所以这就是问题所在。我已经为二进制分隔器编写了代码，该二进制分隔器应输出 7 位 7 段显示二进制代码以进入 8 x 7 段显示。（2 7段，分别表示除数，商，其余部分，并按该顺序排列）。我的开发板上的这个 8 x 7 段显示器有一个 7 位输入（a 到 g）和一个 3 位选择。所以基本的想法是我必须按顺序、连续和足够快地输出除数、除数、商和余数，这样对于人眼来说，输出看起来是恒定的，尽管事实上八个 7 段中的每一个都是根据我的输出逐个启用的。最初，分频器以二进制形式给出所有输出（除数、除数、商、余数），然后由一个函数转换为 8 位 bcd，然后将该 bcd 数字分解为两个 4 位 bcd 数字由另一个函数（现在我有 8 个输出变量：2 个表示除数，2 个表示除数等）。这些 4 位数字由另一个函数转换为 7 段。
以下是完整的代码：
library IEEE;
use IEEE.STD_LOGIC_1164.all;
use IEEE.STD_LOGIC_UNSIGNED.all;
use IEEE.STD_LOGIC_ARITH.all;
entity division is
generic(SIZE: INTEGER := 8);
port(reset: in STD_LOGIC; --reset
en: in STD_LOGIC; --enable
clk: in STD_LOGIC; --clock
num: in STD_LOGIC_VECTOR((SIZE - 1) downto 0); --dividend
den: in STD_LOGIC_VECTOR((SIZE - 1) downto 0); --divisor
whatgoes:out STD_LOGIC_VECTOR(6 downto 0) --output
);
end division;
architecture behav of division is
signal bufreg: STD_LOGIC_VECTOR((2 * SIZE - 1) downto 0); --signal array to hold both accumulator and dividend registers as one i.e bufreg(18 bits)
signal dbuf: STD_LOGIC_VECTOR((SIZE - 1) downto 0); --signal array to hold the divisor
signal count: INTEGER range 0 to SIZE; --count to determine when to stop
signal MYcount: INTEGER range 0 to 100;
signal res: STD_LOGIC_VECTOR((SIZE - 1) downto 0); --result/quotient
signal rm : STD_LOGIC_VECTOR((SIZE - 1) downto 0); --remainder
alias ADreg is bufreg((2 * SIZE - 1) downto SIZE); --ADreg is is alias for top half of bufreg register(17th to 9th bit)
alias DVNDreg is bufreg((SIZE - 1) downto 0); --DVNDreg is is alias for bottom half of bufreg register(8th to 0th bit)
--Function definitions
function to_bcd ( bin : std_logic_vector(7 downto 0) ) return std_logic_vector; --converts 8 bit binary to 8 bit BCD
function m7seg (bin : std_logic_vector(3 downto 0) ) return std_logic_vector; --converts 4 bit BCD to 7 bit 7segment
function breakdown1 ( bin : std_logic_vector(7 downto 0) ) return std_logic_vector; --breaks an 8 bit BCD into a 4 bit BCD with lower bits
function breakdown2 ( bin : std_logic_vector(7 downto 0) ) return std_logic_vector; ----breaks an 8 bit BCD into a 4 bit BCD with higher bits
--this function assigns the first 4 bits of an 8 bit BCD number to a 4-bit vector
function breakdown1 ( bin : std_logic_vector(7 downto 0) ) return std_logic_vector is
variable bint : std_logic_vector(3 downto 0) :=bin(3 downto 0);
begin
return bint;
end breakdown1;
--this function assigns the last 4 bits of an 8 bit BCD number to a 4-bit vector
function breakdown2 ( bin : std_logic_vector(7 downto 0) ) return std_logic_vector is
variable bint : std_logic_vector(3 downto 0) :=bin(7 downto 4);
begin
return bint;
end breakdown2;
--This function converts 8 bit binary to 8 bit BCD
function to_bcd ( bin : std_logic_vector(7 downto 0) ) return std_logic_vector is
variable i : integer:=0;
variable bcd : std_logic_vector(7 downto 0) :=(others => '0');
variable bint : std_logic_vector(7 downto 0) :=bin;
variable bcd2 : std_logic_vector(7 downto 0) :=(others => '0');
begin
for i in 0 to 7 loop -- repeating 8 times.
bcd(7 downto 1) := bcd(6 downto 0); --shifting the bits.
bcd(0) := bint(7);
bint(7 downto 1) := bint(6 downto 0);
bint(0) :='0';
if(i < 7 and bcd(3 downto 0) > "0100") then --add 3 if BCD digit is greater than 4.
bcd(3 downto 0) := bcd(3 downto 0) + "0011";
end if;
if(i < 7 and bcd(7 downto 4) > "0100") then --add 3 if BCD digit is greater than 4.
bcd(7 downto 4) := bcd(7 downto 4) + "0011";
end if;
--if(i < 7 and bcd(11 downto 8) > "0100") then --add 3 if BCD digit is greater than 4.
--bcd(11 downto 8) := bcd(11 downto 8) + "0011";
--end if;
end loop;
bcd2(7 downto 0):=bcd(7 downto 0);
return bcd2;
end to_bcd;
--This function converts 4 bit bcd to 7 segment
function m7seg (bin : std_logic_vector(3 downto 0))return std_logic_vector is
variable bint : std_logic_vector(3 downto 0):=bin(3 downto 0);
variable out7 : std_logic_vector(6 downto 0);
begin
case bint is
when "0000"=> out7:="1111110";
when "0001"=> out7:="0110000";
when "0010"=> out7:="1101101";
when "0011"=> out7:="1111001";
when "0100"=> out7:="0110011";
when "0101"=> out7:="1011011";
when "0110"=> out7:="X011111";
when "0111"=> out7:="1110000";
when "1000"=> out7:="1111111";
when "1001"=> out7:="111X011";
when others=> out7:="0000000";
end case;
return out7;
end m7seg;
begin
--our process begins here (shift and subtract/ Non restoring division)
p_001: process(reset, en, clk, bufreg)
begin
if reset = '1' then
res <= (others => '0');
rm <= (others => '0');
dbuf <= (others => '0');
bufreg <= (others => '0');
count <= 0;
MYcount <= 1;
elsif rising_edge(clk) then
if en = '1' then
case count is
when 0 =>
ADreg <= (others => '0');
DVNDreg <= num;
dbuf <= den;
res <= DVNDreg;
rm <= ADreg;
count <= count + 1;
when others =>
if bufreg((2 * SIZE - 2) downto (SIZE - 1)) >= dbuf then
ADreg <= '0' & (bufreg((2 * SIZE - 3) downto (SIZE - 1)) - dbuf((SIZE - 2) downto 0));
DVNDreg <= DVNDreg ((SIZE - 2) downto 0) & '1';
else
bufreg <= bufreg((2 * SIZE - 2) downto 0) & '0';
end if;
if count /= SIZE then
count <= count + 1;
else
count <= 0;
end if;
end case;
end if;
res <= DVNDreg;
rm <= ADreg;
MYcount<=MYcount+1;
whatgoes<=(others => '0');
case MYcount is
when 2 =>
whatgoes<=m7seg(breakdown1(to_bcd(rm))); --first 7segment(lower bits of remainder)
when 3 =>
whatgoes<=m7seg(breakdown2(to_bcd(rm))); --second 7segment (higher bits of remainder)
when 4 =>
whatgoes<=m7seg(breakdown1(to_bcd(res))); --third 7segment (lower bits of result/quotient)
when 5 =>
whatgoes<=m7seg(breakdown2(to_bcd(res))); --fourth 7segment (higher bits of result/quotient)
when 6 =>
whatgoes<=m7seg(breakdown1(to_bcd(den))); --fifth 7segment (lower bits of divisor)
when 7 =>
whatgoes<=m7seg(breakdown2(to_bcd(den))); --sixth 7segment (higher bits of divisor)
when 8 =>
whatgoes<=m7seg(breakdown1(to_bcd(num))); --seventh 7segment (lower bits of number/dividend)
when 9 =>
whatgoes<=m7seg(breakdown2(to_bcd(num))); --eigth 7segment (higher bits of number/dividend)
when 10 =>
MYcount<=1;
when others =>
NULL;
end case;
end if;
end process;
end behav;
当我尝试运行模拟时，它给了我各种时髦的东西。我希望输出（whatgoes（6 下降到 0））随着时钟（clk）的上升沿而变化。问题是，由于我是VHDL的初学者，我在合成顺序语句时遇到了很多问题。在敏感度列表中具有启用、时钟和重置的进程p_001中，我放置了此 case 语句。它在正边缘条件下执行。代码提取：
case MYcount is
when 2 =>
whatgoes<=m7seg(breakdown1(to_bcd(rm))); --first 7segment(lower bits of remainder)
when 3 =>
whatgoes<=m7seg(breakdown2(to_bcd(rm))); --second 7segment (higher bits of remainder)
when 4 =>
whatgoes<=m7seg(breakdown1(to_bcd(res))); --third 7segment (lower bits of result/quotient)
when 5 =>
whatgoes<=m7seg(breakdown2(to_bcd(res))); --fourth 7segment (higher bits of result/quotient)
when 6 =>
whatgoes<=m7seg(breakdown1(to_bcd(den))); --fifth 7segment (lower bits of divisor)
when 7 =>
whatgoes<=m7seg(breakdown2(to_bcd(den))); --sixth 7segment (higher bits of divisor)
when 8 =>
whatgoes<=m7seg(breakdown1(to_bcd(num))); --seventh 7segment (lower bits of number/dividend)
when 9 =>
whatgoes<=m7seg(breakdown2(to_bcd(num))); --eigth 7segment (higher bits of number/dividend)
when 10 =>
MYcount<=1;
when others =>
NULL;
end case;
我很确定我的问题出在这里，因为我的其余代码工作正常。对于上传如此复杂的代码，我深表歉意。我真的被困住了，我已经在这里呆了好几个小时了。任何帮助将不胜感激。我知道需要一种特殊的奉献精神和耐心来回答这样一个冗长、无聊和无用的问题。但是，对于任何可以帮助或提供指向可以解决我的问题的链接的人，您都会为我提供出色的服务。
我正在使用 ISE 14.3 和 iSim。
所以，多亏了里克，我解决了这个问题。他帮助我意识到我忘记驱动3位选择输出。事实证明，使用 case 语句和计数变量驱动它解决了我按顺序执行代码的问题。我知道代码不是以有组织的方式编写的，但我希望随着时间的推移我会变得更好。
process (clk,tmp,rm,res,den,num)
variable CLR: boolean:=true;
begin
if (CLR=true) then
tmp <= "000";
CLR:=false;
elsif (clk'event and clk='1') then
tmp <= tmp + 1;
if tmp<=8 then
CLR:=true;
end if;
end if;
case tmp is
when "000" =>
whatgoes<=m7seg(breakdown1(to_bcd(rm))); --first 7segment(lower bits of remainder)
when "001" =>
whatgoes<=m7seg(breakdown2(to_bcd(rm))); --second 7segment (higher bits of remainder)
when "010" =>
whatgoes<=m7seg(breakdown1(to_bcd(res))); --third 7segment (lower bits of result/quotient)
when "011" =>
whatgoes<=m7seg(breakdown2(to_bcd(res))); --fourth 7segment (higher bits of result/quotient)
when "100" =>
whatgoes<=m7seg(breakdown1(to_bcd(den))); --fifth 7segment (lower bits of divisor)
when "101" =>
whatgoes<=m7seg(breakdown2(to_bcd(den))); --sixth 7segment (higher bits of divisor)
when "110" =>
whatgoes<=m7seg(breakdown1(to_bcd(num))); --seventh 7segment (lower bits of number/dividend)
when "111" =>
whatgoes<=m7seg(breakdown2(to_bcd(num))); --eigth 7segment (higher bits of number/dividend)
when others =>
NULL;
end case;
sel<=tmp;
end process;
我基本上是在黑暗中拍摄;也许如果您发布模拟图片，它将帮助我们更好地了解您的问题。无论如何，既然我们已经这样做了，为什么不谈论一些随机问题：
-
如果您将代码分成几个块，每个块都有一个目的，那么代码会更容易理解（和使用）。你可以让一个块做除法，只输出商和余数。另一个模块可以接收 8 个 BCD 值，并对其进行多路复用，以便它们正确显示在电路板的显示器上。如果我们一次可以专注于问题的一部分，就会更容易发现任何错误。
-
您在 LCD 上提到了 3 位选择，但我在您的代码中没有看到您在哪里驱动它。也许你应该根据你的信号输出一些东西？
-
为了确保您的函数正常工作，您可以将它们放在一个包中并创建一个自检查测试平台。至少我是这样做的。这将把这个变量从等式中剔除。
请发布一些模拟结果，以便我们为您提供帮助。