Verilog For loop

Question

I have the following code in verilog to test a For loop:

module test1;

reg [2:0] i;

initial
begin
    for(i=0;i<=3;i=i+1) begin
         #10;
         $display("%d",i);
    end
end

endmodule

The printed output shows:

0 1 2 3

Which makes sense to me. But the waveform below confuses me:

How does the reg 'i' take the value '4' here?

This is part of a bigger code I am working on where I have many nested for loops used in a testbench.

I am adding a nested loop which behaves weird:

module test1;

reg signed [2:0] i,j;

initial
begin
    for(i=-3;i<=3;i=i+1) begin
        for(j=-3;j<=3;j=j+1) begin
            #10;
            $display("%d   %d",i,j);
        end
    end
end

endmodule

The output waveform is shown below:

Why is the 2nd loop not working and first loop keep repeating?

Owen Owen · Accepted Answer · 2020-12-11T04:16:23

This has to do with how for loops know when to stop.

When you write

for(i=0;i<=3;i=i+1)

this says three things:

i starts at 0
The loop continues until i<=3 becomes false
Each time, increment i by one

The thing is, i<=3 can't become false until i becomes 4. When i becomes 4, the loop stops looping. But i is still 4.

In your nested loop example, the problem is overflow.

You have declared j as signed three bits:

reg signed [2:0] i,j;

But you are looping until j becomes 4:

for(j=-3;j<=3;j=j+1)

There is no way to represent 4 as a signed three bit integer. Because j<=3 must always be true, the loop can never stop.

Verilog For loop

1 Answers