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[英]Verilog Error: Can't elaborate user hierarchy “counter:counter”
[英]Can't fit settability in counter Verilog
我已经编写了计数器,并为可设置的起点创建了代码。 到目前为止,还不错,但是我想不起来如何将其添加到计数器中。 我必须强调,我是Verilog和类似语言的新手。
//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,如果总是启动always @ posege Clock / 计数 /否则/ *几乎在功能上(当发生更改时立即更改)* /
将其添加到if(start)else中似乎可以完成工作,但仅在相当低频的时钟的上升沿进行。 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,精度为100%(教授代码)。
dek7seg ss1(
.bits(units[3:0]),
.seg(DISP1[6:0])
);
dek7seg ss10(
.bits(tens[3:0]),
.seg(DISP2[6:0])
);
endmodule
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delay
逻辑用作条件语句。
由于要在number
更改时存储一个新值,因此需要存储以前的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
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