Consider the following Verilog testbench code. What will be printed before the simulation ends?
module test;
initial begin
$display("Start of simulation");
$finish;
$display("This will not print");
end
endmoduleThink about what $finish does in a simulation.
The $finish system task ends the simulation immediately. So, the second $display is never executed.
Given this Verilog code, what output will appear before the simulation ends?
module test;
initial begin
fork
$display("Before finish");
$finish;
$display("After finish");
join
end
endmoduleRemember that $finish stops the entire simulation immediately.
Even though the fork-join runs multiple threads, $finish stops the simulation immediately, so only the first $display runs.
Analyze the code below. Why is the last $display statement never printed?
module test;
initial begin
$display("Message 1");
#5 $finish;
$display("Message 2");
end
endmoduleThink about the timing of $finish and when statements execute.
The #5 delay causes $finish to run after 5 time units, ending the simulation before the next $display can execute.
Which option contains a syntax error in using $finish in Verilog?
Check the correct syntax for calling system tasks like $finish.
$finish is a system task and should be called without parentheses. Using parentheses causes a syntax error.
Consider this Verilog code. How many times will "Loop iteration" be printed before the simulation ends?
module test;
integer i;
initial begin
for (i = 0; i < 5; i = i + 1) begin
$display("Loop iteration %0d", i);
if (i == 2) $finish;
end
end
endmoduleThink about when $finish stops the loop.
The loop prints for i=0,1,2. When i==2, $finish ends simulation immediately, so no further iterations.
