题解 | #数据串转并电路#
数据串转并电路
https://www.nowcoder.com/practice/6134dc3c8d0741d08eb522542913583d
//本题最坑的地方在于data_a到data_b相对计数器是两级逻辑,但是valid_b拉高相对计数器只需要一级逻辑 //这就导致在计数器为5的时候data_a要寄存到缓存寄存器中,而data_b此时直接寄存缓存寄存器时将遗漏一位 //当然牛客网的题目和仿真也很垃圾,鬼知道他的TB给的是哪些数据,想自己仿真都不行 `timescale 1ns/1ns module s_to_p( input clk , input rst_n , input valid_a , input data_a , output reg ready_a , output reg valid_b , output reg [5:0] data_b ); reg [2:0] counter; reg [5:0] data_reg; always@(posedge clk or negedge rst_n)begin if(!rst_n) ready_a <= 1'b0; else ready_a <= 1'b1; end always@(posedge clk or negedge rst_n)begin if(!rst_n) counter <= 1'b0; else if(valid_a)begin if(counter == 3'd5) counter <= 1'b0; else counter <= counter + 1'b1; end else counter <= counter; end always@(posedge clk or negedge rst_n)begin if(!rst_n) data_reg <= 1'b0; else if(valid_a) data_reg <= {data_a,(data_reg[5:1])}; else data_reg <= data_reg; end always@(posedge clk or negedge rst_n)begin if(!rst_n)begin data_b <= 1'b0; valid_b <= 1'b0; end else if(counter == 3'd5)begin data_b <= {data_a,(data_reg[5:1])}; valid_b <= 1'b1; end else begin data_b <= data_b; valid_b <= 1'b0; end end endmodule