HDLBits 系列（37）此系列关于独热码的题目的疑问？
【摘要】 目录
背景
我的做法
第一题
第二题
第三题
解决办法
第一题
第二题
第三题
推荐
背景
目前为止，关于状态机独热码的题目，几乎没一个题目能做对，这令我疑惑？是不是题目的答案有问题？在此请大家一试？（已解决，谢谢）
我的做法
第一题
第一题（点击蓝色字体进入题目链接做答）
本人答案：
module top_module ( input ...
目录
背景
目前为止，关于状态机独热码的题目，几乎没一个题目能做对，这令我疑惑？是不是题目的答案有问题？在此请大家一试？（已解决，谢谢）
我的做法
第一题
第一题（点击蓝色字体进入题目链接做答）
本人答案：
-
module top_module (
-
input [6:1] y,
-
input w,
-
output Y2,
-
output Y4);
-
-
localparam A = 6'b0000_01, B = 6'b0000_10, C = 6'b0001_00, D = 6'b0010_00, E = 6'b0100_00, F = 6'b1000_00;
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reg [6:1] next_state;
-
assign Y2 = next_state[2];
-
assign Y4 = next_state[4];
-
always@(*) begin
-
next_state = A;
-
case(y)
-
A: begin
-
if(w) next_state = A;
-
else next_state = B;
-
end
-
B: begin
-
if(w) next_state = D;
-
else next_state = C;
-
end
-
C: begin
-
if(w) next_state = D;
-
else next_state = E;
-
end
-
D: begin
-
if(w) next_state = A;
-
else next_state = F;
-
end
-
E: begin
-
if(w) next_state = D;
-
else next_state = E;
-
end
-
F: begin
-
if(w) next_state = D;
-
else next_state = C;
-
end
-
endcase
-
end
-
-
endmodule
第二题
本人答案：
-
module top_module (
-
input [5:0] y,
-
input w,
-
output Y1,
-
output Y3
-
);
-
-
localparam A = 6'b0000_01, B = 6'b0000_10, C = 6'b0001_00, D = 6'b0010_00, E = 6'b0100_00, F = 6'b1000_00;
-
reg [5:0] next_state;
-
assign Y1 = next_state[1];
-
assign Y3 = next_state[3];
-
always@(*) begin
-
case(y)
-
A: begin
-
if(~w) next_state = A;
-
else next_state = B;
-
end
-
B: begin
-
if(~w) next_state = D;
-
else next_state = C;
-
end
-
C: begin
-
if(~w) next_state = D;
-
else next_state = E;
-
end
-
D: begin
-
if(~w) next_state = A;
-
else next_state = F;
-
end
-
E: begin
-
if(~w) next_state = D;
-
else next_state = E;
-
end
-
F: begin
-
if(~w) next_state = D;
-
else next_state = C;
-
end
-
default: begin
-
next_state = A;
-
end
-
endcase
-
end
-
-
endmodule
第三题
https://blog.csdn.net/Reborn_Lee/article/details/103428895
本人答案：
-
module top_module(
-
input in,
-
input [9:0] state,
-
output [9:0] next_state,
-
output out1,
-
output out2);
-
-
localparam S0 = 10'b0000_0000_01, S1 = 10'b0000_0000_10, S2 = 10'b0000_0001_00, S3 = 10'b0000_0010_00,S4 = 10'b0000_0100_00;
-
localparam S5 = 10'b0000_1000_00, S6 = 10'b0001_0000_00, S7 = 10'b0010_0000_00, S8 = 10'b0100_0000_00,S9 = 10'b1000_0000_00;
-
always@(*) begin
-
case(state)
-
S0: begin
-
if(in) next_state = S1;
-
else next_state = S0;
-
end
-
S1: begin
-
if(in) next_state = S2;
-
else next_state = S0;
-
end
-
S2: begin
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if(in) next_state = S3;
-
else next_state = S0;
-
end
-
S3: begin
-
if(in) next_state = S4;
-
else next_state = S0;
-
end
-
S4: begin
-
if(in) next_state = S5;
-
else next_state = S0;
-
end
-
S5: begin
-
if(in) next_state = S6;
-
else next_state = S8;
-
end
-
S6: begin
-
if(in) next_state = S7;
-
else next_state = S9;
-
end
-
S7: begin
-
if(in) next_state = S7;
-
else next_state = S0;
-
end
-
S8: begin
-
if(in) next_state = S1;
-
else next_state = S0;
-
end
-
S9: begin
-
if(in) next_state = S1;
-
else next_state = S0;
-
end
-
default: begin
-
next_state = S0;
-
end
-
-
endcase
-
end
-
-
assign out1 = (state == S8 | state == S9) ? 1 : 0;
-
assign out2 = (state == S9 | state == S7) ? 1 : 0;
-
-
-
-
endmodule
如有大神知道，还望告知，谢谢。
（崩溃中）
解决办法
更新：
群里的一个同学给我解答了，道理是有的，但是这个网站上有关独热码的题目真的没必要深究了，上面的设计本身就是没有问题的，仅仅为了正确的答案，给出Success的答案：
第一题
第一题：
-
module top_module (
-
input [6:1] y,
-
input w,
-
output Y2,
-
output Y4);
-
-
assign Y2 = y[1]&&(~w);
-
//assign Y4 = y[2]&&w || y[3]&&w || y[5]&&w || y[6]&&w;
-
assign Y4 = w&&(y[2] || y[3] || y[5] || y[6]);
-
endmodule
第二题
第二题：
-
module top_module (
-
input [5:0] y,
-
input w,
-
output Y1,
-
output Y3
-
);
-
assign Y1 = y[0]&& w;
-
assign Y3 = ~w && (y[1] || y[2] || y[4] || y[5]);
-
-
-
endmodule
第三题
第三题：
-
module top_module(
-
input in,
-
input [9:0] state,
-
output [9:0] next_state,
-
output out1,
-
output out2);
-
-
assign next_state[0] = ~in & (state[0] | state[1] | state[2] | state[3] | state[4] | state[7] | state[8] | state[9]);
-
assign next_state[1] = in & (state[0] | state[8] | state[9]);
-
assign next_state[2] = in & state[1];
-
assign next_state[3] = in & state[2];
-
assign next_state[4] = in & state[3];
-
assign next_state[5] = in & state[4];
-
assign next_state[6] = in & state[5];
-
assign next_state[7] = in & (state[6] | state[7]);
-
assign next_state[8] = ~in & state[5];
-
assign next_state[9] = ~in & state[6];
-
-
assign out1 = state[8] | state[9];
-
assign out2 = state[7] | state[9];
-
-
endmodule
推荐
思想见：
https://blog.csdn.net/Reborn_Lee/article/details/103338804
https://blog.csdn.net/Reborn_Lee/article/details/103339538
最后感谢群里的小伙伴，进群见：
https://blog.csdn.net/Reborn_Lee/article/details/99715080
文章来源: reborn.blog.csdn.net，作者：李锐博恩，版权归原作者所有，如需转载，请联系作者。
原文链接：reborn.blog.csdn.net/article/details/103480678
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