ARM9点灯实验

[cang_rong]

友善之臂的ARM9板子做实验发现：

/* key_led.c */

#define GPBCON      (*(volatile unsigned long *)0x56000010)
#define GPBDAT      (*(volatile unsigned long *)0x56000014)

#define GPGCON      (*(volatile unsigned long *)0x56000060)
#define GPGDAT      (*(volatile unsigned long *)0x56000064)

/*
 * LED1~4,GPB5/GPB6/GPB7/GPB8
 */
#define GPB5_out        (1 << (5 * 2))
#define GPB6_out        (1 << (6 * 2))
#define GPB7_out        (1 << (7 * 2))
#define GPB8_out        (1 << (8 * 2))


/*
 * KEY1~4, GPG0/GPG3/GPG5/GPG6
 */
#define GPG0_in     (0 << (0 * 2))
#define GPG3_in     (0 << (3 * 2))
#define GPG5_in     (0 << (5 * 2))
#define GPG6_in     (0 << (6 * 2))

int main()
{
        unsigned long dwDat;

        GPBCON = GPB5_out | GPB6_out | GPB7_out | GPB8_out;

        GPGCON = GPG0_in & GPG3_in & GPG5_in & GPG6_in;
        while(1)
        {
            dwDat = GPGDAT;             

	    /* 对应位同1做与运算，低电平LED亮 */
            if (dwDat & (1<<0)) 
                GPBDAT |= (1 << 5);       // LED1 turn off
            else
                GPBDAT &= ~(1 << 5);      // LED1 turn on

            if (dwDat & (1<<3))
                GPBDAT |= ( 1 << 6);      // LED2 turn off
            else
                GPBDAT &= ~(1 << 6);      // LED2 turn on

            if (dwDat & (1<<5))
                GPBDAT |= (1 << 7);       // LED3 turn off
            else
                GPBDAT &= ~(1 << 7);      // LED3 turn on

            if (dwDat & (1 << 6))
                GPBDAT |= (1 << 8);		// LED4 turn off
            else
                GPBDAT &= ~(1 << 8);      // LED4 turn on
        }

    return 0;
}

这段代码可以很方便准确地实现，按键1、2、3、4对应LED灯1、2、3、4。上述代码详实，直白，我将其中main函数的while循环内容改为如

下：
GPBDAT = ( ((GPGDAT & 0x01 ) << 5)
| ((GPGDAT & 0x08 ) << 6)
| ((GPGDAT & 0x32 ) << 7)
| ((GPGDAT & 0x64 ) << 8));

这段代码显得简洁，但是达不到预期效果，只有按键1对应LED灯1有反应，其余皆为，而且上电后LED2、3是常亮。

不知为何？

[犇犇犇程序猿]

1<<3= 8 , 1<<5 =32 ,1<<6 =64. 你加个16进制，不错能行么？

[Tchely.Pan]

GPBDAT = ( ((GPGDAT & 0x01 ) << 5)
| ((GPGDAT & 0x08 ) << 6)
| ((GPGDAT & 0x32 ) << 7)
| ((GPGDAT & 0x64 ) << 8));

应该为：
GPBDAT = ( ((GPGDAT & 1 ) << 5)
| ((GPGDAT & 8 ) << 6)
| ((GPGDAT & 32 ) << 7)
| ((GPGDAT & 64 ) << 8));

1楼正解

[lengxujun5000]

这样真的能看到效果吗?
中间没有任何延时操作,可能的结果是,如果某个灯一开始是亮的,很可能一直长亮,因为变化太快,根本看不到变化.

[cang_rong]

GPBDAT = ( ((GPGDAT & 0x01 ) << 5)
| ((GPGDAT & 0x08 ) << 6)
| ((GPGDAT & 0x32 ) << 7)
| ((GPGDAT & 0x64 ) << 8));

应该为：
GPBDAT = ( ((GPGDAT & 1 ) << 5)
| ((GPGDAT & 8 ) << 6)
| ((GPGDAT & 32 ) << 7)
| ((GPGDAT & 64 ) << 8));

实验成功，我总是犯一些低级错误。

[cang_rong]

GPBDAT = ( ((GPGDAT & 0x01 ) << 5)
| ((GPGDAT & 0x08 ) << 6)
| ((GPGDAT & 0x32 ) << 7)
| ((GPGDAT & 0x64 ) << 8));

应该为：
GPBDAT = ( ((GPGDAT & 1 ) << 5)
| ((GPGDAT & 8 ) << 6)
| ((GPGDAT & 32 ) << 7)
| ((GPGDAT & 64 ) << 8));

实验成功，我总是犯一些低级错误。