为什么我的程序在条件为假 (Ada) 时进入循环（并停留在那里）？

Question

我的程序中有一个变量控制用户是否要重复名为 "done." 的程序如果 done = '0'，程序重复；如果 done = '1'，程序退出。如果用户输入的值不是“0”或“1”，它会进入一个循环，提示他们输入有效的内容，直到他们输入为止。问题是无论用户输入什么，它都会（无限地）进入这个循环。

我关注的代码从第63行开始；这是完整的 .adb 文件：

with ada.text_io; use ada.text_io;
with ada.strings.bounded;
with unchecked_conversion;
with stack;

procedure main is
   package int_io is new ada.text_io.integer_io(integer);

   stack_type : positive;
   stack_size : natural;
begin
   loop
      -- Select stack type
      put_line("Select stack type by inputting corresponding number: ");
      put_line("1) Character");
      put_line("2) Integer");
      put_line("3) Date");
      int_io.get(stack_type);

      -- Select size of stack
      put_line("Enter the maximum size of your stack: ");
      int_io.get(stack_size);

      case (stack_type) is
     -- Character stack (option C/D)
     when 1 =>
        declare
           package char_stack is new stack(stack_size, character); use char_stack;
           char : character;
           done : integer := 0;
           num_chars : natural := 48;  -- 48 = ASCII '0'
           -- Reflects how many characters (including the number) of the name on top of the stack; used for limit in popping for loop
           num_to_pop : natural := 0;

           -- Pop and print characters from stack
           procedure print_chars is
           begin
          for i in 0 .. (num_to_pop - 1) loop
             put(char_stack.pop);
          end loop;

          -- Reset num_to_pop
          num_to_pop := 0;
           end print_chars;
        begin
           put_line("Enter characters; To finish sequence, terminate with '#': ");
           new_line;

           while (done /= 1) loop
          while (char /= '#') loop
             get(char);
             if (char /= '#') then
            push(char);
            num_chars := num_chars + 1;
             end if;
          end loop;

          -- Push number of characters onto stack; reset num_to_pop; reset num_chars
          push(character'val(num_chars));
          num_to_pop := (character'pos(peek)) - 47;
          num_chars := 48;

          -- Ask user if they wish to continue
          new_line;
          put_line("To repeat, enter '0'; to terminate program, enter '1': ");
          new_line;
          int_io.get(done);

          -------------TEST----------------
          new_line;
          int_io.put(done);
          new_line;
          ---------------------------------
          while (done /= 0 or done /= 1) loop
             put_line("Please enter either a '0' or a '1'.");
             new_line;
             int_io.get(done);
          end loop;
           end loop;
     end;
    -- Integer stack (option B)
    when 2 =>
        begin
           put("asfsF");
        --
     end;
     -- Date stack (option A)
     when 3 =>
        begin
           put("Asfasdf");
        --
        end;

     when others =>
        begin
           put("asdfasdF");
        end;
    end case;
   end loop;
end main;

Answer 1

你写

      while (done /= 0 or done /= 1) loop

但你应该写

      while (done /= 0 and done /= 1) loop

甚至，因为 Ada 在这里不需要括号，

      while done /= 0 and done /= 1 loop

或者，应用德摩根定律，

      while not (done = 0 or done = 1) loop

为什么我的程序在条件为假 (Ada) 时进入循环（并停留在那里）？

Why is my program entering a loop (and staying there) when condition is false (Ada)?

ada

while-loop