关于矩阵键盘扫描程序个别按键卡机

mwh1233 · 发表于 2017-7-30 15:01:15

今天做测试矩阵键盘模块发现第一行的四个按键一按就卡机，，代码如下迫切望大佬解释
#define Key_Hang GPIO_Pin_0 | GPIO_Pin_1 | GPIO_Pin_2 | GPIO_Pin_3
#define Key_Lie  GPIO_Pin_4 | GPIO_Pin_5 | GPIO_Pin_6 | GPIO_Pin_7

u8 Key_Scan(void)
{
u8 hang;
u8 lie;
u8 key,i,final=0xff;

RCC->APB2ENR |= 1<<2;  //使能
GPIOA->CRL &= 0X00000000;
GPIOA->CRL |= 0X88888888;
GPIOA->ODR |= 0XF0<<0;

if(ReKey != 0xf0)
{
delay_ms(10);
if(ReKey != 0xf0)
{
hang = 0x0f&Read_hang();
lie = 0xf0&Read_lie();
key = hang | lie;
for(i=0;i<16;i++)
{
if(key == key_table[i])
{
final = i+1; //此处做了修改
}
}
GPIOA->CRL &= 0X00000000;
    GPIOA->CRL |= 0X88888888;
GPIOA->ODR |= 0XF0<<0;

while(ReKey != 0xf0);
}

}
return final;
}
以下是主函数部分
int main(void)
{
  delay_init();            //延时函数初始化
   uart_init(9600);
   LCD_Init();
   RCC_APB2PeriphClockCmd(RCC_APB2Periph_GPIOA,ENABLE);
   POINT_COLOR=RED;

price=0;
weight=1;
DIS_Init();
LCD_Clear(RED);

while(1)
{

Key_Num=Key_Scan();

POINT_COLOR=RED;
if(Key_Num==255);
else {LCD_Clear(RED);
LCD_ShowxNum(80,40,Key_Num,2,16,0);}

}