Question

我的任务是从键盘输入2个输入，这些输入是从0到9的数字并添加它们。问题在于存储这些数字。我想先保存（输入编号）进入＆＃34; a＆＃34;然后按（输入编号）进入＆＃34; B＆＃34;但是通过使用以下代码，首先在a和b中按商店。第二次按下是没用的。这里 scan_code =按下按钮的扫描码（键盘接口代码的输出） a =二进制数（例如，如果我第一次按＆＃34; 1＆＃34;则代码检查扫描码并将＆＃34; 1＆＃34;的二进制值分配给a）。谁能帮忙？

process (clk, scan_code, cin)
  variable scancode1 : std_logic_vector (7 downto 0) := "00000000";
  variable cin2      : std_logic_vector(2 downto 0);
begin
  if(clk'event and clk = '1') then

    scancode1 := scan_code;
    a         <= "0000";
    b         <="0000";
    if (scancode1 = "00010110") then
      a <= "0001";
    elsif (scancode1 = "00011110") then
      a <="0010";
    elsif (scancode1 = "00100110") then
      a <="0011";
    elsif (scancode1 = "00100101") then
      a <="0100";
    elsif (scancode1 = "00101110") then
      a <="0101";
    elsif(scancode1 = "00110110") then
      a <="0110";
    elsif (scancode1 = "00111101") then
      a <="0111";
    elsif (scancode1 = "00111110") then
      a <="1000";
    elsif (scancode1 = "01000110") then
      a <="1001";
    elsif (scancode1 = "01000101") then
      a <="0000";
    end if;

    if (scancode1 = "01010101") then    --scancode for + sign
      a <=a;
    end if;

    if (scancode1 = "00010110") then
      b <="0001";
    elsif (scancode1 = "00011110") then
      b <="0010";
    elsif (scancode1 = "00100110") then
      b <="0011";
    elsif (scancode1 = "00100101") then
      b <="0100";
    elsif (scancode1 = "00101110") then
      b <="0101";
    elsif(scancode1 = "00110110") then
      b <="0110";
    elsif (scancode1 = "00111101") then
      b <="0111";
    elsif (scancode1 = "00111110") then
      b <="1000";
    elsif (scancode1 = "01000110") then
      b <="1001";
    elsif (scancode1 = "01000101") then
      b <="0000";
    end if;

    sum(0)  <= a(0) xor b(0) xor cin;
    cin2(0) := (a(0) and b(0)) or (cin and (a(0) xor b(0)));

    sum(1)  <= a(1) xor b(1) xor cin2(0);
    cin2(1) := (a(1) and b(1)) or (cin2(0) and (a(1) xor b(1)));

    sum(2)  <= a(2) xor b(2) xor cin2(1);
    cin2(2) := (a(2) and b(2)) or (cin2(1) and (a(2) xor b(2)));

    sum(3) <= a(3) xor b(3) xor cin2(2);
    cout   <= (a(3) and b(3)) or (cin2(2) and (a(3) xor b(3)));

  end if;
end process;

Answer 1

＆＃34;＆＃34;和＆＃34; b＆＃34;正在你这样写的同时进行更新（同一＆＃34;扫描码＆＃34;检查＆＃34; a＆＃34;＆＃34; b＆＃34;在同一个时钟上）。考虑使用事件触发器，如下例所示：

...
if reset='1' then
   a<=(others => '0');
   b<=(others => '0');
   event_last<='0';
   event_nr<=0;

elsif rising_edge(clk) then 
   event_last<=event;

   -- trigger on "rising edge" of event
   if event='1' and event_last='0' then

        case event_nr is
        -- first event is "a"
        when 0 =>
            event_nr<=1;

            if (scancode1 = "00010110") then
                 a <= "0001";
            ...

        -- second event is "b"
        when others => 
            event_nr<=0;

            if (scancode1 = "00010110") then
                b <= "0001";
            ...
        end case;

   ...
   ...

使用斯巴达3（vhdl）从键盘添加2个数字

1 个答案: