SV 類繼承的多態性問題（NVDIA2019筆試）

1、原題

class class_a;
        virtual function void print_name();
            $display("this is class_a");
        endfunction
endclass

class class_b extends class_a;
        virtual function void print_name();
            $display("this is class_b");
        endfunction
endclass

module test;
    initial begin
    begin:class_test1
            class_a a;
            class_b b;
            a = new();
            b = new();
            a.print_name();
            b.print_name();
    end
    begin:class_test2
            class_a a;
            class_b b;
            b = new;
            a = b;
            a.print_name();
            b.print_name();
    end
    begin:class_test3
            class_a a;
            class_a a2;
            class_b b;
            class_b b2;
            b = new();
            a = b;
            a2 = new();
            b2 = b;
            a2.print_name();
            $cast(a2,a);
            a2.print_name();
            b2.print_name();
    end
    end

endmodule

 

 2、執行結果

　　1）使用a=new會創建一個新對象，並且將句柄a指向這個新創建的對象；使用a=b不會創建新對象，只是將句柄a指向已經存在的對象b

　　2）原因：基類中display()為虛函數，具體調用哪個x.dispaly()取決於句柄x所指向的對象類型，而不是取決於x這個句柄本身的類型

# this is class_a
# this is class_b
# this is class_b
# this is class_b
# this is class_a
# this is class_b
# this is class_b

　　3）原因：同2

# this is class_a
# this is class_b
# this is class_b
# this is class_b
# this is class_a
# this is class_b
# this is class_b

　　4）原因：基類中display()不為虛函數，具體調用哪個x.dispaly()取決於x這個句柄本身的類型

# this is class_a
# this is class_b
# this is class_a
# this is class_b
# this is class_a
# this is class_a
# this is class_b