如何在多核环境中从函数中返回一个变量？

huangapple
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英文:

How to return a variable from a function in a multi core environment?

问题

所以我有一个必须在系统中对其他任务可用的变量。
我有一个函数，应该返回这个变量，以便其他任务可以根据这个变量执行它们的操作。

  1. boolean pSystem::isLowPower(){
  2.     return LOW_POWER;
  3. }

如果我直接使用它或通过函数使用它，那么我会从想要使用这个变量的随机任务中得到一个IllegalInstruction错误，我的ESP会崩溃。

所以如果我想用互斥锁保护这个变量，我就不能返回它。

  1. boolean pSystem::isLowPower(){
  2.     xSemaphoreTake( testMutex );
  3.     return LOW_POWER;
  4.     xSemaphoreGive( testMutex  ); // 不会释放？
  5. }

我如何能够保护这个变量并同时返回它？

用例将类似于这样：

  1. void TaskOne(void* parameter){
  2.     for(;;){
  3.         if( Preformance.isLowPower() ){ vTaskDelay(100); return; }
  4.         // 执行其他与任务相关的操作...
  5.         vTaskDelay(1);
  6.     }
  7. }
  9. void TaskTwo(void* parameter){
  10.     for(;;){
  11.         if( Preformance.isLowPower() ){ vTaskDelay(100); return; }
  12.         // 执行其他与任务相关的操作...
  13.         vTaskDelay(1);
  14.     }
  15. }

我得到的错误日志（来自随机任务）

  1. Guru Meditation Error: Core 0 panic'ed (IllegalInstruction). Exception was unhandled.
  2. Memory dump at 0x400f1034: 08e08096 00f01d00 257bfea1
  3. Core 0 register dump:
  4. PC      : 0x400f1039  PS      : 0x00060430  A0      : 0x00000000  A1      : 0x3ffd2b50
  5. A2      : 0x00000000  A3      : 0x00000000  A4      : 0x00000000  A5      : 0x00000000
  6. A6      : 0x00000000  A7      : 0x00000000  A8      : 0x800f1039  A9      : 0x3ffd2b30
  7. A10     : 0x00000064  A11     : 0x3ffbf638  A12     : 0x3ffbdd58  A13     : 0x0000000a
  8. A14     : 0x0000000a  A15     : 0x80000001  SAR     : 0x00000000  EXCCAUSE: 0x00000000
  9. EXCVADDR: 0x00000000  LBEG    : 0x00000000  LEND    : 0x00000000  LCOUNT  : 0x00000000  
  11. Backtrace: 0x400f1036:0x3ffd2b50
  13.   #0  0x400f1036:0x3ffd2b50 in mBusLoopTask(void*) at src/utilities/modBus.cpp:1495 (discriminator 1)
  15. ELF file SHA256: 1c0b237e74fe9d04
  17. Rebooting...
英文:

So i have a variable which must be available to other tasks in the system.
I have a function which should return this variable so other tasks can do their thing according to this variable.

  1. boolean pSystem::isLowPower(){
  2.     return LOW_POWER;
  3. }

If i use it RAW or trought the function, I got an IllegalInstruction error from a random task that wants to use this variable and my ESP crashes.

So if i want to guard this variable with a mutex i couldn't return it.

  1. boolean pSystem::isLowPower(){
  2.     xSemaphoreTake( testMutex );
  3.     return LOW_POWER;
  4.     xSemaphoreGive( testMutex  ); // Will not release?
  5. }

How can i guard this variable and return it at the same time?

The use case would be only something like this:

  1. void TaskOne(void* parameter){
  2.     for(;;){
  3.         if( Preformance.isLowPower() ){ vTaskDelay(100); return; }
  4.         // Do other task related stuffs...
  5.         vTaskDelay(1);
  6.     }
  7. }
  9. void TaskTwo(void* parameter){
  10.     for(;;){
  11.         if( Preformance.isLowPower() ){ vTaskDelay(100); return; }
  12.         // Do other task related stuffs...
  13.         vTaskDelay(1);
  14.     }
  15. }

The error log i got ( from random tasks )

  1. Guru Meditation Error: Core  0 panic'ed (IllegalInstruction). Exception was unhandled.
  2. Memory dump at 0x400f1034: 08e08096 00f01d00 257bfea1
  3. Core  0 register dump:
  4. PC      : 0x400f1039  PS      : 0x00060430  A0      : 0x00000000  A1      : 0x3ffd2b50
  5. A2      : 0x00000000  A3      : 0x00000000  A4      : 0x00000000  A5      : 0x00000000
  6. A6      : 0x00000000  A7      : 0x00000000  A8      : 0x800f1039  A9      : 0x3ffd2b30
  7. A10     : 0x00000064  A11     : 0x3ffbf638  A12     : 0x3ffbdd58  A13     : 0x0000000a
  8. A14     : 0x0000000a  A15     : 0x80000001  SAR     : 0x00000000  EXCCAUSE: 0x00000000
  9. EXCVADDR: 0x00000000  LBEG    : 0x00000000  LEND    : 0x00000000  LCOUNT  : 0x00000000  
  12. Backtrace: 0x400f1036:0x3ffd2b50
  14.   #0  0x400f1036:0x3ffd2b50 in mBusLoopTask(void*) at src/utilities/modBus.cpp:1495 (discriminator 1)
  19. ELF file SHA256: 1c0b237e74fe9d04
  21. Rebooting...

答案1

得分: 0

The problem was that I wanted to return from the task without deleting it, and it produced the IllegalInstruction error.

The solution is to not return but continue the loop.

  1. void TaskOne(void* parameter){
  2.     for(;;){
  3.         if( Preformance.isLowPower() ){ vTaskDelay(100); continue; }
  4.         // Do other task-related stuff...
  5.         vTaskDelay(1);
  6.     }
  7. }

OR

  1. void TaskTwo(void* parameter){
  2.     for(;;){
  3.         if( !Preformance.isLowPower() ){
  4.             // Do other task-related stuff...
  5.         }else{
  6.             vTaskDelay(100);
  7.         }
  8.         vTaskDelay(1);
  9.     }
  10. }
英文:

So the problem was that i wanted to return from the task without deleting it and it produced the IllegalInstruction error.

The solution is to not return but continue the loop

  1. void TaskOne(void* parameter){
  2.     for(;;){
  3.         if( Preformance.isLowPower() ){ vTaskDelay(100); continue; }
  4.         // Do other task related stuffs...
  5.         vTaskDelay(1);
  6.     }
  7. }

OR

  1. void TaskTwo(void* parameter){
  2.     for(;;){
  3.         if( !Preformance.isLowPower() ){
  4.             // Do other task related stuffs...
  5.         }else{
  6.             vTaskDelay(100);
  7.         }
  8.         vTaskDelay(1);
  9.     }
  10. }

huangapple
  • 本文由 发表于 2023年7月6日 19:15:10
  • 转载请务必保留本文链接：https://go.coder-hub.com/76628228.html
  • arduino-esp32
  • freertos
  • multicore
  • mutex
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