将两个单个位转换为向量

如何解决将两个单个位转换为向量

module ALUControl(ALUOp,FuncCode,ALUCtl);
input [1:0] ALUOp;
input [5:0] FuncCode;
output reg [3:0] ALUCtl;
always @(ALUOp,FuncCode) begin
    if ( ALUOp == 2 )
        case (FuncCode)
            32: ALUCtl<=2; // add
            34: ALUCtl<=6; //subtract
            36: ALUCtl<=0; // and
            37: ALUCtl<=1; // or
            39: ALUCtl<=12; // nor
            42: ALUCtl<=7; // slt
            default: ALUCtl<=15; // should not happen
        endcase
    else
        case (ALUOp)
            0:  ALUCtl<=2;
            1: ALUCtl<=6;
            default: ALUCtl<=15; // should not happen
        endcase
end

终端模块

module Control(op0,op1,op2,op3,op4,op5,MemtoReg,RegDst,RegWrite,MemRead,MemWrite,Branch,ALUSrc,ALUOp1,ALUOp2,MemWrite);
input  op0;
input  op1;
input  op2;
input  op3;
input  op4;
input  op5;
output RegDst;
output ALUSrc;
output MemtoReg;
output MemWrite;
output MemRead ;
output RegWrite;
output Branch;
output ALUOp1;
output ALUOp2;

assign RegDst = (~op0)&(~op1)&(~op2)&(~op3)&(~op4)&(~op5);
assign ALUSrc = (((op0)&(op1)&(~op2)&(~op3)&(~op4)&(op5))| ((op0)&(op1)&(~op2)&(op3)&(~op4)&(op5)));
assign MemtoReg = ((op0)&(op1)&(~op2)&(~op3)&(~op4)&(op5));
assign RegWrite = ((~op0)&(~op1)&(~op2)&(~op3)&(~op4)&(~op5))|((op0)&(op1)&(~op2)&(~op3)&(~op4)&(op5));
assign MemRead = ((op0)&(op1)&(~op2)&(~op3)&(~op4)&(op5));
assign MemWrite = ((op0)&(op1)&(~op2)&(op3)&(~op4)&(op5));
assign Branch = ((~op0)&(~op1)&(op2)&(~op3)&(~op4)&(~op5));
assign ALUOp1 = ((~op0)&(~op1)&(~op2)&(~op3)&(~op4)&(~op5));
assign ALUOP2 = ((~op0)&(~op1)&(op2)&(~op3)&(~op4)&(~op5));

终端模块在这些代码中，控件具有名为“ ALUOp1”和“ ALUOp2”的两个输出，而ALUControl具有名为“ ALUOp”的输入是2位向量.. ALUOp的位是ALUOp1，另一个是ALUOp2 ...我该怎么做？

解决方法

代替：

output ALUOp1;
output ALUOp2;

你要：

output [1:0] ALUOp;

wire ALUOp1;
wire ALUOp2;

assign ALUOp = {ALUOp2,ALUOp1};

它使用我在上一个问题的答案中提到的串联运算符。