两个中断响应问题

jackding

因为中断引脚用了GPIO0_9、GPIO0_8,所以每次中断触发的时候bank0都响应了，
现在怎么在中断函数中区分是哪个引脚触发的中断？谢谢指导。

Lewis

不能将 GPIO_BANK_0 中断映射到两个中断服务函数，正确的做法是将GPIO_BANK_0中断映射到一个中断服务函数，使能GPIO0_8和GPIO0_9上升或者下降沿中断，在GPIO_BANK_0产生中断后在中断服务函数里查询是GPIO0_8产生中断还是GPIO0_9产生中断。

jackding

Lewis 发表于 2015-12-29 13:37
不能将 GPIO_BANK_0 中断映射到两个中断服务函数，正确的做法是将GPIO_BANK_0中断映射到一个中断服务函数， ...

谢谢您的回复，现在就是这么做的，但是现在不会查询，是GPIO0_8,还是GPIO0_9产生的中断，具体的说就是不知用那个函数进行查询，能具体说一下嘛？

希望缄默

jackding 发表于 2015-12-29 13:40
谢谢您的回复，现在就是这么做的，但是现在不会查询，是GPIO0_8,还是GPIO0_9产生的中断，具体的说就是不 ...

1. /**
   
2. * \brief    This function determines the status of interrupt on a specified
   
3. *           pin.  
   
4. *
   
5. * \param    baseAdd    The memory address of the GPIO instance being used.
   
6. 
   
7. * \param    pinNumber  The serial number of the GPIO pin to be accessed.
   
8. *                      The 144 GPIO pins have serial numbers from 1 to 144.
   
9. 
   
10. * \return   This returns a value which expresses the status of an interrupt
    
11. *           raised over the specified pin.
    
12. *           1> GPIO_INT_NOPEND, if no interrupts are left to be serviced.
    
13. *           2> GPIO_INT_PEND, if the interrupt raised over that pin is yet
    
14. *              to be cleared and serviced.
    
15. *              
    
16. * \note     If an interrupt over a pin is found to be pending, then the
    
17. *           application can call GPIOPinIntClear() to clear the interrupt
    
18. *           status.  
    
19. *           
    
20. *
    
21. */
    
22. unsigned int GPIOPinIntStatus(unsigned int baseAdd, unsigned int pinNumber)
    
23. {
    
24.     unsigned int intStatus = GPIO_INT_NOPEND;
    
25.     unsigned int regNumber = 0;
    
26.     unsigned int pinOffset = 0;
    
27. 
    
28.     /*
    
29.     ** Each register contains settings for each pin of two banks. The 32 bits
    
30.     ** represent 16 pins each from the banks. Thus the register number must be
    
31.     ** calculated based on 32 pins boundary.
    
32.     */
    
33.     regNumber = (pinNumber - 1)/32;
    
34.    
    
35.     /*
    
36.     ** In every register the least significant bits starts with a GPIO number on
    
37.     ** a boundary of 32. Thus the pin offset must be calculated based on 32
    
38.     ** pins boundary. Ex: 'pinNumber' of 1 corresponds to bit 0 in
    
39.     ** 'register_name01'.
    
40.     */
    
41. 
    
42.     pinOffset = (pinNumber - 1) % 32;
    
43. 
    
44.     if(HWREG(baseAdd + GPIO_INTSTAT(regNumber)) & (1 << pinOffset))
    
45.     {
    
46.         intStatus = GPIO_INT_PEND;
    
47.     }
    
48.         
    
49.     return intStatus;
    
50. }

复制代码

jackding

希望缄默 发表于 2015-12-29 16:53

非常感谢