题解 | 自动售卖机

module sale(
   input                clk   ,
   input                rst_n ,
   input                sel   ,//sel=0,5$dranks,sel=1,10&=$drinks
   input          [1:0] din   ,//din=1,input 5$,din=2,input 10$
 
   output   reg  [1:0] drinks_out,//drinks_out=1,output 5$ drinks,drinks_out=2,output 10$ drinks
   output	reg        change_out   
);

    //生成要给多少钱*5
    wire [1:0]money_pay;
    assign money_pay=sel+1;

    
    reg [1:0]money_get;     //时序逻辑，前面已经收到多少钱
    wire [1:0]money_get_now;//当前收到多少钱
    wire enough;
    always@(posedge clk,negedge rst_n)begin
        if(!rst_n)begin
            money_get<=0;
        end
        else if(enough)
            money_get<=0;
        else begin
            money_get<=money_get+din;
        end
    end

    assign money_get_now=money_get+din;
    assign enough=(money_get_now>=money_pay);

    //输出
    always@(posedge clk,negedge rst_n)begin
        if(!rst_n)begin
            drinks_out<=0;
            change_out<=0;
        end
        else if(enough) begin
            drinks_out<=sel+1;
            change_out<=money_get_now-money_pay;
        end
        else begin
            drinks_out<=0;
            change_out<=0;
        end
    end

endmodule

不用状态机的写法，计算收到的钱和要给的钱，来判断给的钱是否足够enough，进而输出