Need help to understand array method 'unique(expression)' followed by 'with' clause

SystemVerilog
1

Hi,
Can someone help explain why output is 3 1 for the below usage of unique? how to interpret this arr.unique(x) with (x < 3)?
for x < 4, output is '{4, 1}.
for x > 6, output is '{1, 7}.
I am unable to figure out the pattern.

module tb;

  int arr[7] = '{1, 2, 3, 4, 5, 6, 7};

  initial begin
    $display("%0p", arr.unique(x) with (x < 3));
  end 
endmodule;

Output: '{3, 1}

I got the code from here: SystemVerilog Array Manipulation

2

In reply to poojanm2002:

The unique array method returns element not the expression enclosed in with clause.Concider following example


 module top();
  int q[$];
  int result[$];

  initial begin
 
     q={1,2,3,4,4,5};
     $display("queue 'q' : %p", q);

     result.delete();
    
     result = q.unique(item)with(item>3);
     $display("using unique: %p", result);
    
     result.delete();
    
     result = q.unique(item)with((item>3)*item);
     $display("using unique *item : %p", result);
    
    end
endmodule : top

================
queue ‘q’ : '{1, 2, 3, 4, 4, 5}
using unique: '{1, 4} //**Few simulators treats 1,2,3 as 1’s ,others treat them as 0’s
using unique *item : '{1, 4, 5}

3

In reply to poojanm2002:

When using a with_expression as part of a unique() method, the with_expression value is used to calculate uniqueness instead of the element’s value. When the with_expression is x<3, the result of that expression is one of two possible values: 1’b0, or 1’b1.

The with_expression for array elements 1 and 2 have the value 1’b1, and the remaining elements have the value 1’b0. The simulator is free to choose 1 or 2 as the unique element for 1’b1, and 3, 4, 5, 6, or 7 as the unique element for representing the value 1’b0.