ISE写的.vhd用ISim仿真时输入全是‘U’
4位的移位寄存器，编写的代码如下
entity shift is
port(
a,clk,rst:in std_logic;
b:out std_logic
);
end shift;
architecture Behavioral of shift is
signal temp1,temp2,temp3:std_logic;
begin
process
begin
if (rst='1') then
temp1<='0';
temp2<='0';
temp3<='0';
b<='0';
elsif (rising_edge(clk)) then
temp1<=a;
temp2<=temp1;
temp3<=temp2;
b<=temp3;
end if;
end process;
end Behavioral;
编写的TestBench如下
ENTITY shift_tb IS
END shift_tb;
ARCHITECTURE behavior OF shift_tb IS
COMPONENT shift
PORT(
a : IN std_logic;
clk : IN std_logic;
rst:IN std_logic;
b : OUT std_logic
);
END COMPONENT;
--Inputs
signal a : std_logic := '0';
signal clk : std_logic := '0';
signal rst : std_logic := '0';
--Outputs
signal b : std_logic;
-- Clock period definitions
constant clk_period : time := 10 ns;
BEGIN
uut: shift PORT MAP (
a => a,
clk => clk,
rst=>rst,
b => b
);
-- Clock process definitions
clk_process :process
begin
clk <= '0';
wait for clk_period/2;
clk <= '1';
wait for clk_period/2;
end process;
-- Stimulus process
stim_proc: process
begin
a<='0';
rst<='0';
wait for clk_period7;
a<='1';
wait for clk_period;
a<='0';
wait for clk_period7;
a<='1';
wait for clk_period;
a<='0';
wait for clk_period7;
a<='1';
wait for clk_period;
a<='0';
wait for clk_period7;
a<='1';
wait for clk_period;
a<='0';
wait for clk_period7;
a<='1';
wait for clk_period;
a<='0';
rst<='1';
wait for clk_period7;
a<='1';
wait for clk_period;
a<='0';
rst<='0';
wait for clk_period7;
a<='1';
wait for clk_period;
a<='0';
wait for clk_period7;
a<='1';
wait for clk_period;
end process;
END;
检查了很多遍觉得没有问题，而且代码可以综合并查看RTL图
仿真界面如下
不知道是代码的问题还是仿真设置的问题，VHDL小白求帮助
回答
这是一个有点愚蠢的问题，仿真的时候应该是点击编写代码的.vhd进行仿真，我居然是点的testbench仿真的，所以出现了这样的情况。并且在源代码中process中应当加入a,clk,rst三个敏感信号。
以上是 ISE写的.vhd用ISim仿真时输入全是‘U’ 的全部内容， 来源链接： utcz.com/a/37618.html
