楼上正解。



请问，为什么会这样。

难道是将TMR2IE所在的寄存器（8BIT）全部和另外一个的一起做按位于，

可是在以前的编译器中基本所有都是&。

可能这个编译器要求严格一些吧，不要太在意这些，另，写程序最好正规些，可以少走弯路，呵呵

"发现只要T0溢出，以后怎么搞都逛if(TMR2IE&TMR2IF)// .
     {TMR2IE=0; "

because you turned on PEIE, which made TMR2IE possible (which could have been accidentally set) and tmr2 starts to run so you go into it from time to time.

to try it, you can explicitly clear TMR2IE.

another mistake you made is that you never cleared TMR2IF so the isr is stuck processing TMR2 overflow. what you should have done is to clear TMR2IF in the isr, not TMR2IE.

I ran the following code:

========code=========

#include <htc.h>
//#include "Main_heard.h"
#include "gpio.h"

__CONFIG(BORDIS & MCLRDIS & PWRTDIS & WDTDIS & HS & LVPDIS);
__CONFIG(IESODIS & FCMDIS & DEBUGDIS & BORV21);

#define LED_PORT                        PORTC                        //led indicators on portc
#define LED_DDR                                TRISC
#define LED_TMR0                        (1<<0)                        //flip'd if tmr0 interrupt is serviced
#define LED_TMR1                        (1<<1)                        //flip'd if tmr1 interrupt is serviced
#define LED_TMR2                        (1<<2)                        //flip'd if tmr2 interrupt is serviced
void main(void) {

        ANSEL=0x00;                                                                //all pins gpio
        ANSELH=0x00;
        IRCF2=1, IRCF1=1, IRCF0=0;                                //running at 4mhz
        //T0IF=0;
        T0IE=1;//open int  need alawys open  
        PSA=0; PS1=0;  PS0=0;
        T0CS=0;
        TMR0=0;
   
        PEIE=1;
        GIE=1;
       
        IO_CLR(LED_PORT, LED_TMR0 | LED_TMR1 | LED_TMR2);        //led_tmr0/1/2 driven to low
        IO_OUT(LED_DDR, LED_TMR0 | LED_TMR1 | LED_TMR2);        //led_tmr0/1/2 as output

        while(1)
                continue;
}


/****************ISR******************************/
static void interrupt ISR(void)         // Here be interrupt function - the name is unimportant.
{
        if (T0IE&T0IF)// SYSTEMTICK 4.096MS int  need scan led an count or decount  
                {T0IF=0;
                IO_FLP(LED_PORT, LED_TMR0);                //flip led_tmr0
        }

        if (RBIE&RBIF)//partB int need scan key  
                {RBIF=0;
        }
  
        if (TMR1IE&TMR1IF)//ccp INT  read perid
                {TMR1IF=0;
                IO_FLP(LED_PORT, LED_TMR1);                //flip led_tmr1
        }

        if (TMR2IE&TMR2IF)// PWM TIME INT    CHANGE DUTY
                {TMR2IE=0;                                                 //should reset the flag, not the interrupt enabling bit
                IO_FLP(LED_PORT, LED_TMR2);                //flip led_tmr2
        }
}
==========end code=============

it has three indicators, portc.0, portc.1, and portc.2 (on a 16F886), designated as tmr0/1/2 interrupt indicators: when tmr0 interrupt is serviced, LED_TMR0 (portc.0) is flip'd, and so on and so forth.

when I ran the code, the only thing flip'd is portc.0 (=LED_TMR0). that means the only interrupt service being executed is the one related to T0IF, as expected.

和_谐得不好，总有些怪异的问题，坛里有9.8的，bug比较少