tag 标签: 组合环

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  • 热度 26
    2011-4-22 13:39
    5462 次阅读|
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      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 dout;     parameter IDLE  = 3'd0; parameter STA1  = 3'd1;   //一段式写法 reg cstate; reg 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 cstate,nstate; reg 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 cstate,nstate; reg 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。