在这个很小的设计中，手动操作可以完成以上两步。但是如果设计变成了一个大工程，手动删除操作简直是不能忍受的，也许我需要一个星期不吃不喝一直在做删除操作！在我最无助的时候，IC工程师的好朋友Perl，勇敢的站出来帮了一个他看来再简单不过的忙儿。Perl做了一个简单的动作，帮我完成了一个不可能完成的工作！结果如下：

// Library - bcdee_sc, Cell - RS_XOR2_A, View - schematic

// LAST TIME SAVED: Jul  2 09:43:03 2014

// NETLIST TIME: Aug  9 20:10:38 2014

// Library - bcdee_sc, Cell - RS_INV_A, View - schematic

// LAST TIME SAVED: Jul  2 08:55:40 2014

// NETLIST TIME: Aug  9 20:10:38 2014

// Library - bcdee_sc, Cell - RS_NR2_A, View - schematic

// LAST TIME SAVED: Jul  2 08:51:46 2014

// NETLIST TIME: Aug  9 20:10:38 2014

// Library - bcdee_sc, Cell - RS_ND2_A, View - schematic

// LAST TIME SAVED: Jul  2 08:40:16 2014

// NETLIST TIME: Aug  9 20:10:38 2014

// Library - adder, Cell - fulladder, View - schematic

// LAST TIME SAVED: Aug  9 19:32:32 2014

// NETLIST TIME: Aug  9 20:10:38 2014

module fulladder ( Cout, Sum, A, B, Cin );

output  Cout, Sum;

input  A, B, Cin;

/* specify

    specparam CDS_LIBNAME  = "adder";

    specparam CDS_CELLNAME = "fulladder";

    specparam CDS_VIEWNAME = "schematic";

endspecify */

RS_XOR2_A I6 ( .O(Sum), .I1(net9), .I2(Cin));

RS_XOR2_A I5 ( .O(net9), .I1(A), .I2(B));

RS_INV_A I4 ( .O(net12), .I(net14));

RS_NR2_A I3 ( .O(net14), .I1(A), .I2(B));

RS_ND2_A I2 ( .O(net20), .I1(net12), .I2(Cin));

RS_ND2_A I1 ( .O(Cout), .I1(net23), .I2(net20));

RS_ND2_A I0 ( .O(net23), .I1(A), .I2(B));

endmodule

// Library - adder, Cell - twobitadder, View - schematic

// LAST TIME SAVED: Aug  9 20:01:39 2014

// NETLIST TIME: Aug  9 20:10:38 2014

module twobitadder ( Cout, Sum, A, B, Cin );

output  Cout;

input  Cin;

output [1:0]  Sum;

input [1:0]  A;

input [1:0]  B;

/* specify

    specparam CDS_LIBNAME  = "adder";

    specparam CDS_CELLNAME = "twobitadder";

    specparam CDS_VIEWNAME = "schematic";

endspecify */

fulladder I1 ( .Cin(net25), .B(B[1]), .A(A[1]), .Sum(Sum[1]),

     .Cout(Cout));

fulladder I0 ( .Cin(Cin), .B(B[0]), .A(A[0]), .Sum(Sum[0]),

     .Cout(net25));

endmodule