关闭变量自动初始化

这个线程HTTP://www. McCHIP.COM/FUMMS/M798029ASPX可能比您要求的覆盖更多。



      以下为原文

    This thread  http://www.microchip.com/forums/m798029.aspx  might cover more than you requested

就I/O端口而言，I/O端口寄存器（大多数，如果不是全部SFRs）在第一次启动时总是出现在芯片的数据表指定的状态中。这不是编译器正在做的事情，而是芯片的行为。作为程序员，在启动时，将I/O端口初始化为所需的配置。我通常只有一个IntItPosits（）函数，在主叫（）中，我首先调用它来初始化所有的TIS和LAT寄存器。当然，如果我弄错了，还有一些方法来改变I/O端口是如何上机的，请大家纠正我的微笑：



      以下为原文

    As far as I/O ports go, I/O port registers (Most, if not all SFRs for that matter) on first boot will always come up in the state specified by the datasheet of the chip.  That isn't something the compiler is doing, it's the behavior of the chip.  It's on you as the programmer to, on startup, initialize the I/O ports to your desired configuration.  I usually just have an Init_Ports() function that I call first thing in main() to initialize all the TRIS and LAT registers.
 
Of course, if I'm mistaken and there's some way to change how I/O ports come up on power up someone please correct me Smile:
 

AdjixAtEdTyTyx（（持久）”到一个变量定义，以保持编译器在重置后初始化变量。



      以下为原文

    Add __attribute__ ((persistent)) to a variables definition to keep the compiler from initializing the variable after a reset.

你没有错，它们出现在数据表中。



      以下为原文

   
You are not wrong , they come up as listed in the Data sheet.
 

正如其他人所说的，它不是初始化端口，硬件是这样做的。它将所有的全局变量和静态变量都清除为零，因为这是C标准规定的。你的主要问题是什么？如果是因为你有一些变量必须在重置中生存（而不是电源失效），那么正如你所说的，你可以使用“持久化”限定符来阻止编译器初始化相关变量。如果你需要变量在一个功率循环中生存，那么你的工作就是保存和检索。它们来自非易失性存储器。



      以下为原文

   
As others have stated, it is NOT initialising the ports, the hardware does that.
It is clearing all global and static variables to zero, because that is what the C standard specifies it must do.
What are your major problems?
If it is because you have some variables that must survive a reset (but not a power fail), then as stated, you can use the "persistent" qualifier to stop the compiler from initialising the relevant variables.
If you need the variables to survive a power cycle, then it's your job to save and retrieve them from non volatile storage.

如果你希望编译器不初始化一些变量，就使用其他人已经说过的持久属性。如果这与Bootloader /应用程序相关，那么您可能也需要地址属性，并在引导加载程序和应用程序中定义这些变量。如果您知道您在做什么，并且确实不希望编译器初始化任何数据对象，那么可以转到Project属性。-gt；xc16ld，不选中“init data章节”。在这种情况下，如果您有，那么FO将不会被初始化为123。



      以下为原文

   
If you want the compiler to not initialize some variables, use the persistent attribute as others have already said. If this is related to bootloader/application, it is likely you'll need the address attribute as well, and define these variables in both the bootloader and the application.
 
If you know what you doing and really don't want the compiler to initialize any data objects, you can go to Project Properties ->xc16-ld, uncheck "Init data sections". In this case, if you have static int foo = 123; then foo won't be initialized to 123.

谢谢大家。我从来不知道全局必须在C规范下初始化。我认为它们就像普通变量一样，占据了内存位置的值。同样，我知道IO端口在上电时自动清除，但我认为它们是持久的。其他重置。显然，我错了。添加持久力可以帮助我克服我的主要问题。无论如何，讽刺的是，我正在努力追寻我自己的尾巴。我在代码中有一个未知的bug，它偶尔会导致看门狗触发。我们试图找到它，但没有运气。我曾试图将调试信息存储在一些全局信息中，以便给我一个暗示，可能会被卡住，但它永远不会起作用。与此相联系的另一个问题是，微控制器正在执行系统保持辊，而我需要它来保持状态而不是加电。清除所有IO的默认操作会导致所有设备关闭，并且由于各种原因超出了我的控制，所以它不能正常重启。而现在，我希望能找到什么原因导致看门狗在正常操作下触发。后者的操作意味着它几乎毫无意义，因此，我虽然我不会问。不过，假设IO端口初始化为0是有意义的，因为看门狗超时是一个失败的信号，因此假设IO是正确的，这不是一个好的举动。



      以下为原文

    Thanks all.
 
I never knew that globals had to be initialised under the C specifications I thought they were just like normal variables an took on the value left in the memory location.
 
Likewise I knew that The IO ports were automatically cleared on power up, but I thought that they were persistant over other resets. Evidently I'm wrong.
 
Adding persistant should help me get over my major problem at any rate.
 
Ironically I'm trying to chase my own tail a little bit. I've got an unknown bug in the code which sporadically causes the watchdog to trigger. We've tried to find it; but with no luck. I'd attempted to store debug information in some globals to give me a hint as to what might be getting stuck but it never worked. The other problem, linked off of this is that the microcontroller is performing a system houskeeping roll and I need it to retain state on a reset other than power up. The default action of clearing all the IO causes everything to shutdown and for various reasons beyond my control it doesn't restart properly.
 
Whilst I can now, hopefully, find what's causing the watchdog to trigger under normal operation. The latter operation means that it's almost pointless having it enabled hence I though't I'd ask. I suppose however it does make sense for the IO ports to initialise back to 0 as a watchdog timeout is a signal of a failure and thus assuming the IO is correct isn't a good move.
 
Thanks again
 
Russell

一些I/O端口的行为可能不同于@ POR RESP。一些其他的重置（如果有的话）。大多数端口在任何类型的重置时都会有相同的行为，因为在某些重置期间，端口不会被卡住输出/激活。这样的行为可能会让有价值的资产“被炸毁”。另一个豁免将是“重置原因”寄存器（可用的）。启动代码通常清除整个RAM（包括堆栈和堆区域）-仅用声明“持久”变量的区域豁免。



      以下为原文

    Some of the I/O ports may behave differently @ POR resp. some other reset (documented as well if so). Most ports will behave the same way at any kind of reset as you won't e.g. have an port got stuck to output/active during some reset. Such a behavior might get valuable assets "blown". Another exemption would be the "reset reason" register (where available).
The startup code usually clears the whole RAM (including stack and heap areas) - with the only exemption of the area with the variables declared "persistent".

同意。讽刺的是，在我的情况下，把IO变成0的违约导致了有价值的资产被吹走。当一个重置导致电源不完整地被关闭到整个系统，直到PIC回来，代码将电源重新打开。我接受在这种情况下没有对或错，下次我只需要在系统中放置一些锁存器来执行状态存储。手指一旦划过，发现它永远不会触发任何东西，这一切都将是哑巴。罗素。P.S.喜欢签名。多么真实。



      以下为原文

    Agreed.
 
The ironic thing is that in my case the default of turning the IO to 0 is causing valuable assets to get blown.... as a reset causes the power to be switched off ungracegully to the entire system until the pic comes back up and the code powers everything back on.
 
I accept that there is no right or wrong in this situation, and next time I'll just have to put some latches in the system to perform the state storage.
 
Fingers crossed once the fault is found it will never trigger any way and all this will be mute.
 
Russell.
 
P.s. love the signature. how very true.

I/OS重置为输入。我不确定，但假定EVTL。内部拉起/下拉电阻器也可能被禁用。留给你的“选项”使用外部电阻器来定义复位状态行为。



      以下为原文

    The I/Os reset to input. I'm not sure but assume that evtl. internal pullup/pulldown resistors might be disabled as well. Leaves you with the "option" to use external resistors to define the reset state behavior.

即使在复位后LAT寄存器恰好为0，TIS寄存器默认为输入，所以PIC不会将任何驱动器驱动到0。



      以下为原文

    Even if the LAT registers happen to be 0 after a reset, the TRIS registers default to inputs, so nothing will be driven to 0 by the PIC.

现在看到了第二部分…外部锁存器可能会产生“新”的问题，如果没有正确初始化在上电：根据确切的模型，他们可以启动“初始化随机”（初始化状态取决于微小的电荷失衡。有时看起来好像在记忆最后一个状态之前关闭。一种并非完全错误的印象。



      以下为原文

    Just now seen the 2nd part...
External latches might create "new" issues if not initialized properly during power-up: depending on the exact model they may start up "initialized randomly".
(Initialization state depending on tiny charge imbalances. Sometimes looks as if they were remembering the last state prior shutoff. An impression that is not completely wrong.)
 

嗨，大家好！你真是命中注定…我没有考虑到IO返回输入的短暂周期是什么使所有的事情变得混乱。我有下拉电阻器，部分地设置了电源状态，他们当然会把所有的东西都关上。我现在有一些简单的事情想办法让我理解。数据表指示LAT和端口寄存器初始化为XXXX，所以看起来您的关于“维护”状态的评论是正确的，这就是我所想到的发生或放弃的短暂的时间返回到输入。我需要仔细关注他们的手术，但这不是世界末日，我很高兴能适应。IRC有几个具有全局设置或复位输入，可以与RC一起使用来设置上电状态。再次感谢罗素。



      以下为原文

    Hi All.
 
You've hit the nail on the head.... I hadn't taken into account the brief period of the IO returning to inputs is what's messing everything up.
 
I have pull down resistors, in part to set the power up state, and they are of course turning everthing off. I have a few ideas for simple things to try tomorrow now to get me over this. The datasheet indicates that LAT and PORT registers initialise to XXXX so looks like your comment about "maintaing" the state is correct, and it's what I thought happened give or take forgetting about the brief period of them returning to inputs.
 
Yes agreed that latches have their own downsides and I'd need to pay careful attention to their operation but tht's not the end of the world and something I'm perfectly happy accomodating. IIRC there are several with global set or reset inputs which can be used with an RC to set the power up state.
 
Again Thanks.
 
Russell

如果你在硬件中产生清晰的信号，只需要三个IO引脚来驱动它。可以级联到任何长度以获得更多的输出。如果你想直接处理更高的电压或电流（如果开路漏是可接受的），则简单地通过SPI或BIT-Bangin。TPIC6B595来驱动。



      以下为原文

    74HC595.
Only needs three IO pins to drive it if you're generating the clear signal in hardware.
Can be cascaded to any length to get more outputs.
Simple to drive via SPI or bit-banging.
 
TPIC6B595 if you want to handle higher voltage or current directly (if open drain is acceptable).
 
 

你没有错，它们出现在数据表中。在某些PIC设备上，一些引脚的复位状态可以通过配置位来配置。



      以下为原文

   
You are not wrong , they come up as listed in the Data sheet.

On some PIC devices, the Reset state of some pins can be configured via the config bit.

它可能有助于确切地知道你正在使用的设备，但我怀疑“XXXX”的LATX和PORTX意味着“随机”（初始电源）或“无论它是什么之前的重置”（所有其他重置）为所有PIC24和DSIC33设备。当然，“PORTX”寄存器将由任何外部驱动。这是有道理的。我必须承认，我本来希望LATX寄存器有一个更明确的复位状态，但我想它们是你应该总是设置到所要求的状态。无论如何，苏珊。



      以下为原文

    It might help to know exactly which device you are using, but I suspect that the 'XXXX' for the LATx and PORTx means 'random' (initial power up) or 'whatever it was before the reset' (all other resets) for all PIC24 and dsPIC33 devices.
Of course the 'PORTx' registers will be driven by any external circuitry so that makes sense.
I must admit I would have expected the LATx registers to have a more definite reset state, but I suppose that they are something you you (should) always set to the required state regardless.
Susan

LATX是D锁存器的输出，其中电源上的PIN Q的初始状态是不确定的，因此，它是未知的“X”。在配置TrISX输出之前，总是写入LATX。



      以下为原文

   
LATx is the output of a D latch where the initial state of pin Q upon power-on is indeterminate; hence, it's unknown 'x'.  Always write to LATx before configuring its TRISx to output.
 

如果在复位期间设置TRISx，则不需要通过硬件初始化LATX。任何硬件复位动作都需要更多的晶体管-更大的面积&更高的价格。因此，在没有明确要求的情况下，不要期望任何冗余。



      以下为原文

    No need to initialize LATx by hardware if TRISx is set during Reset.
Any hardware reset action requires more transistors -> more area -> higher price. Thus don't expect any redundancy where it is not explicitly required.