带先行进位的 4 位加法器的 Verilog 测试平台
Verilog test bench for 4-bit adder with Carry Lookahead
老实说,考虑到我只创建过一次幼稚的简单项目,我对 Verilog 尤其是测试台还是有点不熟悉。我不确定如何为我制作的 Verilog 文件制作测试台,所以我无法测试它是否有效。这是我的代码:
`timescale 1ns/1ps
module adder_4bit_cla(sum, Cout, A, B, S);
input [3:0] A, B;
input S;
output [3:0] sum;
output Cout;
wire P0, G0, P1, G1, P3, G3;
wire C4, C3, C2, C1;
assign
P0 = A[0] ^ B[0],
P1 = A[1] ^ B[1],
P2 = A[2] ^ B[2],
P3 = A[3] ^ B[3];
assign
G0 = A[0] & B[0],
G1 = A[1] & B[1],
G2 = A[2] & B[2],
G3 = A[3] & B[3];
assign
C1 = G0 | (P0 & S),
C2 = G1 | (P1 & G0) | (P1 & P0 & S),
C3 = G2 | (P2 & G1) | (P2 & P1 & G0) | (P2 & P1 & P0 & S),
C4 = G3 | (P3 & G2) | (P3 & P2 & G1) | (P3 & P2 & P1 & G0) | (P3 & P2 & P1 & P0 & S);
assign
sum[0] = P0 ^ S,
sum[1] = P1 ^ C1,
sum[2] = P2 ^ C2,
sum[3] = P3 ^ C3;
assign Cout = C4;
endmodule
老实说,我真正需要做的是一个使用先行进位的 4 位加减法器,但我不知道如何开始先行进位,所以我来了。如果有人能帮助我,那就太好了:<
编辑:我已经冷静下来,我终于可以查明确切的问题:测试台的 A 和 B 的值。虽然我可以暴力破解它,但我怎样才能利用循环来增加 A 和 B 以便它像这样:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
同时更新 M
?
使用以下方法改进您的测试平台。
`timescale 1ns / 1ps
module adder_4bit_cla_tb();
// inputs - keep them having reg as the data type
reg [3:0] A, B;
reg S;
// outputs - keep them having wire as the data type
wire [3:0] sum;
wire Cout;
adder_4bit_cla adder_4bit_cla_inst
(
.sum(sum), .Cout(Cout), .A(A), .B(B), .S(S)
);
initial begin
A = 4'd1; B = 4'd2; S = 1'd1;
#10 A = 4'd2; B = 4'd5; S = 1'd0;
#10 A = 4'd5; B = 4'd6; S = 1'd0;
#50 $stop;
end
endmodule
波形结果
可以使用 for 循环提供输入,如下所示。
`timescale 1ns / 1ps
module adder_4bit_cla_tb();
// inputs - keep them having reg as the data type
reg [3:0] A, B;
reg S;
// outputs - keep them having wire as the data type
wire [3:0] sum;
wire Cout;
reg [3:0] i;
adder_4bit_cla adder_4bit_cla_inst
(
.sum(sum), .Cout(Cout), .A(A), .B(B), .S(S)
);
initial begin
for(i = 4'd0; i < 4'd15; i = i + 4'd1) begin
A = i; B = i; S = 1'b0;
#10;
end
#200 $stop;
end
endmodule
波形结果
老实说,考虑到我只创建过一次幼稚的简单项目,我对 Verilog 尤其是测试台还是有点不熟悉。我不确定如何为我制作的 Verilog 文件制作测试台,所以我无法测试它是否有效。这是我的代码:
`timescale 1ns/1ps
module adder_4bit_cla(sum, Cout, A, B, S);
input [3:0] A, B;
input S;
output [3:0] sum;
output Cout;
wire P0, G0, P1, G1, P3, G3;
wire C4, C3, C2, C1;
assign
P0 = A[0] ^ B[0],
P1 = A[1] ^ B[1],
P2 = A[2] ^ B[2],
P3 = A[3] ^ B[3];
assign
G0 = A[0] & B[0],
G1 = A[1] & B[1],
G2 = A[2] & B[2],
G3 = A[3] & B[3];
assign
C1 = G0 | (P0 & S),
C2 = G1 | (P1 & G0) | (P1 & P0 & S),
C3 = G2 | (P2 & G1) | (P2 & P1 & G0) | (P2 & P1 & P0 & S),
C4 = G3 | (P3 & G2) | (P3 & P2 & G1) | (P3 & P2 & P1 & G0) | (P3 & P2 & P1 & P0 & S);
assign
sum[0] = P0 ^ S,
sum[1] = P1 ^ C1,
sum[2] = P2 ^ C2,
sum[3] = P3 ^ C3;
assign Cout = C4;
endmodule
老实说,我真正需要做的是一个使用先行进位的 4 位加减法器,但我不知道如何开始先行进位,所以我来了。如果有人能帮助我,那就太好了:<
编辑:我已经冷静下来,我终于可以查明确切的问题:测试台的 A 和 B 的值。虽然我可以暴力破解它,但我怎样才能利用循环来增加 A 和 B 以便它像这样:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
同时更新 M
?
使用以下方法改进您的测试平台。
`timescale 1ns / 1ps
module adder_4bit_cla_tb();
// inputs - keep them having reg as the data type
reg [3:0] A, B;
reg S;
// outputs - keep them having wire as the data type
wire [3:0] sum;
wire Cout;
adder_4bit_cla adder_4bit_cla_inst
(
.sum(sum), .Cout(Cout), .A(A), .B(B), .S(S)
);
initial begin
A = 4'd1; B = 4'd2; S = 1'd1;
#10 A = 4'd2; B = 4'd5; S = 1'd0;
#10 A = 4'd5; B = 4'd6; S = 1'd0;
#50 $stop;
end
endmodule
波形结果
可以使用 for 循环提供输入,如下所示。
`timescale 1ns / 1ps
module adder_4bit_cla_tb();
// inputs - keep them having reg as the data type
reg [3:0] A, B;
reg S;
// outputs - keep them having wire as the data type
wire [3:0] sum;
wire Cout;
reg [3:0] i;
adder_4bit_cla adder_4bit_cla_inst
(
.sum(sum), .Cout(Cout), .A(A), .B(B), .S(S)
);
initial begin
for(i = 4'd0; i < 4'd15; i = i + 4'd1) begin
A = i; B = i; S = 1'b0;
#10;
end
#200 $stop;
end
endmodule
波形结果