Why i am getting output as a=0, b=0 and c=0, in below posted system verilog code?

SystemVerilog
1

class packet;
rand bit [3:0] a;
rand bit [3:0] b;
rand bit [3:0] c;

constraint a_value { a == b + c; }
constraint b_value { b > 15; }
constraint c_value { c < 10; }
endclass

module soft_constr;
initial begin
packet pkt;
pkt = new();
repeat(5) begin
pkt.randomize();
$display(“Value of a = %0d \tb = %0d \tc =%0d”,pkt.a,pkt.b,pkt.c);
end
end
endmodule

2

In reply to swapnilsrf9:
Because of the following constraint failure. the max value of b is 4’b1111(15). But constraint is saying > 15.
constraint b_value { b > 15; }

if you modify b width to 5 bit, you will get the correct values

3

In reply to murali.gubba:

class packet;
rand bit [3:0] a;
rand bit [3:0] b;
rand bit [3:0] c;
constraint b_value { b > 3; }
constraint c_value { c < 10; }

constraint a_value { a == b + c; }

endclass

module soft_constr;
initial begin
packet pkt;
pkt = new();
repeat(5) begin
pkt.randomize();
$display(“Value of a = %0d \tb = %0d \tc =%0d”,pkt.a,pkt.b,pkt.c);
end
end
endmodule

i am not getting the satisfied result of a==b+c; when b=13, c=9; what is the calculation behind it?

4

In reply to swapnilsrf9:

Again, the size of your variables is leading to overflow conditions. You need to size your variables appropriately to handle all valid conditions.

Also, the randomize() function will return a success/failure code. You should always check this:


class packet;
  rand bit [4:0] a;
  rand bit [3:0] b;
  rand bit [3:0] c;

  constraint b_value { b > 3; }
  constraint c_value { c < 10; }

  constraint a_value { a == b + c; }
endclass

module soft_constr;
  initial begin
    packet pkt;
    pkt = new();
    repeat(5) begin
      if (!pkt.randomize()) $display("Randomization failed...");
      else $display("Value of a = %0d \tb = %0d \tc =%0d",pkt.a,pkt.b,pkt.c);
    end
  end
endmodule
5

In reply to swapnilsrf9:

Please modify width of a to 6, because when b = 13 & c = 9, a is constrained to (b + c), according to your constraint adjusted to 4 bits.

i,e b = 4’b1101b & c = 4’b1001. So, when you add b & c, you will get 10110b, but since a is only 4 bits, the higher bits are truncated. So, you may be getting a as 6 for this randomisation. Am I correct???

Changing a to rand int, or rand bit [5:0] might solve…

6

In reply to BHEEMA THANGAVELU:

Thanks i got my solution