原创 两段式状态机不可能完成的任务

两段式状态机不可能完成的任务（特权）

         最近折腾状态机，发现一个小任务对于两段式状态机写法是不可能完成的。这个小任务很简单，先看用一段式状态机实现的代码：

module test(

            clk,rst_n,

            din,dout

        );

input clk;

input rst_n;   

input din;

output[3:0] dout;  

parameter IDLE  = 3'd0;

parameter STA1  = 3'd1;

//一段式写法

reg[2:0] cstate;

reg[3:0] cnt;

always @(posedge clk or negedge rst_n)

    if(!rst_n) cstate <= IDLE;

    else begin

        case(cstate)

            IDLE: begin

                    cnt <= 4'd0;

                    if(din) cstate <= STA1;

                    else cstate <= IDLE;       

                end

            STA1: begin

                    cnt <= cnt+1'b1;

                    if(cnt == 4'd10) cstate <= IDLE;

                    else cstate <= STA1;

                end

            default: cstate <= IDLE;

        endcase

    end

assign dout = cnt;

endmodule

         同样的，用三段式状态机也能够实现这个功能：

//三段式写法

reg[2:0] cstate,nstate;

reg[3:0] cnt;

always @(posedge clk or negedge rst_n)

    if(!rst_n) cstate <= IDLE;

    else cstate <= nstate;

always @(cstate or din or cnt) begin

    case(cstate)

        IDLE:   if(din) nstate = STA1;

                else nstate = IDLE;    

        STA1:   if(cnt == 4'd10) nstate = IDLE;

                else nstate = STA1;

        default: nstate = IDLE;

    endcase

end

always @(posedge clk or negedge rst_n)

    if(!rst_n) cnt <= 4'd0;

    else begin

        case(nstate)

            IDLE:   cnt <= 4'd0;

            STA1:   cnt <= cnt+1'b1;

            default: ;

        endcase

    end

         严格来看，上面的三段式状态机相比于一段式会滞后一个时钟周期。但是我们的重点不在这里，大家大可以不必钻这个牛角尖。另外，这个实例实现的功能本身也没 有什么意义，当然也是可以用别的更简单（不需要状态机）的方式实现，但是你可以想象成这是实际应用中状态机各种复杂输出的一部分。

         而如果大家希望用两段式状态机实现这个功能，或许会这么写：

//两段式写法

reg[2:0] cstate,nstate;

reg[3:0] cnt;

always @(posedge clk or negedge rst_n)

    if(!rst_n) cstate <= IDLE;

    else cstate <= nstate;

always @(cstate or din or cnt) begin

    case(cstate)

        IDLE: begin

                cnt = 4'd0;

                if(din) nstate = STA1;

                else nstate = IDLE;    

            end

        STA1: begin

                cnt = cnt+1'b1;

                if(cnt == 4'd10) nstate = IDLE;

                else nstate = STA1;

            end

        default: nstate = IDLE;

    endcase

end

         如果大家有兴趣对三中代码方式都做一下仿真，会发现一些有意思的问题，尤其两段式状态机最终根本无法退出STA1，计数器cnt也会死在那里。究其根本原因，可大有学问。在编译工程后，出现了数条类似下面的warning：

Warning: Found combinational loop of 2 nodes

    Warning: Node "Add0~2"

    Warning: Node "cnt~9"

        何为combinational loop？让handbook来解释吧，看不懂英文的可别怪我~_~

Combinational loops are among the most common causes of instability and unreliability in digital designs. They should be avoided whenever possible. In a synchronous design, feedback loops should include registers. Combinational loops generally violate synchronous design principles by establishing a direct feedback loop that contains no registers. For example, a combinational loop occurs when the left-hand side of an arithmetic expression also appears on the right-hand side in HDL code. A combinational loop also occurs when you feed back the output of a register to an asynchronous pin of the same register through combinational logic, as shown in Figure 5–1.

        没有寄存器打一拍的这种combinational loop（组合环）是一种不推荐的设计方式，就如两段式状态机所实现的效果，甚至最终无法实现功能要求。同样的功能，一段式和三段式状态机之所以能够解决 这个问题，就是避免了在纯组合逻辑中涉及这个反馈逻辑。在初学verilog时，我们常提的latch（锁存器），其实也是combinational loop的一个特例。