ShiftRegister Verilog HDL 输出给出 xxxxxxx
ShiftRegister Verilog HDL Output giving xxxxxxx
我正尝试在 Verilog HDL 中创建一个 64 位移位寄存器。当我在测试平台中尝试代码时,我只得到 xxxxxx 作为输出,直到所有位都被移位。我不知道是什么问题。
这是我的测试平台代码和结果:
module ShiftRegister (shift_out, clk, shift_in); //module ports
parameter n = 64; //Parameter n declared to store 64
input [n-1:0] shift_in; //64-bit input shift_in
input clk; //Input clock
output [n-1:0] shift_out; //64-bit output shift_out
reg [n-1:0] ff; //64-bit flipflop
assign shift_out = ff [n-1:0]; //give the output of the 64th bit
//The operation of verilog:
always @ (posedge clk) //Always at the rising edge of the clock
begin
ff <= ff << 1; //Shift bits to the left by 1
ff[0] <= shift_in; //Take the input bits and give it to the first flipflop
end
endmodule //ShiftRegister module
///Testbench\\
module ShiftRegister_tb; //Module shiftRegister_tb
parameter n = 64; //Parameter n declared to store 64
reg [n-1:0] shift_in; //64-bit register input shift_in
reg clk, rst; //register clock
wire [n-1:0] shift_out; //64-bit wire output shift_out
ShiftRegister DUT(shift_out, clk, shift_in); //Calling the module
initial
begin
clk = 0; //clock = 0 initally
shift_in = 64'd34645767785344; //Random decimal number to test the code
#100;
end
always #50 clk =~clk; //invert the clock input after 50ps
endmodule //ShiftRegister testbench
您将 ff
声明为 reg
,reg
的默认值为 x
。在时钟的第一个姿势之前,ff
的所有 64 位都是 x
(未知)。在时钟的第一个 posedge 之后,ff[0]
变为 0,因为 shift_in[0]
为 0。依此类推,直到达到 64 个时钟,然后所有 ff
位都为 0。 shift_out
紧随 ff
.
通常,您的设计也会有一个复位信号。如果你有一个,你可以在开始时断言重置,并在重置期间将 ff
分配给 0。这是重置后的样子:
module ShiftRegister (shift_out, clk, shift_in, rst); //module ports
parameter n = 64; //Parameter n declared to store 64
input rst;
input [n-1:0] shift_in; //64-bit input shift_in
input clk; //Input clock
output [n-1:0] shift_out; //64-bit output shift_out
reg [n-1:0] ff; //64-bit flipflop
assign shift_out = ff [n-1:0]; //give the output of the 64th bit
always @ (posedge clk or posedge rst) //Always at the rising edge of the clock
begin
if (rst) begin
ff <= 0;
end else begin
ff <= ff << 1; //Shift bits to the left by 1
ff[0] <= shift_in; //Take the input bits and give it to the first flipflop
end
end
endmodule
module ShiftRegister_tb; //Module shiftRegister_tb
parameter n = 64; //Parameter n declared to store 64
reg [n-1:0] shift_in; //64-bit register input shift_in
reg clk, rst; //register clock
wire [n-1:0] shift_out; //64-bit wire output shift_out
ShiftRegister DUT(shift_out, clk, shift_in,rst); //Calling the module
initial
begin
clk = 0; //clock = 0 initally
rst = 1;
shift_in = 64'd34645767785344; //Random decimal number to test the code
#100;
rst = 0;
#50_000 $finish;
end
always #50 clk =~clk; //invert the clock input after 50ps
endmodule
我正尝试在 Verilog HDL 中创建一个 64 位移位寄存器。当我在测试平台中尝试代码时,我只得到 xxxxxx 作为输出,直到所有位都被移位。我不知道是什么问题。 这是我的测试平台代码和结果:
module ShiftRegister (shift_out, clk, shift_in); //module ports
parameter n = 64; //Parameter n declared to store 64
input [n-1:0] shift_in; //64-bit input shift_in
input clk; //Input clock
output [n-1:0] shift_out; //64-bit output shift_out
reg [n-1:0] ff; //64-bit flipflop
assign shift_out = ff [n-1:0]; //give the output of the 64th bit
//The operation of verilog:
always @ (posedge clk) //Always at the rising edge of the clock
begin
ff <= ff << 1; //Shift bits to the left by 1
ff[0] <= shift_in; //Take the input bits and give it to the first flipflop
end
endmodule //ShiftRegister module
///Testbench\\
module ShiftRegister_tb; //Module shiftRegister_tb
parameter n = 64; //Parameter n declared to store 64
reg [n-1:0] shift_in; //64-bit register input shift_in
reg clk, rst; //register clock
wire [n-1:0] shift_out; //64-bit wire output shift_out
ShiftRegister DUT(shift_out, clk, shift_in); //Calling the module
initial
begin
clk = 0; //clock = 0 initally
shift_in = 64'd34645767785344; //Random decimal number to test the code
#100;
end
always #50 clk =~clk; //invert the clock input after 50ps
endmodule //ShiftRegister testbench
您将 ff
声明为 reg
,reg
的默认值为 x
。在时钟的第一个姿势之前,ff
的所有 64 位都是 x
(未知)。在时钟的第一个 posedge 之后,ff[0]
变为 0,因为 shift_in[0]
为 0。依此类推,直到达到 64 个时钟,然后所有 ff
位都为 0。 shift_out
紧随 ff
.
通常,您的设计也会有一个复位信号。如果你有一个,你可以在开始时断言重置,并在重置期间将 ff
分配给 0。这是重置后的样子:
module ShiftRegister (shift_out, clk, shift_in, rst); //module ports
parameter n = 64; //Parameter n declared to store 64
input rst;
input [n-1:0] shift_in; //64-bit input shift_in
input clk; //Input clock
output [n-1:0] shift_out; //64-bit output shift_out
reg [n-1:0] ff; //64-bit flipflop
assign shift_out = ff [n-1:0]; //give the output of the 64th bit
always @ (posedge clk or posedge rst) //Always at the rising edge of the clock
begin
if (rst) begin
ff <= 0;
end else begin
ff <= ff << 1; //Shift bits to the left by 1
ff[0] <= shift_in; //Take the input bits and give it to the first flipflop
end
end
endmodule
module ShiftRegister_tb; //Module shiftRegister_tb
parameter n = 64; //Parameter n declared to store 64
reg [n-1:0] shift_in; //64-bit register input shift_in
reg clk, rst; //register clock
wire [n-1:0] shift_out; //64-bit wire output shift_out
ShiftRegister DUT(shift_out, clk, shift_in,rst); //Calling the module
initial
begin
clk = 0; //clock = 0 initally
rst = 1;
shift_in = 64'd34645767785344; //Random decimal number to test the code
#100;
rst = 0;
#50_000 $finish;
end
always #50 clk =~clk; //invert the clock input after 50ps
endmodule