LATCH Primitive 禁用输出？答案

【问题标题】：LATCH Primitive disables outputs?LATCH Primitive 禁用输出？
【发布时间】：2015-06-01 23:09:25
【问题描述】：

所以我理解了闩锁的概念，但我没有看到我在这里如何推断一个，因为我的 else 条件应该涵盖通过这个过程的所有可能路径。 Quartus 告诉我，由于 LATCH 原语，它正在禁用 greenLEDS 和 redLEDs，并且它们上有一个推断的锁存器。我在这里遗漏了什么吗？

process(current,advance,playerWins,dealerWins) begin
  if(advance) then
        case current is
          when START =>
                deal           <= '1';
                dealTo         <= '1';
                dealToCardSlot <= "00";
                greenLeds <= (others => '0');
                redLeds <= (others => '0');
          when PT1 =>
                deal           <= '1';
                dealTo         <= '1';
                dealToCardSlot <= "01";
                greenLeds <= (others => '0');
                redLeds <= (others => '0');
          when DT1 =>
                deal           <= '1';
                dealTo         <= '0';
                dealToCardSlot <= "00";
                greenLeds <= (others => '0');
                redLeds <= (others => '0');
          when PT2 =>
                deal           <= '1';
                dealTo         <= '1';
                dealToCardSlot <= "10";
                greenLeds <= (others => '0');
                redLeds <= (others => '0');
          when PT3 =>
                deal           <= '1';
                dealTo         <= '1';
                dealToCardSlot <= "11";
                greenLeds <= (others => '0');
                redLeds <= (others => '0');
          when DT2 =>
                deal           <= '1';
                dealTo         <= '0';
                dealToCardSlot <= "01";
                greenLeds <= (others => '0');
                redLeds <= (others => '0');
          when DT3 =>
                deal           <= '1';
                dealTo         <= '0';
                dealToCardSlot <= "10";
                greenLeds <= (others => '0');
                redLeds <= (others => '0');
          when DT4 =>
                deal           <= '1';
                dealTo         <= '0';
                dealToCardSlot <= "11";
                greenLeds <= (others => '0');
                redLeds <= (others => '0');
          when WIN =>
                deal           <= '0';
                dealTo         <= '0';
                dealToCardSlot <= "00";
                if(playerWins) then
                    greenLeds <= (others => '1');
                elsif(dealerWins) then
                    redLeds <= (others => '1');
                else
                    greenLeds <= (others => '0');
                    redLeds <= (others => '0');
                end if;
          when ENDGAME =>
                deal           <= '0';
                dealTo         <= '1';
                dealToCardSlot <= "00";
                greenLeds <= (others => '0');
                redLeds <= (others => '0');
          when others =>
                deal           <= '0';
                dealTo         <= '0';
                dealToCardSlot <= "00";
                greenLeds <= (others => '0');
                redLeds <= (others => '0');
          end case;
  else
    deal           <= '0';
    dealTo         <= '0';
    dealToCardSlot <= "00";
    greenLeds <= (others => '0');
    redLeds <= (others => '0');
  end if;
end process;

【问题讨论】：

标签： vhdl synthesis quartus

【解决方案1】：

      when WIN =>
            deal           <= '0';
            dealTo         <= '0';
            dealToCardSlot <= "00";
            if(playerWins) then
                greenLeds <= (others => '1');
            elsif(dealerWins) then
                redLeds <= (others => '1');
            else
                greenLeds <= (others => '0');
                redLeds <= (others => '0');
            end if;

如果playerWins 为假且dealerWins 为真，则不分配redLeds。同样，如果playerWins 为真，则不要分配greenLeds。

【讨论】：