*我正在Xilinx 14.3中编写VHDL代码并且正在针对Nexys 2主板。*
从我读过的内容来看，锁存器来自于if / case语句不完整或者没有在所有可能的路径中设置输出。
我多次查看我的代码，但仍然有两个锁存器。
WARNING:Xst:737 - Found 8-bit latch for signal <DR>. Latches may be generated from incomplete case or if statements. We do not recommend the use of latches in FPGA/CPLD designs, as they may lead to timing problems.
WARNING:Xst:737 - Found 8-bit latch for signal <P>. Latches may be generated from incomplete case or if statements. We do not recommend the use of latches in FPGA/CPLD designs, as they may lead to timing problems.
现在，我得到了一些其他的错误，因为没有使用某些信号而且有一大块被注释掉等等......但是我还没有使用我的代码，现在这不是问题。< / p>
那么，当每个路径设置每个输出时，为什么我仍然在此代码中获取锁存器？
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;
-- Uncomment the following library declaration if using
-- arithmetic functions with Signed or Unsigned values
--use IEEE.NUMERIC_STD.ALL;
-- Uncomment the following library declaration if instantiating
-- any Xilinx primitives in this code.
--library UNISIM;
--use UNISIM.VComponents.all;
entity project2vhd is
Port ( CE_s : in STD_LOGIC;
A0 : in STD_LOGIC;
RD_s : in STD_LOGIC;
WR_s : in STD_LOGIC;
RESET : in STD_LOGIC;
ACK_s : inout STD_LOGIC;
Y1, Y2, Y3 : inout STD_LOGIC;
-- Din : in STD_LOGIC_VECTOR (7 downto 0);
-- Dout : out STD_LOGIC_VECTOR (7 downto 0);
D : inout STD_LOGIC_VECTOR (7 downto 0);
EN_s : out STD_LOGIC;
P : inout STD_LOGIC_VECTOR (7 downto 0));
end project2vhd;
architecture Behavioral of project2vhd is
signal CR : STD_LOGIC_VECTOR (1 downto 0);
signal SR : STD_LOGIC_VECTOR (2 downto 0);
signal DR : STD_LOGIC_VECTOR (7 downto 0);
begin
process (WR_s, ACK_s, RESET, A0, RD_s)
begin
if (RESET = '1') then --if reset is high
Y1 <= '0';
Y2 <= '0';
Y3 <= '0';
D <= "ZZZZZZZZ";
EN_s <= '0'; --check EN_s's value at reset
P <= P;
DR<= DR;
else
D <= D;
DR <= DR;
P <= P;
-- if (CR(0) = '1') then --Mode 1
--
-- Y1 <= ((Y1 and (not Y2) and Y3) or
-- (WR_s and ACK_s and (not Y2) and Y3));
-- Y2 <= '0';
-- Y3 <= (((not Y1) and (not Y2) and Y3) or
-- (WR_s and ACK_s and (not Y2) and Y3) or
-- ((not Y1) and (not Y2) and (not WR_s) and ACK_s) or
-- ((not WR_s) and (not Y2) and Y3));
-- SR(2) <=(((not ACK_s) and (not Y2) and (not Y3)) or --obf
-- (WR_s and (not ACK_s) and (Y1) and (not Y2)) or
-- (WR_s and (not ACK_s) and (not Y2) and Y3) or
-- (WR_s and (not Y1) and (not Y2)));
-- SR(0) <= (((not Y2) and (not Y3)) or --INTR_enable
-- (WR_s and (not Y1) and (not Y2)) or
-- ((not WR_s) and (not ACK_s) and (not Y1) and (not Y2) and Y3));
--
--
-- if (CE_s = '1') then
-- D <= "ZZZZZZZZ";
--
-- else
-- if (WR_s = '0' and A0 = '0') then --Write Data (MODE 1)
-- EN_s <= '0'; -- enable buffer
--
-- DR <= D;
-- P <= D;
--
-- elsif (WR_s = '0' and A0 = '1') then --control Reg Mode (MODE 1 and 0)
-- EN_s <= '0'; -- enable buffer
--
-- CR(0) <= D(0);
-- CR(1) <= D(1);
---- SR(0)
---- SR(1)
---- SR(2)
--
--
-- elsif (RD_s = '0' and A0 = '1') then -- Read Status (MODE 1)
-- EN_s <= '1'; -- disable buffer
--
-- D <= DR;
---- D <= "10101010"
--
-- else
-- EN_s <= '0'; -- enable buffer
-- D <= "ZZZZZZZZ";
--
-- end if;
-- end if;
-- else --Mode 0
SR <= "111";
if (CE_s = '0' and WR_s = '0' and A0 = '1') then
EN_s <= '0'; -- enable buffer
DR <= D;
CR(1) <= D(1);
CR(0) <= D(0);
P <= P;
elsif (CE_s = '0' and WR_s = '0' and A0 = '0') then
EN_s <= '1'; -- disable buffer
P <= DR;
DR <= DR;
else
EN_s <= '0'; -- enable buffer
D <= "ZZZZZZZZ";
DR <= DR;
P <= P;
end if;
-- end if;
end if;
end process;
end Behavioral;
答案 0 :(得分：3)
您是否删除了DR <= DR;的两个副本？摆脱开始时的默认值和“else”值。
并且不要只删除“else”，将其替换为DR <= (others => '0');或其他一些，并且闩锁应该消失。有两种创建锁存器的方法：DR <= DR;和通过该过程的路径，使DR保持未分配状态。因此，删除或评论它将无法完成任务。
如果您不想要锁存器，但是在某些情况下您希望DR保持其先前的值，则需要一个时钟进程，在这种情况下，您可以将DR保存在寄存器中。
在这种情况下，如果读取保持异步，通常只需要灵敏度列表中的时钟和复位，以及读取访问所需的任何信号。但完全时钟设计是更好的做法。
答案 1 :(得分：2)
因为在各种条件下DR和P信号被分配给它们自己（即：DR＆lt; = DR;）所以它们不会改变，或者换句话说保持它们的最后值。这是一个闩锁。