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我已经编写了向上/向下计数器并为可设置的起点创建了代码。到目前为止一切顺利，但我想不出如何将它添加到柜台。我必须强调我对 Verilog 和类似语言完全陌生。

//UTILS

reg  [2:0] delay;
wire clock;


reg[3:0] tens;
reg[3:0] units;


wire[5:0] number; 
reg[13:0] shift;    
integer i;


//ASSIGNS
assign number[5:0] = SW[5:0];
assign up = SW[7];
assign start = SW[6];



//PRESCALER
always@ (posedge MCLK)
 begin
    delay <= delay + 1;
 end

assign clock = &delay;


//MAIN COUNTER 
always@ (posedge clock)
begin
     if (start)
        begin
            if (up) //going up
                begin
                    if (units == 4'd3 && tens == 4'd6)
                        begin       //63 reached
                            units <= 0;
                            tens <=0;
                        end
                    if (units==4'd9) 
                       begin        //x9 reached                
                            units <= 0;
                            tens <= tens + 1;
                       end
                    else
                        units <= units + 1; //typical case
                end
            else    //goin down
                begin
                    if (units == 4'd0)      
                        if ( tens ==4'd0)   //00 reached back to 63
                            begin
                                units <= 4'd3;
                                tens <= 4'd6;
                            end
                        else
                            begin           //x0 reached
                                tens <= tens-1;
                                units <= 4'd9;
                            end
                    else
                        begin               //typical case
                            units <= units -1;
                        end
                end
        end 
end             //MAIN COUNTER END

在这里我不知道如何合并这两个部分，如果 start always@posedge clock / counting / else /* 几乎功能性地更改数字（当发生更改时立即）*/

将其添加到 if(start) else似乎可以完成工作，但仅在频率相当低的时钟的正边沿上。据我所知，我不能在两个不同的 ALWAYS@ 中使用一个 reg。 

 /* // Clear previous number and store new number in shift register
          shift[13:6] = 0;
          shift[5:0] = number;

          //BINARY TO BCD                               
          for (i=0; i<6; i=i+1) 
            begin
                 if (shift[9:6] >= 5)
                    shift[9:6] = shift[9:6] + 3;

                 if (shift[13:10] >= 5)
                    shift[13:10] = shift[13:10] + 3;
                 shift = shift << 1; 
            end
        units <=  shift[9:6];
        tens <= shift[13:10];

*/

dek7seg 是 7 段显示，100% 没问题（教授的代码）。

dek7seg ss1(
 .bits(units[3:0]),
 .seg(DISP1[6:0])
);

dek7seg ss10(
 .bits(tens[3:0]),
 .seg(DISP2[6:0])
);



endmodule
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1 回答 1

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您正在使用派生时钟来控制您的MAIN COUNTER。而是使用主时钟MCLK并将逻辑delay用作条件语句。

由于您想在 的更改上存储一个新值number，那么您将需要存储以前的number值并进行比较。

根据您的描述，您的代码应如下所示：

//MAIN COUNTER 
always@ (posedge MCLK)
begin
  if (start && &delay)
  begin
      /* your up/down logic here */
  end
  else if (number != prev_number)
  begin // Clear previous number and store new number
    prev_number <= number;
    units <= new_units;
    tens  <= new_tens;
  end
end

// Calculate new units and tens from number
always @* begin
  shift[13:6] = 0;
  shift[5:0] = number;

  //BINARY TO BCD                               
  for (i=0; i<6; i=i+1) begin
    if (shift[9:6] >= 5)
      shift[9:6] = shift[9:6] + 3;

    if (shift[13:10] >= 5)
      shift[13:10] = shift[13:10] + 3;

    shift = shift << 1; 
  end

  new_units = shift[9:6];
  new_tens  = shift[13:10];
end
于 2016-01-03T15:40:42.210 回答