HDLBits Day12 count clock 做一个钟表

Posted 者乎之类的

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了HDLBits Day12 count clock 做一个钟表相关的知识,希望对你有一定的参考价值。


1.BCD码进位时,判断条件是 if(m < 8’h59),这里是用16进制数表示,即4位二进制数表示5,四位二进制数表示9,BCD数实际上就是十六进制数,不过是人为设置满10进1.
自己写的:

module top_module(
    input clk,
    input reset,
    input ena,
    output pm,
    output [7:0] hh,
    output [7:0] mm,
    output [7:0] ss);
    
    always@(posedge clk )
        if (reset) ss <= 8'h00;
   		else if(ena) begin
        if(ss == 8'h59)
           ss <= 8'h00;
        else   begin
            if(ss[3:0] < 4'h9)
               ss[3:0] <= ss[3:0] + 1'h1; 
             else
              begin
               ss[3:0] <= 0;
               ss[7:4] <= ss[7:4] + 1'h1;
              end 
        end
        end
    always@(posedge clk)
        if (reset) mm <= 8'h00;
   		else if(ena) begin
            if(ss == 8'h59)
                if(mm == 8'h59)
            mm <= 8'h00;
            else  
                if(mm[3:0] < 4'h9)
                                        begin
                                            mm[3:0] <= mm[3:0] + 1'h1; 
                                        end
                                    else
                                        begin
                                           mm[3:0] <= 0;
                                            mm[7:4] <= mm[7:4] + 1'h1;
                                        end 
                                end
       
    always@(posedge clk )
        if (reset) hh <= 8'h12;
   		else if(ena) begin
            if(mm == 8'h59 && ss == 8'h59) begin
                if(hh == 8'h12)
            hh <= 8'h01;
            else  
                if(hh[3:0] < 4'h9)
                                        begin
                                            hh[3:0] <= hh[3:0] + 1'h1; 
                                        end
                                    else
                                        begin
                                            hh[3:0] <= 0;
                                            hh[7:4] <= hh[7:4] + 1'h1;
                                        end 
            end
                                
        end
    always@(posedge clk )
             if (reset) pm <= 0;
    else if(hh ==  8'h11 && mm == 8'h59 && ss == 8'h59)
        pm =!pm;
        

endmodule

时分秒分开写的,以为这样理解起来简单。至于为什么用同步复位,异步复位时序不对,同步复位时序对了,就这样,题面里没有说明白,答案是这样。
人家写的:

module top_module 
    (
        input clk,
        input reset,
        input ena,
        output pm,
        output [7:0] hh,
        output [7:0] mm,
        output [7:0] ss
    );

    reg p;	//0 was am, 1 was pm
    reg [7:0] h;
    reg [7:0] m;
    reg [7:0] s;

    always @ (posedge clk)
        begin
            if(reset)
                begin
                    p <= 0;
                    h <= 8'h12;
                    m <= 8'h00;
                    s <= 8'h00;
                end
            else
                begin
                    if(ena)
                        begin
                            if(s < 8'h59)
                                begin
                                    if(s[3:0] < 4'h9)
                                        begin
                                            s[3:0] <= s[3:0] + 1'h1; 
                                        end
                                    else
                                        begin
                                            s[3:0] <= 0;
                                            s[7:4] <= s[7:4] + 1'h1;
                                        end 
                                end
                            else
                                begin
                                    s <= 0;
                                    if(m < 8'h59)
                                        begin
                                            if(m[3:0] < 4'h9)
                                                begin
                                                    m[3:0] <= m[3:0] + 1'h1; 
                                                end 
                                            else
                                                begin
                                                    m[3:0] <= 0;
                                                    m[7:4] <= m[7:4] + 1'h1;
                                                end
                                        end
                                    else
                                        begin
                                            m <= 1'h0;
                                            if(h == 8'h11)
                                                p = !p;
                                            if(h < 8'h12)
                                                begin
                                                    if(h[3:0] < 4'h9)
                                                        h[3:0] <= h[3:0] + 1'h1;
                                                    else
                                                        begin
                                                            h[3:0] <= 4'h0;
                                                            h[7:4] <= h[7:4] + 1'h1;
                                                        end
                                                end
                                            else
                                                begin
                                                   h <= 1'h1; 
                                                end
                                        end
                                end
                        end
                end
        end

    assign pm = p;
    assign hh = h;
    assign mm = m;
    assign ss = s;

endmodule 

tb文件:

`timescale 1ns / 1ns
module tb();

	reg clk,reset,ena;
	wire pm;
	wire[7:0] hh,mm,ss;
	syn_fifo syn_fifo 
    (
         clk,
        reset,
        ena,
         pm,
         hh,
         mm,
        ss
    );
	initial begin 
		clk = 1'b0;
		forever #10 clk = ~clk;
	end
	
	initial begin
		reset = 1'b0;
		ena = 1'b1;
		#100 reset = 1'b1;
		#100 reset = 1'b0;
	end
	
endmodule

以上是关于HDLBits Day12 count clock 做一个钟表的主要内容,如果未能解决你的问题,请参考以下文章

HDLBits --- Count clock

HDLBits——Procedures

HDLBits——Vectors

DAY1_homework

std::chrono::system_clock.now().time_since_epoch().count() 的值是不是单调增加?

day1-python基础