我正在Verilog中创建一个Basys2板,以在二进制中显示七个段显示。我使用的是开关0-3,并想使用4个按钮,以便按下按钮[0]时,开关的二进制数(向上1,向下显示0)。但是,我坚持如何将按钮与二进制和LED集成。我是Verilog的新手,遇到了一些麻烦。有什么建议么?
module final(btn,clk,sw,cathodes,anodes,led,rst);
input clk,rst;
output [6:0] cathodes;
output [7:0] led;
output [3:0] anodes;
input [3:0] sw;
input [3:0] btn;
reg [6:0] cathodes;
reg [15:0] dig;
reg [3:0] anodes;
reg slow_clock;
integer count;
reg [7:0] led;
always @ (posedge clk)
create_slow_clock(clk, slow_clock);
卡在这里如何分配LED:
always @(posedge clk)
begin
if (btn[0:3]);
begin
led[0]
led[1]
led[2]
led[3]
也不确定这是正确的:
always @ (posedge slow_clock)
begin
led=~led;
if (rst == 0) anodes = 4'b1111;
else
begin
case (btn)
0: anodes = 4'b0111;
1: anodes = 4'b1011;
2: anodes = 4'b1101;
3: anodes = 4'b1110;
default: anodes = 4'b1111;
endcase
cathodes = calc_cathode_value(dig);
end
end
。
function[6:0] calc_cathode_value;
input [15:0] dig;
begin
case (dig)
0: calc_cathode_value = 8'b00000011;
1: calc_cathode_value = 8'b10011111;
2: calc_cathode_value = 8'b00100101;
3: calc_cathode_value = 8'b00001101;
4: calc_cathode_value = 8'b10011001;
5: calc_cathode_value = 8'b01001001;
6: calc_cathode_value = 8'b01000001;
7: calc_cathode_value = 8'b00011111;
8: calc_cathode_value = 8'b00000001;
9: calc_cathode_value = 8'b00001001;
'hA: calc_cathode_value = 8'b00010001;
'hb: calc_cathode_value = 8'b11000001;
'hC: calc_cathode_value = 8'b01100011;
'hd: calc_cathode_value = 8'b10000101;
'hE: calc_cathode_value = 8'b01100001;
'hF: calc_cathode_value = 8'b01110001;
default: calc_cathode_value = 8'b0000001;
endcase
end
endfunction
task create_slow_clock;
input clock;
inout slow_clock;
integer count;
begin
if (count > 250000)
begin
count = 0;
slow_clock = ~slow_clock;
end
count = count + 1;
end
endtask
endModule
您的信号btn定义为4位,这意味着案例语句需要表示从0到f的所有16个可能性。考虑一下,当将一个按钮推到以上时,需要发生什么同一时间。如果您仅关心按下一个按钮中的1个案例,则您的案例值需要为0,1,2,4,因为每个位位置代表另一个2^n值。