c - 使用 MCU-ATMega 1280 进行多中断实时数据记录

Question

我的问题是关于实时数据记录和多中断。 我正在尝试通过 winAVR 对 MCU-ATMega 1280 进行编程，使其从正交编码器(20um/pitch)读取脉冲并将数据存储到闪存(Microchip SST25VF080B，串行 SPI 协议(protocol))中。编码器完成操作后(大约2分钟)，MCU将数据从内存导出到屏幕。我的代码如下。

但是我不知道为什么它不能正确运行。有两种错误:一种错误是某些点突然偏离趋势，另一种错误是尽管编码器运行缓慢但值突然跳跃。跳跃似乎只有在转弯时才会出现。

我认为问题可能仅在于数据存储，因为除了跳跃之外，趋势的发生与我的预期一致。我只是想问我是否像我在程序中所做的那样运行两个 ISR。一个 ISR 运行时是否会被另一个 ISR 干预？根据atmega 1280数据表，似乎当一个ISR发生时，在前一个中断完成其例程后不允许发生其他中断。

    #include <stdlib.h>
    #include <stdio.h>
    #include <avr/interrupt.h>
    #include <util/delay.h>
    #include "USART.h"  // this header is for viewing the data on the computer
    #include "flashmemlib.h"  // this header contains the function to read n 
        //write on the memory

    #define MISO     PB3
    #define MOSI     PB2
    #define SCK      PB1
    #define CS       PB0
    #define HOLD     PB6
    #define WP       PB7
    #define sigA     PD0        //INT0
    #define sigB     PD2        //INT2
    #define LED      PD3


        uint8_t HADD,MADD,LADD, HDATA, LDATA,i; //HADD=high address, MADD-medium address, LADD-low address
        volatile int buffer[8]; //this buffer will store the encoder pulse
        uint32_t address = 0;
        uint16_t DATA16B = 0;

        int main(void)
        {
        INITIALIZE();   //initialize the IO pin, timer CTC mode, SPI and USART protocol 
            for(i=0;i<8;i++)
                buffer[i]=0;

        sei();

        //AAI process- AAI is just one writing mode of the memory 
        AAIInit(address,0);
        while (address < 50)        //just a dummy loop which lasts for 5 secs (appox)
        {
        _delay_ms(100);
        address++;
        }
        AAIDI();//disable AAI process
        cli(); //disable global interrupt
        EIMSK &= ~(1<<INT0);
        TIMSK1 &= ~(1<<OCIE1A);

    //code for reading procedure. i thought this part is unnecessary because i am quite //confident that it works correcly
    return (0);
    }


    ISR(INT0_vect) // this interrupt is mainly for counting the number of encoder's pulses
    { // When an interrupt occurs, we only have to check the level of
         // of pB to determine the direction
        PORTB &= ~(1<<HOLD);
        for(i=0;i<8;i++)
            buffer[i+1]=buffer[i];

         if (PIND & (1<<sigB))
             buffer[0]++;
         else buffer[0]--;
        PORTB |= (1<<HOLD);
    }

    ISR(TIMER0_COMPA_vect)   //after around 1ms, this interrupt is triggered. it is for storing the data into the memory.
    {
        HDATA =(buffer[7]>>8)&0xFF;
        LDATA = buffer[7]&0xFF;
        PORTB &= ~(1<<CS);
        SEND(AD);
        SEND(HDATA);
        SEND(LDATA);
        PORTB |=(1<<CS);
    }

void SEND(volatile uint8_t data)
{
  SPDR = data;                    // Start the transmission
  while (!(SPSR & (1<<SPIF))){}   // Wait the end of the transmission
}
uint8_t  SREAD(void)
{
  uint8_t data;
  SPDR = 0xff;                    // Start the transmission
  while (!(SPSR & (1<<SPIF))){}    // Wait the end of the transmission
    data=SPDR;
    return data;
}

Answer 1

在 INT0 的中断服务程序中，您编写:

    for(i=0;i<8;i++)
        buffer[i+1]=buffer[i];

意味着当i=7时，您正在数组的预定空间之外写入，并且可能会覆盖另一个变量。所以你必须这样做:

    for(i=0;i<7;i++)
        buffer[i+1]=buffer[i];

AVR 将根据 ATmega1280 数据表按照您的描述管理中断。或者，如果您希望允许另一个中断中断 ISR vector ，您需要执行以下操作(示例):

ISR(INT0_vect, ISR_NOBLOCK)
{...
...}