【OK210试用体验】按键控制LED点亮

经过一个星期对交叉编译环境的学习，已经懂得如何编写简单的裸机程序。
今天分享的程序是通过按键控制LED，按键按下LED亮，松开LED灭。

首先看原理图中GPIO的描述：

   配置按键的GPIO管脚为输入，并且设置为上拉，然后通过判断管脚的电压值来控制LED。
   下面是程序：
   #define MP0_4CON (*(volatile unsigned long *)0xE0200340)
#define MP0_4DAT (*(volatile unsigned long *)0xE0200344)
#define MP0_4PUD  (*(volatile unsigned long *)0xE0200348)
#define GPH0CON (*(volatile unsigned long *)0xE0200C00)
#define GPH0DAT    (*(volatile unsigned long *)0xE0200C04)
#define GPH0PUD (*(volatile unsigned long *)0xE0200C08)
static void Delay(int count)
{
      volatile int i, j = 0;
      volatile static int loop = 800000000/10000;
      for(;count > 0;count--)
         for(i=0;i < loop; i++) { j++; }
}

int main(void)
{
int i=0;
MP0_4CON |= (MP0_4CON & ~0xFFFF0000) | 0x11110000; //设置高4为管脚为输出方式
GPH0CON |= (GPH0CON &~0xFFFF0000) | 0X00000000;//设置高4为管脚为输入方式
GPH0PUD |=0xAA00;

MP0_4DAT |= 0xff;
GPH0DAT |= 0xff;

while(1)
{
      if((GPH0DAT & 0x80) == 0)
         MP0_4DAT &= ~( 1<< 4);
      else
         MP0_4DAT |= 1<< 4;

      if((GPH0DAT & 0x40) == 0)
         MP0_4DAT &= ~( 1<< 5);
      else
         MP0_4DAT |= 1<< 5;

   if((GPH0DAT & 0x20) == 0)
         MP0_4DAT &= ~( 1<< 6);
   else
         MP0_4DAT |= 1<< 6;

   if((GPH0DAT & 0x10) == 0)
         MP0_4DAT &= ~( 1<< 7);
   else
         MP0_4DAT |= 1<< 7;
}

return 0;
}

   演示：