Help me understand the following forks

UVM
1

I have seen codes like the following one in a project in my company. Does it make sense to fork a single process/statement like the one in the following code? If it does, how these fork behave?

fork
 forever begin
    fork 
       function_1();
    join_none
       
    fork 
       function_2();
    join_none
 end  
join
2

In reply to RB87:
The two functions1,2 will execute in a parallel manner…

3

In reply to RB87:

There is no difference between

 forever begin
  fork
   statement_1;
  join_none
  fork
   statement_2;
  join_none
end

and

 forever begin
  fork
   statement_1;
   statement_2;
  join_none
end

or

forever 
  fork
   statement_1;
   statement_2;
  join_none

Can only guess this was from cut&paste.

BTW, it makes no sense to have a function call, or any other non-time consuming statement as a fork’ed process.

4

In reply to dave_59:

Thank you Dave_59

5

In reply to dave_59:

In reply to RB87:
There is no difference between

 forever begin
fork
statement_1;
join_none
fork
statement_2;
join_none
end

and

 forever begin
fork
statement_1;
statement_2;
join_none
end

or

forever 
fork
statement_1;
statement_2;
join_none

Can only guess this was from cut&paste.
BTW, it makes no sense to have a function call, or any other non-time consuming statement as a fork’ed process.

Hi Dave,

Correct me, if I am wrong.

I feel in the first snippet of code, function_1 will always starts first and then, function_2 concurrently. But in case of snippet 2 and snippet 3 either of the function can start first based on the simulator.
In other words, in snippet 1 there won’t be any race condition, but snippet 2 and 3 will have race condition.

6

In reply to Narendra Sharma:

Correct me, if I am wrong.

You are wrong, I’m correcting you. :)

The process starting order is a race condition with both code snippets. The processes spawned by a fork/join_none gets queued and does not start executing until the parent process blocks or terminates. In this particular example that is not shown, so each iteration of the forever loop spawns two processes and gets into an infinite loop spawning more processes.

7

In reply to dave_59:

In reply to Narendra Sharma:
You are wrong, I’m correcting you. :)

Thanks a lot Dave. :)