I want to make test application which shows round-robin scheduling.

i created three tasks.
xTaskCreate( vErrorChecks, "Task1", mainPRINT_STACK_SIZE, NULL, 5, &xHandle );
xTaskCreate( vErrorChecks1, "Task2", mainPRINT_STACK_SIZE, NULL, 5, &xHandle1 );
xTaskCreate( vErrorChecks2, "Task3", mainPRINT_STACK_SIZE, NULL, 5, &xHandle2 );

static void vErrorChecks( void *pvParameters )
{
for( ;; )
{
for(i=0; i<1000; i++)
{
printf("i am in task1\n");

vTaskDelay( ( portTickType ) 10 );
}
}
}

other two task have same code in their entry functions.

But CPU is not sharing same time for all three tasks.

Can anyone suggest what should i do? if anyone have code then please send on my email id.

Thanks in advance.

If task 1 runs for 3/4 of a time slice then block, task two will start running. It will then only have 1/4 of a time slice before it gets swapped out. Expand this type of execution pattern for the whole application and you will see that the tasks are not necessarily all going to get the same processing time.

Some things to note. One, your tasks are calling printf, do you have any mutual exclusion on the port to which the messages are being sent? If not then your messages might get garbled. Take a look at some of the LCD gatekeeper tasks, for example in the LM3S6965 demo, for a solution to this.

If your tasks were of the form:

for(;;)
{
printf( "Task x" );
taskYIELD();
}

And you were not using preemptive scheduling then three tasks would get much closer to sharing the CPU time equally.

one more issue in this; if is same priority for three tasks, then first task should print first but controlled directly goes to three number task " i am in Task3".

is it some abnormal behaviour?

one more issue in this; if is same priority for three tasks, then first task should print first but controlled directly goes to three number task " i am in Task3".

is it some abnormal behaviour?