我正在使用Vivado 2014.2将BCD的VHDL代码写入可用于计算器或组合锁的二进制输入缓冲区。
我的方法很简单。做x * 10它与x（2 + 8）= x * 2 + x * 8相同。
x * 2 = 1左移（2 ^ 1 = 2）
x * 8 =左移3（2 ^ 3 = 8）
在添加输入之前移位并添加输出缓冲区（tempC）。这样做是为了从null开始，这样输入的第一个数字就不会出现，再乘以10。
我的代码在artix 7 fpga上编译并运行，但是我遇到了确保输出缓冲区（tempC）正常工作的问题。它拒绝输出任何数据，但我不确定原因。
我可能会错误地添加这些值，但我不这么认为。也许我正在使用错误的数据类型？
非常感谢任何帮助。
-- Engineer: greatgamer34
--
-- Create Date: 01/25/2017 04:57:02 PM
-- Design Name:
-- Module Name: buff - Behavioral
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use ieee.numeric_std.all;
entity buff is
Port ( Data : in STD_LOGIC_VECTOR (3 downto 0); ----4bit BCD value input
Clock : in STD_LOGIC;
Reset : in STD_LOGIC;
Output : out STD_LOGIC_VECTOR (15 downto 0);
aout : out STD_LOGIC_VECTOR (6 downto 0));-- 7 segment display output for current state.
end buff;
architecture Behavioral of buff is
type states is (state0, state1, state2, state3);
signal currentstate, nextstate: states;
signal tempA: STD_LOGIC_VECTOR (15 downto 0);---used to store 'Data' for addition.
signal tempB: STD_LOGIC_VECTOR (15 downto 0);---used for x2('Data').
signal tempC: STD_LOGIC_VECTOR (15 downto 0);---used as output register.
signal tempD: STD_LOGIC_VECTOR (15 downto 0);---used for sending data to LED's.
signal tempE: STD_LOGIC_VECTOR (15 downto 0);---used for x8('Data')
begin
Process(Reset,clock)
Begin
if(Reset = '1') then
tempC <= "0000000000000000"; --clear tempC
tempA <= "0000000000000000"; --clear tempA
currentstate <= state0; -------reset state to 0
elsif(clock'event and clock = '1') then
output <= (tempD);--dispaly the output of the buffer
currentstate<=nextstate; -- advance states
end if;
end process;
process(currentstate)
begin
case currentstate is
when state0 =>
tempA(3 downto 0) <= Data; -- load in 4 bit data intoi 16 bit register
tempD <= (tempA); --output the input data(used for debugging)
nextstate <= state1;
aout <= not "1111110"; -- output on the 7 seg the number 0
when state1 =>
tempB <= tempC(14 downto 0) & '0'; --left shift tempC(the output register) save to tempB; this is the x2 multiplication
tempD <= (tempA); -- output the input data(used for debugging)
nextstate <= state2;
aout <= not "0110000"; -- output on the 7 seg the number 1
when state2 =>
tempE <= tempC(12 downto 0) & "000"; --left shift tempC(the output register) three times save to tempE; this is the x8 multiplication
--tempC <=std_logic_vector( unsigned(tempE) + unsigned(tempD)); (TESTING)
tempC <=std_logic_vector( ('0' & unsigned(tempE(14 downto 0))) + ('0' & unsigned(tempD(14 downto 0)))); --add the first 15 bits of tempD and tempE(this is how we multiply by 10)
tempD <= (tempC); -- output the x10 output register
nextstate <= state3;
aout <= not "1101101" ; -- output on the 7 seg the number2
when state3 =>
-- tempC <= ('0' & tempC(14 downto 0)) + ('0' & tempA(14 downto 0)); (TESTING)
tempC <= std_logic_vector( ('0' & unsigned(tempC(14 downto 0))) + ('0' & unsigned(tempA(14 downto 0)))); --add the 'Data' to the x10 shifted number.
tempD <= (tempC);
nextstate <= state0;
aout <= not "1111001"; -- output on the 7 seg the number3
end case;
end process;
end behavioral;
答案 0 :(得分：0)
<强>答案：
tempC在时钟进程中重置，然后获取组合过程中分配的新值。 不允许在两个不同的进程中为信号分配值。此外，组合过程的灵敏度列表缺少信号。
<强>观察：
快乐调试
答案 1 :(得分：0)
好的，在一些评论和答案的帮助下，我能够让它发挥作用。以下是使用的代码。
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use ieee.numeric_std.all;
entity buff is
Port ( Data : in STD_LOGIC_VECTOR (3 downto 0); ----4bit BCD value input
Clock : in STD_LOGIC;
Reset : in STD_LOGIC;
Output : out STD_LOGIC_VECTOR (15 downto 0);
aout : out STD_LOGIC_VECTOR (6 downto 0));-- 7 segment display output for current state.
end buff;
architecture Behavioral of buff is
type states is (state0, state1, state2, state3, state4);
signal currentstate, nextstate: states;
signal tempA: STD_LOGIC_VECTOR (3 downto 0);---used to store 'Data' for addition.
signal tempB: STD_LOGIC_VECTOR (15 downto 0);---used for x2('Data').
signal tempC: STD_LOGIC_VECTOR (15 downto 0);---used as output register.
signal tempD: STD_LOGIC_VECTOR (15 downto 0);---used for sending data to LED's.
signal tempE: STD_LOGIC_VECTOR (15 downto 0);---used for x8('Data')
signal tempF: STD_LOGIC_VECTOR (15 downto 0);
begin
Process(Reset,clock)
Begin
if(Reset = '1') then
currentstate <= state4;
Output <= "0000000000000000"; -------reset state
elsif(clock'event and clock = '1') then
Output <= tempD ;--display the output of the buffer
currentstate <= nextstate; -- advance states
end if;
end process;
process(currentstate)
begin
case currentstate is
when state0 =>
tempA <= Data; -- load in 4 bit data intoi 16 bit register
tempD(3 downto 0) <= tempA; --output the input data(used for debugging)
nextstate <= state1;
aout <= not "1111110"; -- output on the 7 seg the number 0
when state1 =>
tempB <= (tempC(14 downto 0) & "0"); --left shift tempC(the output register) save to tempB; this is the x2 multiplication
tempD <= (tempB); -- output the input data(used for debugging)
nextstate <= state2;
aout <= not "0110000"; -- output on the 7 seg the number 1
when state2 =>
tempE <= tempC(12 downto 0) & "000"; --left shift tempC(the output register) three times save to tempE; this is the x8 multiplication
--tempF <=std_logic_vector( unsigned(tempE) + unsigned(tempB)); --(TESTING)
tempF <=std_logic_vector( ('0' & unsigned(tempE(14 downto 0))) + ('0' & unsigned(tempB(14 downto 0)))); --add the first 15 bits of tempD and tempE(this is how we multiply by 10)
tempD <= tempE; -- output the x10 output register
nextstate <= state3;
aout <= not "1101101" ; -- output on the 7 seg the number2
when state3 =>
--tempC <=std_logic_vector( unsigned(tempC) + unsigned(tempA));
tempC <= std_logic_vector( ('0' & unsigned(tempF(14 downto 0))) + ("000000000000" & unsigned(tempA))); --add the 'Data' to the x10 shifted number.
tempD <= tempC;
nextstate <= state0;
aout <= not "1111001"; -- output on the 7 seg the number3
when state4 =>
tempC <= "0000000000000000";
tempA <= "0000";
tempB <= "0000000000000000";
tempD <= "0000000000000000";
tempE <= "0000000000000000";
tempF <= "0000000000000000";
nextstate <= state0;
aout <= not "0110011";
end case;
end process;
end behavioral;