将多个值散列/编码为单个整数值的算法

问题描述

此算法可通过将指数递增的数字值分配给各个值来“散列”或将多个值编码为单个整数。这种方法尤其在Windows DLL中使用。

可能的用例可以是客户端应用程序从API请求与某些状态代码匹配的项列表。

例如，如果我们具有以下值：

* open
* assigned
* completed
* closed

...我们为每个数字分配一个数值：

* open - 1
* assigned - 2
* completed - 4
* closed - 8

等其后每个值是前一个值的2倍。

编码

当我们需要传递这些值中的任何一个的组合时，我们会将相应的数值相加。例如，对于“打开，已分配”，它是3，对于“已分配，已完成，关闭”，它是14。这涵盖了所有独特的组合。如我们所见，“编码”部分非常简单。

解码

要解码该值，我唯一想到的方法是switch..case语句，例如（伪代码）：

1 = open
2 = assigned
3 = open + assigned
4 = completed
5 = open + completed
6 = assigned + completed
7 = open + assigned + completed
8 = closed
9 = open + closed
10 = assigned + closed
11 = open + assigned + closed
12 = completed + closed
13 = open + completed + closed
14 = assigned + completed + closed
15 = open + assigned + completed + closed

此算法显然可以在以下假设下工作：

• 仅在每个值仅使用一次时起作用
• 仅在双方都知道匹配的数值时才起作用

问题：

1. “解码”值而不是非常复杂的switch..case语句的最佳方法/算法是什么？
2. 此算法有名称吗？

注意：该问题用winapi标记的主要是为了发现问题。该算法相当通用。

解决方法

您所描述的内容被正式称为bit mask，其中整数中的每个位都分配有一个含义。这些位被分配为二进制2的幂（bit0 = 2 ⁰ = 1，bit1 = 2 ¹ = 2，bit2 = 2 ² = 4，bit3 = 2 ³ = 8等）。

您可以使用OR和AND logical bitwise operators来设置/查询整数中的各个位，例如：

const DWORD State_Open = 1;
const DWORD State_Assigned = 2;
const DWORD State_Completed = 4;
const DWORD State_Closed = 8;

void DoSomething(DWORD aStates)
{
  ...

  if (aStates & State_Open)
    // open is present
  else
    // open is not present

  if (aStates & State_Assigned)
    // assigned is present
  else
    // assigned is not present

  if (aStates & State_Completed)
    // completed is present
  else
    // completed is not present

  if (aStates & State_Closed)
    // closed is present
  else
    // closed is not present

  ...
}

DWORD lState = State_Open | State_Assigned | State_Completed | State_Closed;
// whatever combination you need ...
DoSomething(lState);

在Delphi / Pascal中，最好使用Set处理此问题，它在内部实现为位掩码，例如：

type
  State = (State_Open,State_Assigned,State_Completed,State_Closed);
  States = Set of State;

procedure DoSomething(aStates: States);
begin
  ...

  if State_Open in aStates then
    // open is present
  else
    // open is not present

  if State_Assigned in aStates then
    // assigned is present
  else
    // assigned is not present

  if State_Completed in aStates then
    // completed is present
  else
    // completed is not present

  if State_Closed in aStates then
    // closed is present
  else
    // closed is not present

  ...
end;

var
  lState: States;
begin
  ...
  lState := [State_Open,State_Closed];
  // whatever combination you need ...
  DoSomething(lState);
  ...
end;