// 显示输入的密码 if (delete_count == 2'b00 && password_count > 0) begin // 将输入的密码存储到display寄存器中 display <= {input_password,4'b0000}[(count4)+3:(count4)]; count <= count + 2'b01;last_password_count <= password_count; end else if (delete_count == 2'b01 && last_password_count > 0) begin // 删除输入的密码 display <= display-4'b0001; count <= count - 2'b01; last_password_count <= last_password_count - 1; end;// Error (10170): Verilog HDL syntax error at keshe9.v(60) near text "["; expecting ";"还是这个错误改了之后的代码是什么
时间: 2023-07-16 12:13:08 浏览: 79
你需要在 `display` 寄存器的赋值语句中加入一个分号,代码如下所示:
```
// 显示输入的密码
if (delete_count == 2'b00 && password_count > 0) begin
// 将输入的密码存储到display寄存器中
display <= {input_password,4'b0000}[(count4)+3:(count4)];
count <= count + 2'b01;
last_password_count <= password_count;
end else if (delete_count == 2'b01 && last_password_count > 0) begin
// 删除输入的密码
display <= display-4'b0001;
count <= count - 2'b01;
last_password_count <= last_password_count - 1;
end; // 加入分号
```
这样就可以解决 Verilog HDL syntax error at keshe9.v(60) near text "["; expecting ";" 的问题了。
相关问题
// 显示输入的密码if (delete_count == 2'b00 && password_count > 0) begin // 将输入的密码存储到display寄存器中 display <= {input_password,4'b0000}[(count4)+3:(count4)]; count <= count + 2'b01; last_password_count <= password_count;end else if (delete_count == 2'b01 && last_password_count > 0) begin // 删除输入的密码 display <= display-4'b0001; count <= count - 2'b01; last_password_count <= last_password_count - 1;end; // 加入分号 Error (10170): Verilog HDL syntax error at keshe9.v(60) near text "["; expecting ";"这个错误改完之后的代码
如果你想在Verilog中使用花括号来创建一个选择器,你需要在使用选择器前加上一个反斜杠,如下所示:
```
// 显示输入的密码
if (delete_count == 2'b00 && password_count > 0) begin
// 将输入的密码存储到display寄存器中
display <= {input_password,4'b0000}[((count4)+3):(count4)];
count <= count + 2'b01;
last_password_count <= password_count;
end else if (delete_count == 2'b01 && last_password_count > 0) begin
// 删除输入的密码
display <= display-4'b0001;
count <= count - 2'b01;
last_password_count <= last_password_count - 1;
end;
```
这样,Verilog就会将花括号视为选择器,而不是一个新的语法结构。
module top_module; wsh wsh_inst(); efg efg_inst(); divider divider_inst(); assign divider_inst.dividend = wsh_inst.subdivision; assign divider_inst.divisor = efg_inst.count; endmodule module wsh (clk,rst_n,A,B,subdivision); input wire A,B; input clk; input rst_n; output reg [15:0] subdivision; reg [1:0] pre_state; reg [1:0] cur_state; always @(posedge clk or negedge rst_n) begin if(!rst_n) subdivision <=1'b0; else begin if (pre_state == 2'b00 && cur_state == 2'b01) subdivision <= subdivision + 1'b1; else if(pre_state == 2'b01 && cur_state == 2'b11) subdivision <= subdivision + 1'b1; else if(pre_state == 2'b11 && cur_state == 2'b10) subdivision <= subdivision + 1'b1; else if(pre_state == 2'b10 && cur_state == 2'b00) subdivision <= subdivision + 1'b1; else if(pre_state == 2'b00 && cur_state == 2'b10) subdivision <= subdivision - 1'b1; else if(pre_state == 2'b10 && cur_state == 2'b11) subdivision <= subdivision - 1'b1; else if(pre_state == 2'b11 && cur_state == 2'b01) subdivision <= subdivision - 1'b1; else if(pre_state == 2'b01 && cur_state == 2'b00) subdivision <= subdivision - 1'b1; end end endmodule module efg (A,B,count,clk,rst_n); input wire A; input wire B; input clk; input rst_n; output reg [15:0] count; always @(posedge clk) begin if(!rst_n) count <= 0; else begin count <= count + 1; end end endmodule module divider(clk, subdivision, count, quotient); input clk; input [15:0] subdivision; input [15:0] count; output [15:0] quotient; reg [15:0] dividend; reg [15:0] divisor; reg [15:0] quotient; integer i; always @(posedge clk) begin dividend <= subdivision; divisor <= count; quotient <= 0; for (i = 0; i < 16; i = i + 1) begin dividend <= dividend - divisor; quotient <= {quotient[14:0], dividend[15]}; dividend <= dividend << 1; end end endmodule根据所给代码写一个testbench
以下是一个简单的testbench示例,使用Verilog语言:
```
module top_module_tb;
// Inputs
reg clk;
reg rst_n;
reg A;
reg B;
// Outputs
wire [15:0] quotient;
// Instantiate the design under test
top_module dut (
.clk(clk),
.rst_n(rst_n),
.A(A),
.B(B),
.quotient(quotient)
);
// Generate clock signal
always #5 clk = ~clk;
// Initialize inputs
initial begin
clk = 0;
rst_n = 0;
A = 0;
B = 0;
end
// Reset the DUT
initial begin
#10 rst_n = 1;
end
// Test case 1
initial begin
#20 A = 4'b0000; // Dividend = 0
B = 4'b0000; // Divisor = 0
#20 A = 4'b1010; // Dividend = 10
B = 4'b0010; // Divisor = 2
#100; // Wait for division to complete
$display("Quotient = %d", quotient);
#20 $finish;
end
endmodule
```
在测试用例中,我们将输入A和B设置为一些特定的值,并在一段时间后检查输出商的值。注意,在testbench中,我们使用了一个时钟信号来驱动DUT,并在每个上升沿之后对输入进行更改。我们还使用了一些延迟语句来模拟各种等待时间。
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