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標題: VHDL輸入判斷 [打印本頁]

作者: w10789173 時間: 2008-2-10 01:18 AM
標題: VHDL輸入判斷
各位大大,想請問一下,我現在要做,當我鍵盤連續輸入,結果只取最後一個鍵盤輸入的值

這是一個自動控制鍵盤的控制器,我可以先輸入數字,然後再輸入指令,我的機器就會動那個指令n次

例如,先按5再按前進,自東車就會前進5次,如果只按前進,則自動車會自動前進1次(預設值)

當我遇到連續輸入字數時,我不管要輸入多少數字,永遠取最後一個的數字當動作次數

現在我卡在不知道如何判別這種機制,希望有高手可以給我一點提示幫我解決,謝謝!

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;
ENTITY key_controller IS
PORT(
   clk    : IN    std_logic;
   col    : OUT std_logic_vector ( 3 DOWNTO 0 );
   key_out : OUT std_logic;
   row    : IN    std_logic_vector ( 3 DOWNTO 0 );
   digit : OUT std_logic_vector ( 5 DOWNTO 0 )
);
END key_controller ;
architecture behavioral of key_controller is
signal clkm       : std_logic;
signal key_pressed : std_logic;
signal key_valid : std_logic := '0';
signal dbnq       : std_logic_vector(  6 downto 0 ) := ( others => '0' );  --debounce
signal scan_cnt : std_logic_vector(  3 downto 0 ) := ( others => '0' );
signal cmd_buf    : std_logic_vector(  2 downto 0 ) := ( others => '0' );
signal num_buf    : std_logic_vector(  2 downto 0 ) := ( others => '0' );
begin
count : process( clk )
  begin
if rising_edge( clk ) then
   count_t <= count_t + 1;
end if;
clkm <= count_t( 15 ); --scaning clock generaterd
  end process;
-----------------------------------------------
keyboard: process( clkm, scan_cnt, key_pressed )
begin
--scan_cnt is a 4 bit up scan counter
case scan_cnt( 3 downto 2 ) is --keypad column scaning, decoder functional
when "00" => col <= "0111";
  when "01" => col <= "1011";
  when "10" => col <= "1101";
  when "11" => col <= "1110";
  when others => col <= "1111";
  end case;
---------------------------------------------
  case scan_cnt( 1 downto 0 ) is --keypad row scaning, mux functional
when "00" => key_pressed <= row(0);
when "01" => key_pressed <= row(1);
  when "10" => key_pressed <= row(2);
when "11" => key_pressed <= row(3);
  when others => null;
  end case;
----------------------------------------------------------------
  if  rising_edge( clkm )  then
if key_pressed = '1' then --no key_pressed in, continue to scan
   scan_cnt <= scan_cnt + 1;
dbnq <= "1111111";
elsif key_pressed = '0' then --key_pressed in
      dbnq <= dbnq - 1;
end if;
----- debounce
if dbnq = "1001111" then --key_pressed debounce
   key_valid <= '1';
else key_valid <= '0';
end if;
end if;
-- scan_out <= scan_cnt;
  end process;
---------------------------------------------------------------------------
keycode : process( clk, scan_cnt, key_valid )
  begin
------------- asynchronous expression
if key_valid'event and key_valid = '1' then
if scan_cnt( 3 ) = '1' then
   cmd_flag <= '1';
   cmd_buf <= scan( 2 downto 0 ); --command store into ram
elsif scan_cnt( 3 ) = '0' then
      num_flag <= '1';
      num_buf <= scan( 2 downto 0 ); --numeric store into ram
else num_buf <= "000", --numeric store into ram
         cmd_buf <= "000"; --numeric store into ram
         num_fleg <= '0';
         cmd_fleg <= '0';
   end if;
   digit <= cmd_buf & num_buf; -- out of key controller
end if;
  end process;
-------------------
  key_out <= key_valid;
  digit <= seg4bit;
end behavioral;

作者: yhchang 時間: 2008-2-10 11:39 AM
說說看我的想法

設定一個前進次數的變數
default給1

接著去抓你鍵盤的Input
只要有數字出現,  就把前進次數的 value 用現在的input  取代掉

直到你按下前進啟動了  前進這個副程式
這個副程式就會去抓  你目前  前進次數的值
開始做遞減的運算

每前進一次  這個變數就會減1  (或是先減一再前進完全看你設定迴圈停止的條件是什麼?)
直到那個前進的次數到達時
前進的副程式就會停止動作

作者: w10789173 時間: 2008-2-10 02:41 PM
真是非常感謝你的意見

希望以後也還請多多的指教!

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