RB0外部中断怎么配置

为项目做一些特定的逻辑：逻辑有多少？脉搏有多宽？这是你的开关吗？如果（IntNo.In0＆I＆Ap＆OncTun.Itn0IE）/ /您不需要IncNIT.It0IE，如果它从来没有禁用{//清除中断旗语。In0IF＝0；如果（ItNCO2BITS，ITEDG0＝＝1）/ /中断在下降沿触发。Time.InCON2BIT.IndigG0＝1；//在时间关键性的东西之后执行此操作，并且为项目& GT做一些特定的逻辑；/ /如果需要很长的时间，则不要在中断中执行。//启用Int1外部中断/ /内插。如果没有启用，你就不会在这里。}



      以下为原文

     
How Much Logic?
How Wide is the Pulse? Is it do befor you switch?


if((INTCONbits.INT0IF && INTCONbits.INT0IE)) //you do not need INTCONbits.INT0IE if it is never disabled

    {
        
        // Clear the interrupt flag
        INTCONbits.INT0IF = 0;

         if(INTCON2bits.INTEDG0 == 1)
            //interrupt is triggered on falling edge.
            INTCON2bits.INTEDG0=0;
        else
            //interrupt is triggered on faising edge.
            INTCON2bits.INTEDG0=1;
 
      // do this after the time critical stuff, 
        
      // If it takes a long time, then do not do in in the interrupt.
        
        //Enable INT1 External Interrupt
        //INTCONbits.INT0IE=1; // why are you enabling this? if it was not enabled you would not be here.
    }

在你试图做的事情中有一个竞赛条件：如果在PIC到达ISR代码之前边缘再次切换，ISR将ItIdg设置为错误的值。您可以通过使用中断优先级来缩小窗口，而不是测试ItnIF，在ISR结束时清除ItnIF，并执行ItnC2BIT.IndigG0=！ItnC2BIT.IntIG0而不是“IF”语句（或任何变体给您一个BTG指令）。



      以下为原文

    There's a race condition in what you're trying to do:  if the edge toggles again before the PIC gets to your ISR code, the ISR sets INTEDG to the wrong value.  You can narrow the window by using interrupt priorities, not testing INT0IF, clearing INT0IF at the end of the ISR, and do INTCON2bits.INTEDG0 = !INTCON2bits.INTEDG0 instead of the "if" statement (or whatever variant gives you a BTG instruction).

非常感谢你的回应，我将使用3个标志。非常小的指令。I ActNoBix.It00IE＝1在上电并在功能结束时再次启用（IncNord.Itn0IE＝1），因为配置从上升到下降。不管这个声明和逻辑对我的项目是什么，我也有同样的问题。



      以下为原文

    Thank you very much for your response.
 
The logic i am going to use setting 3 flags. very minimal instructions. 
I am INTCONbits.INT0IE=1  at power up and enabling again (INTCONbits.INT0IE=1 ) at the end of function as the configuration changed from rising to falling. irrespective of this statement and logic specific to my project also I have the same problem.

除非你禁用它，否则你不需要重新启用密码。脉宽有多大？按你的时钟速率有多少个指令？



      以下为原文

    You do not need to re-enable INTCONbits.INT0IE unless you disable it.
How Wide is the Pulse?  How many instructions is that at your Clock rate?

那不是必要的。永远不会改变自己的价值。你缺少的一件事是改变Itdig0可以设置In0f，因此你应该在改变ItIdG0之后清除It0f。



      以下为原文

   
 
That isn't necessary.  INT0IE will never change value on its own.  One thing you are missing is that changing INTEDG0 may set INT0IF, so you should clear INT0IF after changing INTEDG0.

非常感谢你不生脉。具体的设备功率信号是连接RB0。其他设备的功率损耗和上电应该很快被检测。我的时钟配置为32mhz。在这些事件上递增现有变量并设置标志。在启动时，外部中断被初始化为TRISBbit.TRISB0=1;INTCON2bits.INTEDG0=1;INTCONbits.INT0IF=0;INTCONbits.INT0IE=1;ANCON1bits.ANSEL10=0;ISR代码类似于//如果设置了Timer标志并且启用了中断DIF（（InCONTITS.IT0IF）{ItNCO2BIT.ITCDG0=！INTCON2bits.INTEDG0Flag=1;//清除中断标志INTCONbits.INT0IF=0;}检测上升沿但不检测下降沿。尝试给出建议但仍面临相同问题。



      以下为原文

    Thank you very much for your 
not generating pulse. To be specific power signal of device is connected RB0.
Power loss and power up of another device should be detected very quickly.
My clock is configured for 32mhz. On these events increment existing variable and setting flag will be done. 
 
At power up external interrupt is initialized as 
TRISBbits.TRISB0=1;
INTCON2bits.INTEDG0=1;
INTCONbits.INT0IF=0;
INTCONbits.INT0IE=1;
ANCON1bits.ANSEL10=0;
 
The ISR code is similar to
// If Timer flag is set & Interrupt is enabled
if((INTCONbits.INT0IF )
{
INTCON2bits.INTEDG0 =!INTCON2bits.INTEDG0
Flag =1; 
// Clear the interrupt flag
INTCONbits.INT0IF = 0;
}
Detecting the rising edge but not falling edge.
Tried with given suggestions but still facing same problem.

我认为您想要的是二进制补码，而不是逻辑非运算符“！”你是使用模拟器还是实际硬件？可能在过渡波形上有噪声。



      以下为原文

    I think you want ~INTCON2bits.INTEDG0 for the bitwise complement, rather than the logical NOT operator "!". Are you using the simulator, or actual hardware? Perhaps there is noise on the transition waveform.

为什么？ItNCO2BITIDENGDG0是一位。



      以下为原文

   
 
Why?  INTCON2bits.INTEDG0 is one bit.

^＝1最好，最后



      以下为原文

    ^=1 is best, in the end

是的，这通常会生成更小的代码。但是没有功能上的差异。



      以下为原文

   
 
Yes, that typically generates smaller code.  But there's no functional difference.