我为加法器设计但结果是错误的。
module FMUL(CLK, St, F1, E1, F2, E2, F, V, done);
input CLK;
input St;
input [3:0] F1;
input [3:0] E1;
input [3:0] F2;
input [3:0] E2;
output[6:0] F;
output V;
output done;
reg[6:0] F;
reg done;
reg V;
reg[3:0] A;
reg[3:0] B;
reg[3:0] C;
reg[4:0] X;
reg[4:0] Y;
reg Load;
reg Adx;
reg SM8;
reg RSF;
reg LSF;
reg AdSh;
reg Sh;
reg Cm;
reg Mdone;
reg[1:0] PS1;
reg[1:0] NS1;
reg[2:0] State;
reg[2:0] Nextstate;
initial
begin
State = 0;
PS1 = 0;
NS1 = 0;
Nextstate=0;
end
always @(PS1 or St or Mdone or X or A or B)
begin : main_control
Load = 1'b0;
Adx = 1'b0;
NS1 = 0;
SM8 = 1'b0;
RSF = 1'b0;
LSF = 1'b0;
V = 1'b0;
F = 7'b0000000;
done = 1'b0;
case (PS1)
0 :
begin
F = 7'b0000000;
done = 1'b0;
V = 1'b0;
if(St == 1'b1)
begin
Load = 1'b1;
NS1 = 1;
end
end
1 :
begin
Adx = 1'b1;
NS1 = 2;
end
2 :
begin
if(Mdone == 1'b1)
begin
if(A==0)
begin
SM8 = 1'b1;
end
else if(A == 4 & B == 0)
begin
RSF = 1'b1;
end
else if (A[2] == A[1])
begin
LSF = 1'b1;
end
NS1 = 3;
end
else
begin
NS1 = 2;
end
end
3 : begin
if(X[4] != X[3])
begin
V = 1'b1;
end
else
begin
V = 1'b0;
end
done = 1'b1;
F = {A[2:0],B};
if(St==1'b0)
begin
NS1 = 0;
end
end
endcase
end
always @(State or Adx or B)
begin : mul2c
AdSh = 1'b0;
Sh = 1'b0;
Cm = 1'b0;
Mdone = 1'b0;
Nextstate = 0;
case(State)
0 :
begin
if(Adx==1'b1)
begin
if((B[0]) == 1'b1)
begin
AdSh = 1'b1;
end
else
begin
Sh = 1'b1;
end
Nextstate = 1;
end
end
1,2 :
begin
if((B[0])==1'b1)
begin
AdSh = 1'b1;
end
else
begin
Sh = 1'b1;
end
Nextstate = State + 1;
end
3:
begin
if((B[0])==1'b1)
begin
Cm = 1'b1;
AdSh = 1'b1;
end
else
begin
Sh = 1'b1;
end
Nextstate = 4;
end
4:
begin
Mdone = 1'b1;
Nextstate = 0;
end
endcase
end
wire [3:0] addout;
assign addout = (Cm == 1'b0)? (A+C) : (A-C);
always @(posedge CLK)
begin : update
PS1 <= NS1;
State <= Nextstate;
if(Load == 1'b1)
begin
X <= {E1[3], E1};
Y <= {E2[3], E2};
A <= 4'b0000;
B <= F1;
C <= F2;
end
if(Adx == 1'b1)
begin
X <= X+Y;
end
if(SM8 == 1'b1)
begin
X <= 5'b11000;
end
if(RSF == 1'b1)
begin
A <= {1'b0, A[3:1]};
B <= {A[0], B[3:1]};
X <= X+1;
end
if(LSF == 1'b1)
begin
A <= {A[2:0], B[3]};
B <= {B[2:0], 1'b0};
X <= X+31;
end
if(AdSh == 1'b1)
begin
A <= {(C[3]^Cm), addout[3:1]};
B <= {addout[0], B[3:1]};
end
if(Sh == 1'b1)
begin
A <= {A[3], A[3:1]};
B <= {A[0], B[3:1]};
end
end
endmodule
测试台。
module tb_FMUL();
wire[6:0] F;
wire done;
wire V;
reg[3:0] A;
reg[3:0] B;
reg[3:0] C;
reg[4:0] X;
reg[4:0] Y;
reg Load;
reg Adx;
reg SM8;
reg RSF;
reg LSF;
reg AdSh;
reg Sh;
reg Cm;
reg Mdone;
reg[1:0] PS1;
reg[1:0] NS1;
reg[2:0] State;
reg[2:0] Nextstate;
reg CLK;
reg St;
reg [3:0] F1;
reg [3:0] E1;
reg [3:0] F2;
reg [3:0] E2;
FMUL u0(CLK, St, F1, E1, F2, E2, F, V, done);
always
begin
#10 CLK <= ~CLK;
end
initial
begin
#100 F1 = 2.125;
E1 = 5; F2 = 5.1; E2 = 1; St=0;
#100 F1 = 1.125;
E1 = 5; F2 = 2.1; E2 = 2; St=0;
#100 F1 = 5.125;
E1 = 5; F2 = 3.1; E2 = 3; St=0;
end
endmodule
模拟结果波形。 enter image description here
我参考了这本书。没有代码测试台。 所以我做了。但是没有运作。
也不会改变CLK。
请查看测试平台代码。
答案 0 :(得分：2)
你有（至少）两个问题：
您的时钟需要初始化（例如1'b0）：
初始CLK = 1＆＃39; b0;
Verilog中wire或reg的初始值为1'bx; ~1'bx是1'bx;因此CLK仍在1'bx。
$finish块中添加了对initial的调用。