如何解决我究竟做错了什么?测试台未正确更新
我正在尝试实施测试平台,以便在 3 个滴答后,每个输入都获得新的指定值。例如,前 3 秒(或滴答声)A = 10,B = 0,然后接下来的 3 秒(3 秒到 6 秒)A = 10,B = 16 等等。但是,在我的实际测试平台上,这些值没有按照我想要的方式更新。我的语法错了吗?
这是我得到的当前值的屏幕截图。我在顶部画了红线代表每 3 个刻度。
https://gyazo.com/f2c0cddc192d0d6734c98334cd377f12
module ALU_tb();
reg [63:0] A,B;
reg [4:0] FS;
reg cin;
wire cout;
wire [63:0] result;
wire [3:0] status;
Final_ALU dut (
.A(A),.B(B),.FS(FS),.cin(cin),.cout(cout),.result(result),.status(status)
);
initial begin //A+1 //A=10 B=0
A <= 64'b0000000000000000000000000000000000000000000000000000000000001010;
B <= 64'b0000000000000000000000000000000000000000000000000000000000000000;
FS <= 5'b01000;
cin <= 1'b1;
end
always begin //A+B //A=10 B=16
#3
A <= 64'b0000000000000000000000000000000000000000000000000000000000001010;
B <= 64'b0000000000000000000000000000000000000000000000000000000000010000;
FS <= 5'b01000;
cin <= 1'd0;
#3;
end
always begin //A-B //A=10 B=16
#6
A <= 64'b0000000000000000000000000000000000000000000000000000000000001010;
B <= 64'b0000000000000000000000000000000000000000000000000000000000010000;
FS <= 5'b01001;
cin <= 1'd1;
#6;
end
always begin //A-1 //A=10,B=1
#9
A <= 64'b0000000000000000000000000000000000000000000000000000000000001010;
B <= 64'b0000000000000000000000000000000000000000000000000000000000000001;
FS <= 5'b01001;
cin <= 1'd1;
#9;
end
always begin //-A //A=10,B=0 (just twos complement of A)
#12
A <= 64'b0000000000000000000000000000000000000000000000000000000000001010;
B <= 64'b0000000000000000000000000000000000000000000000000000000000000000;
FS <= 5'b01010;
cin <= 1'd1;
#12;
end
initial begin
#30 $finish;
end
endmodule
解决方法
您不应对来自多个 always
块的同一信号进行分配。
解决此问题的一种方法是在 fork/join
块中使用 initial
,并去掉 always
关键字。这对您的代码进行了最少的更改:
initial begin
fork
begin //A+1 //A=10 B=0
A <= 64'b0000000000000000000000000000000000000000000000000000000000001010;
B <= 64'b0000000000000000000000000000000000000000000000000000000000000000;
FS <= 5'b01000;
cin <= 1'b1;
end
begin //A+B //A=10 B=16
#3
A <= 64'b0000000000000000000000000000000000000000000000000000000000001010;
B <= 64'b0000000000000000000000000000000000000000000000000000000000010000;
FS <= 5'b01000;
cin <= 1'd0;
#3;
end
begin //A-B //A=10 B=16
#6
A <= 64'b0000000000000000000000000000000000000000000000000000000000001010;
B <= 64'b0000000000000000000000000000000000000000000000000000000000010000;
FS <= 5'b01001;
cin <= 1'd1;
#6;
end
begin //A-1 //A=10,B=1
#9
A <= 64'b0000000000000000000000000000000000000000000000000000000000001010;
B <= 64'b0000000000000000000000000000000000000000000000000000000000000001;
FS <= 5'b01001;
cin <= 1'd1;
#9;
end
begin //-A //A=10,B=0 (just twos complement of A)
#12
A <= 64'b0000000000000000000000000000000000000000000000000000000000001010;
B <= 64'b0000000000000000000000000000000000000000000000000000000000000000;
FS <= 5'b01010;
cin <= 1'd1;
#12;
end
join
end
更传统的方法是去掉 fork
,只在 initial
块内按顺序驱动所有信号组。然后,您必须相应地调整所有 #
延迟。
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