR1 = 1;
ET1 = 1;
EA = 1;
}

void timer1() interrupt 3 using 3
{
static u16 i = 0;
static u8 key = 0;
TH1 = 0xff;
TL1 = 0xff;
GPIO_KEY = 0x0f;

if (GPIO_KEY != 0x0f)
{
      i++;
      if (i >= 10000)
      {
         key = KeyDown();
         i = 0;
      }
}

P0 =  ~smgduan[key];
}

int mian(void)
{
timer1Init();
while (1) ;
}

ID:89515 · 发表于 2018-11-9 12:31

#include "reg51.h"

#define GPIO_KEY P1

typedef unsigned int u16;
typedef unsigned char u8;

u8 code smgduan[17]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,
                  0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};

u8 KeyDown(void)//注意：只是提供思路
{
   u8 KeyValueH = 0;
u8 KeyValueL = 0;
static u8 state_k；

      GPIO_KEY=0x0f;
   if(GPIO_KEY!=0x0f&&!state_k){state_k=1;return;}//消抖
   if(GPIO_KEY!=0x0f)
   {
      KeyValueH=GPIO_KEY；
      GPIO_KEY = 0xf0;
KeyValueL=GPIO_KEY；
   }
GPIO_KEY=0xff;
while(GPIO_KEY!=0xff) ;//可以照着按下消抖的方法自己改
   return  (KeyValueH |KeyValueL);
}
u8 keyscan()
{
      switch(KeyDown())
   {
            键值处理……
   }
   return 键值；
}
void timer1Init(void)
{
   TMOD |= 0x10;

   TH1 =( 65536-20000)/256;
   TL1 =( 65536-20000)%256;

   TR1 = 1;
   ET1 = 1;
   EA = 1;
}

void timer1() interrupt 3 using 3
{
   static u16 i = 0;
   static u8 key = 0;
TH1 =( 65536-20000)/256;
   TL1 =( 65536-20000)%256;
   GPIO_KEY = 0x0f;

   if (GPIO_KEY != 0x0f)
   {
         key = keyscan()；
   }

   P0 =  ~smgduan[key];
}

int mian(void)
{
   timer1Init();
   while (1) ;
}

ID:213173 · 发表于 2018-11-9 20:34

按键扫描程序里有这条语句while(GPIO_KEY!=0xff) 待松手，不适合放在中断中使用。而且65ms中断时间太长，数码管动态扫描玩不转。

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