如何只更新端口的低字节？

你的代码看起来像什么？



      以下为原文

    What does your code look like?

有一个特殊的寄存器LATXIN。当你给它写时，它反转了写TOTO一个的所有比特。所以你可以做这样的事情：



      以下为原文

    There is a special register LATxINV. When you write to it, it inverts all bits which are written to to one. So you can do something of that sort:
 
void WriteBus(char new_state_of_the_bus) {
  static char previous_state_of_the_bus = 0;
  LATDINV = previous_state_of_the_bus ^ new_state_of_the_bus;
  previous_state_of_the_bus = new_state_of_the_bus;
}

LATC^＝（NeXiLuxByth-^ LATC）和0xFF；



      以下为原文

   
LATC ^= (new_low_byte ^ LATC) & 0xFF;
 

LATC^＝（NeXixLuxByth-LATC）和0xFF；或者，因为这是针对PIC32 LATCVI=（NeXiLuxByth-^ LATC）和0xFF；



      以下为原文

   
LATC ^= (new_low_byte ^ LATC) & 0xFF;

Or, as this is for a Pic32
LATCINV=(new_low_byte ^ LATC) & 0xFF;

数据是一个简单的像素LCD。我刚刚在宏中使用了一个“LATC=数据”，用于显示的所有写入。我希望我能避免额外的阅读和额外的代码混合在上面的位。感谢优秀的例子！/古恩瑟



      以下为原文

    The data is going to a simple pixel LCD. I just had a 'LATC =  data' in a macro used for all writes to the display. I was hoping I can avoid the extra read and extra code to blend in the upper bits.
 
Thanks for the excellent examples!
/Guenther

我运行了一个快速测试整个屏幕200次：LATC=数据；24秒StLC^＝（数据^ LATC）和0xFF；32秒LATCVI=（数据^ LATC）和0xFF；28秒/ GueNETH。



      以下为原文

    I ran a quick test painting the whole screen 200 times:
 
LATC = data;                               24 seconds
LATC ^= (data ^ LATC) & 0xFF;    32 seconds
LATCINV=(data ^ LATC) & 0xFF;   28 seconds
 
/Guenther

试试我的。它不做“和；0xFF”。



      以下为原文

    Try mine. It doesn't do "& 0xff".

Hi，PIC32 MX微控制器在端口控制寄存器中有额外的硬件门，以便能够对寄存器中的一个比特进行原子更新。对于不同的操作有3组门：或门，在外围SFR寄存器XOR门中设置任何位选择，以反转A s。IMPARAR选择位和门来清除寄存器中的任何位选择。然而，设置一些位和清除其他位来更新寄存器的任意部分，将需要进行2次写入操作。事实上，这不仅对端口控制寄存器，几乎每一个外围SFR寄存器都是如此。具有这些位操作寄存器。与端口锁存寄存器：LATC，有：LATCLR、LATCSET和LATCVIN，它们都有独立的总线地址。因为PIC32 CPU是流水线的，而PIC32微控制器中的总线系统具有桥路，并且可能有不同的操作频率，PIC32。无法读取SFR寄存器，计算更新并将结果存储在同一指令中。此外，阅读SFR寄存器需要更长的时间比写作。（PIC32 MZ在访问SFR寄存器时甚至比MX还要慢）所有上述示例代码都可以在不干扰寄存器的其他部分中的内容的情况下工作，但是引用寄存器的以前内容的所有代码都必须进行读修改写操作，这不是原子的，我们将更多的时间比需要…你可以尝试：问候，Mysil



      以下为原文

    Hi,
PIC32MX microcontrollers have extra hardware gates in in Port Control registers,
to be able to do atomic updates to a selection of bits in the register.
There are 3 sets of gates for different operations:
OR gates, to SET any selection of bits in the peripheral SFR register,
XOR gates, to INVert a similar selection of bits,
and gates to CLeaR  any selection of bits in the register.
 
However, Setting some bits, and Clearing other bits, to update an arbitrary part of the register, will take 2 write operations.
In fact, this is not only for port control registers, nearly every peripheral SFR register have these bit manipulation registers.
 
Together with Port Latch register: LATC, there are: LATCCLR, LATCSET and LATCINV,
they all have a separate bus address.
 
Since the PIC32  CPU is pipelined, and the bus system in PIC32 microcontrollers have bridges, and possibly different operating frequency, PIC32 cannot Read a SFR register, calculate a update and store the result in the same instruction. Also, Reading a SFR register take longer time than writing. (PIC32MZ is even slower than MX in accessing SFR registers.)
 
All of the above example codes, may work without disturbing contents in other parts of the register,
but all codes that refer to previous content of the register, will have to do a Read-Modify-Write operation,
that is not atomic, and will use more time than needed..
You may try this: 
   LATCSET = new_low_byte & 0xFF;
    LATCCLR = (~new_low_byte) & 0xFF; 
 
Regards,
   Mysil

诺斯盖尔，你的版本计时了54秒。主要是因为我想-调用子程序。我可以改变你的例子，不要使用子程序。但最后还有额外的任务。



      以下为原文

    NorthGuy, your version clocked 54 seconds. Mostly due - I think - to the subroutine call. I could change your example to not use a subroutine. But there is still this extra assignment at the end.
 
/Guenther

当然，您需要删除函数调用——这需要很长时间，也会阻止编译器在寄存器中保持“PyviuyBuSuthStand”。在删除函数调用之后，如果编译器自己无法理解它，则可以给出“PrviviuBuSyStand”变量“寄存器”指定。保持“PyviuyBuSuthStand”只需要在寄存器（新到先前）之间进行一次移动。CPU暂停，因为它不能缓存）加上一个额外的“& 0xFFF”操作。



      以下为原文

   


Of course, you need to remove the function call - it takes long time, and also precludes the compiler from keeping the "previous_bus_state" in a register. After removing the function call you can give the "previous_bus_state" variable a "register" designation if the compiler cannot figure that out by itself.
 
Maintaining the "previous_bus_state" requires only one move between registers (new to previous).
 
Reading LATC requires SFR reading (which causes a CPU stall because it cannot be cached) plus an extra "&0xff" operation.
 

LATCSET = NexyLuxByth＆0xFF；LATCLR= =（~NeXiLuxByb）和0xFF；…用我的测试花费了32秒。



      以下为原文

    LATCSET = new_low_byte & 0xFF;
LATCCLR = (~new_low_byte) & 0xFF; 
 
...took 32 seconds with my test.
 
/Guenther

或者一个更好的解决方案，或者两者都应该编译成一个简单的字节存储。在模拟器和实际硬件中，这都不干扰端口的其余部分。编辑：第一个版本的地址是MZ，对于MX，它是用于LATC的0xBF8660A0。可能第二个版本更通用，因为第一个版本只与BYTE0一起工作，因为对齐问题（不能指定未对齐的地址）和不可移植性。



      以下为原文

    Or an even better solution //Global var
uint8_t __attribute__((section("sfrs"),address(0xBF860230))) LATCLOBYTE;


void SetLATCLoByte(uint8_t newvalue)
{
    LATCLOBYTE=newvalue;
}ortypedef struct __attribute__((packed))
{
    uint8_t BYTE0;
    uint8_t BYTE1;
    uint8_t BYTE2;
    uint8_t BYTE3;
} SFRBYTE;
extern volatile SFRBYTE LATCBytes __asm__ ("LATC") __attribute__((section("sfrs")));
 
void SetLATCLoByte(uint8_t newvalue)
{
    LATCBytes.BYTE0=newvalue;
}
Both the above should compile to a simple byte store. This is confirmed to not disturb the rest of the port, both in the simulator and on actual hardware.

Edit:- The address for the first version is for an MZ, for MX it is 0xBF8660A0 for LATC. Probably the second version is more universal, as the first version only works with BYTE0 due to alignment issues (you cannot specify an un-aligned address) and non-portability.

当然，我用我的帖子把这个线程移到错误的方向：（完全是我的错）。



      以下为原文

   


Of course. I derailed the thread into the wrong direction with my post :( Totally my fault.

使用“影子寄存器”怎么样？这将消除读取寄存器的需要——只需要一个写入来更新，并且您总是可以同时更新整个端口。改变的比特，改变的，不改变的。



      以下为原文

    How about using a "Shadow Register"? This would eliminate the need to read the register - only a write is needed to update and you can always update the entire port at once. Bits that change - change and those that don't - don't.

那也行得通。但会采取更多的指令周期。



      以下为原文

    That would work too.
But would take more instruction cycles.