VHDL语言几段键盘去抖程序有些疑惑，拜托大家解释
Library ieee;Use ieee.std_logic_1164.all;
Entity q_dou is
Port(d_in,clk:in std_logic;
d_out:out std_logic
);
End;
Architecture bhv_q_dou of q_dou is
Component dcfq is
Port(clk,clrn,prn,d:in std_logic;
q:out std_logic);
end component dcfq;
signal vcc,inv_d:std_logic;
signal q0,q1:std_logic;
signal d1,d0：std_logic;
begin
vcc<=’1’;
inv_d<= not d_in;
u1:dcfq port map(clk=>,clrn=>inv_d,prn=>vcc,d=>vcc,q=>q0);
u2:dcfq port map(clk=>clk,clrn=>q0,prn=>vcc,d=>vcc,q=>q1);
process(clk)
begin
if clk’event and clk=’1’ then
d0<= not q1;
d1<= d0;
end if;
end process;
d_out<=not(d1 and not d0);
end;
以上程序是键盘输入去抖程序，主要是用2个D触发器，加与-非门，再经RS触发器输出，原理图如下：很不解的是那个元件例化语句中U1,U2端口连接，拜托哪位看懂的朋友解释下（dcfq:为D触发器元件，此处省略）
http://cache.amobbs.com/bbs_upload782111/files_28/ourdev_545428.JPG
(原文件名:弹跳消除电路拍照.JPG)
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