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Design Languages >> Verilog-AMS >> Order of execution at a timestep in Verilog
https://designers-guide.org/forum/YaBB.pl?num=1092599412

Message started by sutapanaki on Aug 15^th, 2004, 12:50pm

Title: Order of execution at a timestep in Verilog
Post by sutapanaki on Aug 15^th, 2004, 12:50pm

Hi,

I need some help with the following Verilog code. If someone understands this, please respond. My question is about the order of execution during one time step.
In the code I have it goes like this:

1: always
2: begin
3: #50 clk <= 1'b0;
4: #50 clk <= 1'b1;
$display($time, "clk%b", clk) ;
//here clk will be clk == 1'b0 because of the non blocking assignment
5: end

6: always @(posedge clk)
7: begin
$display($time, "clk%b", clk)
/*I expected different behavior here: I expect clk == 1'b0. As it is above. That is what I expect from a non blocking assignment when the timestep (100) has not finished yet. The simulation shows clk == 1'b1. So the non blocking assignment is executed not at the end of the time step but somewhere in the midle - before the event it trigers. This surprises me*/

8: if(clk === 1'b1)
9: clk1 <= clk;

10: end