中断端口位冲突？

事实上，原子性和RMW UsLATABIT.LATA1^＝1用于切换。



      以下为原文

    Indeed, atomicity and RMW
 
Use
LATAbits.LATA1 ^= 1

for toggling

如果用限流电阻正确地驱动LED，则端口I/O电压将始终在规范内，并且不会出现RMW问题。RMW问题只在外部电路过载I/O电流规范并导致电压从它应该在的位置被拉出时发生。是。



      以下为原文

    If the LEDs are being driven correctly with a current limit resistor then the port I/O voltages will always be within spec and there will be no RMW issue.
 
RMW issues only occur when external circuitry is overloading the I/O current spec and causing the voltage to be pulled away from where it should be.

谢谢你，Cinzia，这是个骗局。Atomicity才是问题所在。



      以下为原文

    Thank you Cinzia, that did the trick.
Atomicity was the issue.

另外那些引脚设置为数字模式？还是在模拟模式下留下？



      以下为原文

    Additional are those pin set to digital mode? Or left in analog mode?

嗨，Crosland，谢谢你的回答。LED有一个270欧姆串联电阻（当然）。答案来自Cinzia，原子性是我的问题。问候，Willem



      以下为原文

    Hi Crosland,
Thank you for replying.
The LED's have a 270 ohm resistor in series (of course). The solution came from Cinzia, atomicity was my problem.
Regards,
Willem

这是一个经典的读修改写案例（RMW）。写入LAT而不是端口。这是因为写入端口将在内部首先读取实际的PIN逻辑电平，改变要更改的PIN，然后回写端口中的所有位。当设置LED输出时，在实际设置到端口引脚电压达到逻辑1或0电平时需要一些时间。因此，在设置一个端口引脚高之后，它将被读取为低的时间，之后如果端口内的另一个比特被写入，那么端口引脚将返回低电平。这不会发生在LAT位上，这就是它存在的原因。



      以下为原文

    This is a classical Read-Modify-Write case (RMW).
Write to LAT instead of PORT.
 
This happens because writing to port will internally first read the actual pin logical levels, change the pin you intend to change and then write back all of the bits in the port. When setting the LED output it will take some time between the actual setting to when the port pin voltage has reach logical 1 or 0 level. So after setting a port pin high, it will be read as low some time after which will bring the port pin back low if another bit in the port is written within this time.
 
This will not happen for the LAT bits and is why it exists.
 
/Ruben

嗨，一种解决办法是，分别在主和中断处理程序中将每个LED的所需状态保持在单独的私有变量中，并在单向操作中将所需值分配给位，而在具有单独变量的中断例程中类似的。t个SFR地址来反转一个单位似乎有些麻烦。PIC24指令集有切换寄存器中的单位的指令，但是如果涉及两个不同的寄存器地址，则不能使用这个指令，编译器也不可能理解硬件将映射不同的加法器。可以尝试：LATAbits.LATA1^=1；LATAbits.LATA0^=1；看看编译器是否能够将此理解为同一寄存器中的位切换。



      以下为原文

    Hi,
One way to get around, could be to keep the wanted state of each LED in separate private variables in main and in the interrupt handler respectively, and assign the wanted value to the bit in a one-way operation.
static bool onoff;
 
    LATAbits.LATA0 = onoff;
    onoff = ! onoff;
and similar in the interrupt routine with a separate variable.
 
Accessing two different SFR addresses to invert a single bit seem somewhat cumbersome.
The PIC24 instruction set have instruction to Toggle a single bit in a register,
but this cannot be used if there are 2 different register addresses involved,
and compiler is not likely to understand that hardware will map different addresses to the same register.
You may try: LATAbits.LATA1 ^= 1; 
and        LATAbits.LATA0 ^= 1;
to see if compiler may be able to understand this as a bit toggle in the same register.
 
Regards,
   Mysil

不，它总是需要一些时间来升高或降低引脚上的电压。如果输出电流高和/或负载电容高但仍在电流输出规格内，则需要更多的时间。



      以下为原文

   


No, it always takes some time to rise or lower the voltage on a pin. More time if the output current is high and/or the load capacitance is high but still within the current output specs.
 
/Ruben

这不应该是必要的。只需在原始文章中将PORT更改为LAT。问题不在于原子性，因为主代码不查看中断在LAT中更改的位，反之亦然。为PIC24写入单个位是一个原子操作。/ Ruben



      以下为原文

   


That should not be necessary. Just change PORT to LAT in the original post.
The problem is not atomicity since main code doesn't look at the bit that the interrupt changes in LAT and vice versa. Writing to a single bit for PIC24 is an atomic operation.
 
/Ruben

不，它总是需要一些时间来升高或降低引脚上的电压。如果输出电流高和/或负载电容高但仍然在电流输出规格内，则需要更多的时间。



      以下为原文

   


No, it always takes some time to rise or lower the voltage on a pin. More time if the output current is high and/or the load capacitance is high but still within the current output specs.
 
/Ruben


That's the WRMW problem :)
 

谢谢Ruben，这是我的第二个错误，写信给端口而不是锁存器。问题固定了，学到了一些东西；-W）



      以下为原文

    Thanks  Ruben,
 
That was my second mistake, writing to the port instead of the latch.
Problem fixed and something learned ;-)
/W

在ASM中，有一个指令。BTG LATB，γ0。



      以下为原文

    In asm, there is an instruction for that.
 
btg latb,#0
 
;sensor active?
    mov     M_Sensor_Port,w0
    mov     M_Sensor_Pin,w1
    btst.c  [w0],w1
    btsc    M_Flag,#MFLAG_SENSOR_LOW
    btg     SR,#C
    bra     c,hot    ;active

如果C代码是单位XORing，那么我希望它使用BTG指令。在任何一种语言中，它仍然无法寻址PORTx寄存器，并且无法寻址LATx寄存器。



      以下为原文

   
If the C code was XORing with a single bit, then I would expect it to use the BTG instruction.
In either language, it would still fail addressing the PORTx register, and work addressing the LATx register.

M_Sensor_Port包含一个到portx加法器的ptr，并且不会失败，因为trisx是输入的，并且它没有读取锁存器。如果设备是活动的低电平，则使用该代码将马达对中到霍尔效应传感器。，无符号int PIN，BOL ActuvLoW；



      以下为原文

    btw. M_Sensor_Port contains a ptr to a portx addr and will not fail
since trisx is input and it is not reading a latch.

I use that code to center a motor to a hall effect sensor.

btg is used if the device is active low.
Using the zero flag to invert the carry flag.

addsensor   (unsigned int volatile *Port, unsigned int Pin, bool ActiveLow);