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Question

看到以下示例代码：

architecture arch of disp_mux is
  constant N:integer :=18;
  signal q_reg, q_next: unsigned(N-1 downto 0);
  signal sel: std_logic_vector(1 downto 0);
begin
  process(clk, reset)
  begin
    if reset='1' then
      q_reg <= (others=>'0');
    elsif (clk'event and clk='1') then
      q_reg <= q_next;
    end if;
  end process;
  q_next <= q_reg + 1;
  sel <= std_logic_vector(q_reg(N-1 downto N-2));
  process(sel, ...)
  begin
    case sel is...
    end case;
  end process;
end arch;

如果您从未将1写入reset，q_reg的价值是多少？如果q_next <= q_reg + 1;是q_reg UUUUUUUUUUUUUUUUUU，a.nav-toggle会发生什么，这是我怀疑的。

Answer 1

如果您从未向reset写1，则q_reg为"UUUUUUUUUUUUUUUUUU"，执行q_next <= q_reg + 1后q_next成为"XXXXXXXXXXXXXXXXXX"。