题解 | #可置位计数器#
可置位计数器
https://www.nowcoder.com/practice/b96def986e29475e8100c213178b73a8
`timescale 1ns/1ns
module count_module(
input clk,
input rst_n,
input set,
input [3:0] set_num,
output reg [3:0]number,
output reg zero
);
reg [3:0]num_reg;
always@(posedge clk or negedge rst_n)
if (rst_n == 1'b0)
num_reg <= 4'b0;
else if (num_reg == 4'd15)
num_reg <= 4'b0;
else if (set == 1'b1)
num_reg <= set_num;
else
num_reg <= num_reg + 1'b1;
always@(posedge clk or negedge rst_n)
if (rst_n == 1'b0)
number <= 4'b0;
else
number <= num_reg;
always@(posedge clk or negedge rst_n)
if (rst_n == 1'b0)
zero <= 1'b0;
else if (num_reg == 4'd0)
zero <= 1'b1;
else
zero <= 1'b0;
endmodule
题中给出的计数器信号为啥要再打一拍啊,感觉有顶啊不合乎常理。
