In reply to vk7715:
ON "Can you please explain why the values of i are 9 down to 0 when the loop is incrementing in an ascending fashion?"
task automatic t();
for (int j = 0; j <= 9; j++) begin
int k = j;
// #10;
fork
begin
$display("j %0d k %0d ",j,k);
A(k);
end
join_none
end
endtask
Per rag123's response "fork/join_none does not start any processes until the parent process suspends or terminates". Thus, the A(k) are queued up by the simulator from 0 to 9
until t() ends and the initial reached the @ (done == 10'b11_1111_1111);
Following that, the simulator processes those A(k) tasks from the queue. I believe that would be up to the tool to decide how it wants to pull the job off the queu. Looks like it did in a LIFO fashion.
On But how would I truly know if all the 10 iterations executed in parallel at the same time? I ask this because since the task is called inside an initial block which takes a simulation time of 0, wouldn't all 10 iterations complete in time 0 irrespective of the fork join_none? The tasks did complete at time 0. If you put a #3 just after the initial begin the time would be 3.
On task B, the formal needs to be defined as an int, it defaulted to a bit.
Made changes in https://www.edaplayground.com/x/ak_H
module m;
// event a;
bit [9:0] done;
task automatic A(input int i);
$display("Task A. Time = %0t, i = %0d", $time, i);
done[i] = 1;
endtask
task automatic B(int p);
// int w;
// w=p;
$display("Task B. Time = %0t, p = %0d", $time, p);
endtask
/* task automatic t();
for (int j = 0; j <= 1; j++) begin
#10;
fork
A(j);
join_none
end
endtask */
task automatic t();
for (int j = 0; j <= 9; j++) begin
int k = j;
// #10;
fork
begin
$display("j %0d k %0d ",j,k);
A(k);
end
join_none
end
endtask
initial begin
#3;
t();
$display("sent t()");
// wait (done == 10'b00_0000_0011);
@ (done == 10'b11_1111_1111);
$display("done=%b", done);
// @(a);
B(2);
end
endmodule
//
# sent t()
# j 10 k 9
# Task A. Time = 3, i = 9
# j 10 k 8
# Task A. Time = 3, i = 8
# j 10 k 7
# Task A. Time = 3, i = 7
# j 10 k 6
# Task A. Time = 3, i = 6
# j 10 k 5
# Task A. Time = 3, i = 5
# j 10 k 4
# Task A. Time = 3, i = 4
# j 10 k 3
# Task A. Time = 3, i = 3
# j 10 k 2
# Task A. Time = 3, i = 2
# j 10 k 1
# Task A. Time = 3, i = 1
# j 10 k 0
# Task A. Time = 3, i = 0
# done=1111111111
# Task B. Time = 3, p = 2
# exit
