用verilog HDL语法编写4

Question

请编写程序,可以给出仿真图，题目如下图，请按题目要求编写，并给出程序。

Answer 1

module led_play
(
    input    clk,            //0.5MHz 时钟
    input    key0,            //按键，低电平按下，高电平抬起
    input    key1,            //按键，低电平按下，高电平抬起
    input    key2,            //按键，低电平按下，高电平抬起
    output    [7:0]    ledr    //8 个流水灯，左边高位 ledr[7] ，右边低位 ledr[0] 
);
//三个按键，假定优先级 key0>key1>key2
parameter CLK_IN_FREQ    = 500*1000;    //0.5MHz

reg            clk1s    = 0;        //1Hz 秒脉冲
reg            clk1_ck    = 0;        //1Hz 方波，闪烁脉冲
reg    [31:0]    cnt        = 0;
reg    [7:0]    led_cnt = 0;
reg    [1:0]    led_st    = 0;        //按键状态

assign    ledr = clk1_ck == 0 ? led_cnt:~led_cnt;
 
always @(posedge clk)
begin
    if(cnt < CLK_IN_FREQ-1)
    begin
        cnt        <= cnt + 1;
        clk1s    <= 0;
    end
    else
    begin
        cnt        <= 0;
        clk1s    <= 1;
    end
end

always @(posedge clk)
begin
    if(cnt == CLK_IN_FREQ/2-1 || cnt == CLK_IN_FREQ-1)
    begin
        clk1_ck    <= ~clk1_ck;
    end
end

always @(posedge clk or negedge key0)
begin
    if( key0 == 0 )            //key0 按下
        led_st <= 0;
    else if( key1 == 0 )    //key1 按下
        led_st <= 1;
    else if( key2 == 0 )    //key1 按下
        led_st <= 2;
end

always @(posedge clk or negedge key0)
begin
    if( key0 == 0 )            //key0 按下
        led_cnt <= 8'b1111_1111;
    else if(key1 == 0)
        led_cnt    <= 8'b1111_1110;
    else if(key2 == 0)
        led_cnt    <= 8'b0111_1111;
    else if(clk1s == 1)        //1Hz 秒脉冲
    begin
        if( led_st == 1 )            //key1 按下
            led_cnt    <= {led_cnt[6:0],led_cnt[7]};    //左循环
        else if( led_st == 2 )        //key2 按下
            led_cnt    <= {led_cnt[0],led_cnt[7:1]};    //右循环
    end
end
endmodule