My if and else statements in verilog are not giving expected results

Question

I am trying a simple code to compare an input to some value(threshold) and tried using the if and else condition statements to assign a variable to either 0 or 1 depending of the variable crossing the threshold or not. But, in this case I am always getting same single value for different inputs. I am new to verilog and trying to write an algorithm for a different application, in which this is my first step. Thanks in advance.

module makebig(a,z);
input signed [15:0]  a;
output reg z;
initial
begin
if (a<16'b0000000000000000) begin
 z<=1;
end
else  begin
 z<=0;
end

end
endmodule

Now the test bench of it.

module makebig_tb();

reg  signed [15:0] a;

wire z;

makebig uut(.a(a),.z(z));
initial
begin
a=16'b0100000000000000;
#5 a=16'b000000000000101;
#5 a=-2;
end
initial
begin
$monitor("a=%b,z=%b",a,z);
end
end module

I don't understand why it is not working. The output for this is something like this...

a=0100000000000000,z=0;
a=0000000000000101,z=0;
a=1111111111111110,z=0;

Not my area, but.... Are you sure the constant in if is signed? May be the comparison is evaluated in unsigned numbers despite input signed declaration.... — CiaPan
– CiaPan, Commented Jul 14, 2021 at 5:46
Yes, in the assignment of the variable a, I have mentioned 'signed'. I even tried taking the half of 2^16 as the threshold but, still not the expected results. — Abhinav kumar
– Abhinav kumar, Commented Jul 14, 2021 at 5:58
Does this: Why is this verilog relational statement returning true? answer your question? — CiaPan
– CiaPan, Commented Jul 14, 2021 at 11:32

sjw · Accepted Answer · 2021-07-19 11:23:56Z

2

There are two errors here, and both of them have been pointed out in the comments:

You are only running the comparison once (initial begin). Replace this with always @(*) begin to run it each time one of the inputs changes
A comparison with an unsigned constant will result in the whole comparison being unsigned. Replace 16'b0000000000000000 with $signed(16'b0000000000000000)

This produces the expected output

a=0100000000000000,z=0;
a=0000000000000101,z=0;
a=1111111111111110,z=1;

edited Jul 19, 2021 at 11:23

answered Jul 14, 2021 at 15:57

sjw

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