Semaphore?

Kernel
1

nobody wrote on Thursday, February 15, 2007:

Hi,

I am using a Semaphore to ensure that two tasks are not accessing the same SPI port at the same time. But I am not sure if it is okay to it in this way.

TaskOne()  // priority 0

{
for (;:wink:
{
    xSemaphoreTake( xSemaphoreSPI0, portMAX_DELAY  );

    ReadSPI0();

    xSemaphoreGive( xSemaphoreSPI0 );

    vTaskDelay(1000);
}
}

TaskTwo() // priority 1

{
   xSemaphoreTake( xSemaphoreSPI0, portMAX_DELAY  );

    ReadSPI0();

    xSemaphoreGive( xSemaphoreSPI0 );

    vTaskDelay(2000);
}

}

What happens if TaskTwo gets ready to run, but the semaphore is obtained by TaskOne, does TaskOne then get time to finish?

Best regards,

Frank Andersen

2

embeddedc wrote on Thursday, February 15, 2007:

Presuming that the semphore is only given once successfully taken then your code will be ok.  An alternative is to have one task that controls the SPI port, and any other task that wants to send or receive send a message to the SPI controlling task with the info to be sent/received.

3

embeddedc wrote on Thursday, February 15, 2007:

That did not read very well, try again.  An alternative is to have one task that controls the SPI port.  If another task want to send data on the SPI port it sends the data to the task in charge of the SPI rather than directly to the SPI peripheral itself.  The message can be a structure, with for example the message destination, length and data bytes.  The message can be send to the controlling task on a queue to ensure serial access is maintained.

4

nobody wrote on Monday, February 19, 2007:

Well, it does not work.

Could someone please come with a working example, where to tasks are sharing a resource, controlled by a semaphore?

Best regards,

Frank Andersen

5

nobody wrote on Monday, March 12, 2007:

{
if ( xSemaphoreTake( xSemaphoreSPI0, portMAX_DELAY )!= pdFALSE);
{
   ReadSPI0();

   xSemaphoreGive( xSemaphoreSPI0 );
}
vTaskDelay(1000);
}