HDLBits Count clock-12hour clock

二刷HDLBits发现自己可以独立做出这道题，来分享一下自己的解法。

这道题主要难点就是计数器的十位和个位需要分开计数，如果直接按照八位一起计数就会产生十六进制的结果。

module top_module(
    input clk,
    input reset,
    input ena,
    output pm,
    output [7:0] hh,
    output [7:0] mm,
    output [7:0] ss); 
	wire [2:0]num2;
    assign num2[0] = ena;
    assign num2[1] = (ss[7:4] == 4'd5) && (ss[3:0] == 4'd9) && ena;
    assign num2[2] = (mm[7:4] == 4'd5 && mm[3:0] == 4'd9) && (ss[7:4] == 4'd5 && ss[3:0] == 4'd9) && ena;
    always @ (posedge clk)begin
        if(reset)begin
            pm <= 1'b0;
            hh[7:4] <= 4'd1;
            hh[3:0] <= 4'd2;
        end
        else if(num2[2])begin
            if(hh[7:4] == 4'd1 && hh[3:0] == 4'd2)begin
                hh[7:4] <= 4'd0;
                hh[3:0] <= 4'd1;
            end
            else begin
                if(hh[7:4] == 4'd1 && (hh[3:0] == 4'd1 || hh[3:0] == 4'd0))begin
                    hh[3:0] <= hh[3:0] + 1'b1;
                end
                else if(hh[7:4] == 4'd0 && hh [3:0] == 4'd9)begin
                    hh[7:4] <= 4'd1;
                    hh[3:0] <= 4'd0;
                end
                else begin
                    hh [3:0]<= hh[3:0] + 1'b1;
                end
                if(hh[7:4] == 4'd1 && hh[3:0] == 4'd1 && num2[2])begin
                    pm <= ~pm;
                end
            end
        end
    end
    fncount s1(.clk(clk), .reset(reset), .ena(num2[0]), .num1(ss));
    fncount m1(.clk(clk), .reset(reset), .ena(num2[1]), .num1(mm));
    
endmodule
module fncount(
    input clk,
    input reset,
    input ena,
    output reg[7:0]num1);
    always @ (posedge clk)begin
        if(reset)begin
            num1 <= 8'd0;
        end
        else if(ena)begin
            if(num1[7:4] == 4'd5 && num1[3:0] == 4'd9)begin
                num1[7:4] <= 4'd0;
                num1[3:0] <= 4'd0;
            end
            else begin
                if(num1[3:0] == 4'd9)begin
                    num1[3:0] <= 4'd0;
                    num1[7:4] <= num1[7:4] + 1'b1;
                end
                else begin
                    num1[3:0] <= num1[3:0] + 1'b1;
                end
               
            end
        end
    end
endmodule

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