Question

我需要在Ada中创建一个哈希函数，它接受一个字符串并返回一个整数。我到目前为止所做的是：

function hash(Word: unbounded string) return Integer is
    h := 5381
    c := (first charater of "Word")
begin
    while c /= (end of "Word") loop
         h := h*33 + c;
         c := (next character of "Word");
    end while;
return h mod 20;
end hash;

我不知道如何在Ada中选择一个角色，怎么说“我想要这个词的第三个字”字“，这是r。

感谢您的帮助，

吨。

Answer 1

查看Ada.Strings.Unbounded包提供的服务。

Answer 2

正如Marc所说，您需要使用Ada.Strings.Unbounded来获取Word的内容。

您还需要解决另外两个问题：

在Ada中，Unbounded_String（或String中没有终止字符，来到那里）。相反，将循环写为for J in 1 .. (length of Word) loop。
一旦你提取了J个字符，它仍然是Character，你无法将其添加到h（我认为h是不可或缺的）。 Character'Pos (Ch)返回字符的等效数字，并且可以添加。

Answer 3

function hash(Word: Ada.Strings.Unbounded.Unbounded_String) return Integer is
-- First, because there's no manipulation of the string's
-- contents, doing the work on an unbounded-string is
-- rather pointless... so let's do our work on a regular --' fix for formatting
-- [static-length] string.
Working : String := Ada.Strings.Unbounded.To_String(Word);
-- Second, you need types in your declarations.
h : Integer := 5381;
c : Character := 'e'; --(first charater of "Word");
begin
-- Why use a 'while' loop here? Also, what if the 'word' is
-- abracadabra, in that case c [the first letter] is the 
-- same as the last letter... I suspect you want an index.

  for Index in Working'Range loop -- Was: while c /= EOW loop --'
    declare
    -- This is where that 'c' should actually be.
    This  : Character renames Working(Index);
    -- Also, in Ada characters are NOT an integer.
    Value : constant Integer := Character'Pos( This ); --'
    begin
    h := h*33 + value; -- PS: why 33? That should be commented.
    -- We don't need the following line at all anymore. --'
    --c := (next character of "Word");
    end;        
  end loop;
return h mod 20;
end hash;

当然，这也可以重写，以利用Ada 2012中新的循环结构。

function hash_2012(Word: Ada.Strings.Unbounded.Unbounded_String) return Integer is
    -- Default should be explained.
    Default : Constant Integer := 5381;
    Use Ada.Strings.Unbounded;
begin
    -- Using Ada 2005's extended return, because it's a bit cleaner.
    Return Result : Integer:= Default do
        For Ch of To_String(Word) loop
            Result:= Result * 33 + Character'Pos(Ch); --'
        end loop;

        Result:= Result mod 20;
    End return;
end hash_2012;

...我必须问，格式化程序发生了什么？这太可怕了。

选择字符串中的字母

3 个答案: