The SpinalMap parameter type passed in Spinalhdl is different, resulting in meaningless generated code

Question

The first parameter transfer I wrote generated the code I wanted spinalHDL CODE

    io.data := SpinalMap(io.funct3, B"3'b0" -> B"2'b1",
        B"3'b1" -> B"2'b10",
        B"3'b10" -> B"2'b11",
        B"3'b11" -> B"2'b00",
        default -> B"2'b00")

generate verilog code

  always @(*) begin
    case(io_funct3)
      3'b000 : begin
        _zz_io_data = 2'b01;
      end
      3'b001 : begin
        _zz_io_data = 2'b10;
      end
      3'b010 : begin
        _zz_io_data = 2'b11;
      end
      3'b011 : begin
        _zz_io_data = 2'b00;
      end
      default : begin
        _zz_io_data = 2'b00;
      end
    endcase
  end

The second parameter transfer I wrote generated the code

spinalHDL CODE

    val table:Seq[(Any, Bits)] = Seq(B"3'b0" -> B"2'b1", B"3'b1" -> B"2'b10", B"3'b10" -> B"2'b11", B"3'b11" -> B"2'b00", spinal.core.default -> B"2'b00")
    io.data := SpinalMap.list(io.funct3, table)

verilog code

  always @(*) begin
    if((io_funct3 == table_0_0)) begin
        _zz_io_data = table_0_1;
    end else if((io_funct3 == table_1_0)) begin
        _zz_io_data = table_1_1;
    end else if((io_funct3 == table_2_0)) begin
        _zz_io_data = table_2_1;
    end else if((io_funct3 == table_3_0)) begin
        _zz_io_data = table_3_1;
    end else begin
        _zz_io_data = table_4_1;
    end
  end

How do I use the second type of spinalHDL code to generate the first type of verilog code?Why do the two writing methods generate different verilog code?

It seems that I only generate case statements by directly passing (Any, T) * types, and I generate if statements using other set types