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 [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;reg [6:1] next_state;assign Y2 = next_state[2];assign Y4 = next_state[4];always@(*) beginnext_state = A;case(y)A: beginif(w) next_state = A;else next_state = B;endB: beginif(w) next_state = D;else next_state = C;endC: beginif(w) next_state = D;else next_state = E;endD: beginif(w) next_state = A;else next_state = F;endE: beginif(w) next_state = D;else next_state = E;endF: beginif(w) next_state = D;else next_state = C;endendcaseendendmodule
第二题
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@(*) begincase(y)A: beginif(~w) next_state = A;else next_state = B;endB: beginif(~w) next_state = D;else next_state = C;endC: beginif(~w) next_state = D;else next_state = E;endD: beginif(~w) next_state = A;else next_state = F;endE: beginif(~w) next_state = D;else next_state = E;endF: beginif(~w) next_state = D;else next_state = C;enddefault: beginnext_state = A;endendcaseendendmodule
第三题
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@(*) begincase(state)S0: beginif(in) next_state = S1;else next_state = S0;endS1: beginif(in) next_state = S2;else next_state = S0;endS2: beginif(in) next_state = S3;else next_state = S0;endS3: beginif(in) next_state = S4;else next_state = S0;endS4: beginif(in) next_state = S5;else next_state = S0;endS5: beginif(in) next_state = S6;else next_state = S8;endS6: beginif(in) next_state = S7;else next_state = S9;endS7: beginif(in) next_state = S7;else next_state = S0;endS8: beginif(in) next_state = S1;else next_state = S0;endS9: beginif(in) next_state = S1;else next_state = S0;enddefault: beginnext_state = S0; endendcaseendassign 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
