verilog - 如何识别同步复位（在 verilog 中）

Question

我是 EDA 的新手，我有以下 verilog 代码，我需要清楚地识别同步复位。

module test(clk,d,rst,a);
  input clk,d,rst;
  output reg a;

  always @(posedge clk)
  begin
    if(rst)
      a <= 1'b0;
    else
      a <= 1'b1; // assigned to a constant
  end
endmodule

'rst' 是重置（同步）吗？

Answer 1

接受的答案是错误的，因为第二个代码示例实际上是一个组合代码并且根本不使用时钟，我们需要实现一个顺序代码。第一个代码示例是同步复位：

//Synchronous Reset
module test(clk,d,rst,a);
  input clk,d,rst;
  output reg a;

  always @(posedge clk) 
  begin
    if(rst) // In order to execute this line, clk and reset both have to be in posedge.  
      a <= 1'b0;
    else
      a <= 1'b1; // assigned to a constant
  end
endmodule

第二个代码示例是异步复位：

//Asynchronous Reset
module test(clk,d,rst,a);
  input clk,d,rst;
  output reg a;

  always @(posedge clk, posedge rst) 
  begin    
    if(rst) // In order to execute this line, rst has to be in posedge(clk's value doesn't
            // matter here, it can be either posedge or negedge, what's important is rst's value). 
      a <= 1'b0;
    else
      a <= 1'b1; // assigned to a constant
  end
endmodule

Answer 2

以下是您的同步和异步重置代码。

//Synchronous Reset
module test(clk,d,rst,a);
  input clk,d,rst;
  output reg a;

  always @(posedge clk) //This clock makes the reset synchronized to a clock signal.
  begin
    if(rst)
      a <= 1'b0;
    else
      a <= 1'b1; // assigned to a constant
  end
endmodule

//Asynchronous
module test(clk,d,rst,a);
  input clk,d,rst;
  output reg a;

  always @* //No clock to synchronize with. 
  begin     //Reset signal will drive anytime a input value changes
    if(rst)
      a <= 1'b0;
    else
      a <= 1'b1; // assigned to a constant
  end
endmodule