TIM1寄存器读取勘误表

你好，
在您所指的三个代码片段中，唯一的多循环指令是您在Read seq_2中的INCW X.
因此，读取seq_1和读取seq_3是正常的，但不应使用Read seq_2进行16位TIMer访问。
这也意味着RM中的建议是正确的。
如果要检查指令是否在多个周期内执行，请参阅
http://www.st.com/content/ccc/resource/technical/document/programming_manual/43/24/13/9a/89/df/45/ed/CD00161709.pdf/files/CD00161709.pdf/jcr：内容/翻译/ en.CD00161709.pdf
表42第64至74页的表格显示了循环次数。
如果这还不清楚，请告诉我



以下为原文




Hello,
In the three code snippets you are referring too, the only multiple cycle instruction is the INCW X you have in the Read seq_2
As a consequence Read seq_1 and Read seq_3 are OK, but you should not use Read seq_2 for 16-bit TIMer access.
This also means that the recommendation in the RM is correct.
If you want to check if an instruction execute in multiple cycles, please refer to
http://www.st.com/content/ccc/resource/technical/document/programming_manual/43/24/13/9a/89/df/45/ed/CD00161709.pdf/files/CD00161709.pdf/jcr:content/translations/en.CD00161709.pdf
.The Table 42 page 64 to 74 has a column showing the number of cycles.
Let me know if this is still not clear

这是来自您链接的doc：
而对于Seq_3：ldw是2循环指令。它用于将TIM1-> CNTRL加载到索引寄存器。
在Seq_1我犯了错误。 LD无法从内存加载到XL。
因此，这三个读取都不会组合或返回正确的结果。
目前我正在使用''mov _store_mem，_ read_mem
它得到了这份工作吗？因为我不知道外围功能如何受到指令大小或执行时间的影响。如果在读取高位寄存器和低位寄存器之间有5-10-20个单周期指令怎么办？



以下为原文




This is from doc  linked by you:


And for Seq_3: ldw is 2 cycle instruction. It's used to load TIM1->CNTRL to index register.
in Seq_1 I've made mistake. LD could not load from memory to XL.
So, none of this three reads will assemble or return correct result.
Currently I'm using ''mov _store_mem, _read_mem

• mov _cnt_hi, TIM1_CNTRH
  

• 
  

• mov _cc_hi, TIM2_CCR1H
  

• 
  

• mov _cnt_low, TIM1_CNTRL
  

• 
  

• mov _cc_low, TIM2_CCR1L
  

• ?
  

• ?
  

• ?
  

• ?
  

Does it get the job? Because I don't know how peripheral feature could be impacted by instruction size or exec time. And what if I have 5-10-20 single cycle instructions between reading High and Low registers?

我在STM8S103F3P6 rev Y上尝试过，无论我做什么，只要在TIM1_CNTRL之前读取TIM1_CNTRH，我就无法读取定时器以返回错误的答案。我甚至尝试读取CNTRH，调用函数将该值写入UART，读取CNTRL，它仍然是正确的。
我的测试程序：停止计数器，将0xffxx写入CNTR，启动计数器，少数NOP，然后以各种方式读取CNTRH / CNTRL。重复增加起始值。在某些时候，你会看到计时器翻转。



以下为原文




I tried it out on a STM8S103F3P6 rev Y, and no matter what I do, I cannot get the timer read to return the wrong answer, as long as I read TIM1_CNTRH before TIM1_CNTRL. I even tried reading CNTRH, calling a function to write that value to the UART, reading CNTRL, and it was still correct.
My test procedure: stop counter, write 0xffxx to CNTR, start counter, handful of NOPs, and then read CNTRH/CNTRL in various ways. Repeat with incrementing start value. At some point, you'll see the timer roll over.