stm32 位带操作
2014-08-12 19:19
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1.
![](http://img.blog.csdn.net/20140812191817005?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQveWluc2h1bmp1bjEyMw==/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast)
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2.
程序代码中定义的位带操作为:
#define BITBAND(addr, bitnum) ((addr & 0xF0000000)+0x2000000+((addr &0xFFFFF)<<5)+(bitnum<<2))
======================================================================================
其实1中的计算方法我们是很好理解的,下面我们看下2是怎么和1对应上的:
![](http://img.blog.csdn.net/20140812191639109?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQveWluc2h1bmp1bjEyMw==/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast)
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#define BITBAND(addr, bitnum) ((addr & 0xF0000000)+0x2000000+((addr &0xFFFFF)<<5)+(bitnum<<2))
#define GPIOA_ODR_Addr (GPIOA_BASE+12) //0x4001080C
#define PAout(n) BIT_ADDR(GPIOA_ODR_Addr,n) //输出
以GPIOA.5为例:
PAout(5) = BIT_ADDR(0x4001080C,5)
= (0x4001080C & 0xF0000000) + 0x2000000 + ((0x4001080C &0xFFFFF)<<5) + (5<<2)
对应40000000 对应2000000 (对应地址中低20位)<<5
对应5<<2
对应1中的42000000 对应1中的(A-40000000)*32
对应1中的n*4
为什么说绿色部分也是对应的呢,因为位带区除了最高位之外,就只有低20位在变化了!!!
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2.
程序代码中定义的位带操作为:
#define BITBAND(addr, bitnum) ((addr & 0xF0000000)+0x2000000+((addr &0xFFFFF)<<5)+(bitnum<<2))
======================================================================================
其实1中的计算方法我们是很好理解的,下面我们看下2是怎么和1对应上的:
![](file:///C:/Documents%20and%20Settings/Administrator/Local%20Settings/Application%20Data/YNote/Data/yinshunjun@126.com/0ffb35a635c341c987c65139b6bf00f1/clipboard.png)
#define BITBAND(addr, bitnum) ((addr & 0xF0000000)+0x2000000+((addr &0xFFFFF)<<5)+(bitnum<<2))
#define GPIOA_ODR_Addr (GPIOA_BASE+12) //0x4001080C
#define PAout(n) BIT_ADDR(GPIOA_ODR_Addr,n) //输出
以GPIOA.5为例:
PAout(5) = BIT_ADDR(0x4001080C,5)
= (0x4001080C & 0xF0000000) + 0x2000000 + ((0x4001080C &0xFFFFF)<<5) + (5<<2)
对应40000000 对应2000000 (对应地址中低20位)<<5
对应5<<2
对应1中的42000000 对应1中的(A-40000000)*32
对应1中的n*4
为什么说绿色部分也是对应的呢,因为位带区除了最高位之外,就只有低20位在变化了!!!
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