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algorithm - 将多个值散列/编码为单个整数值的算法

问题描述

有这种算法用于“散列”或将多个值编码为单个整数，通过将指数增加的数值分配给各个值。这种方法特别用于 Windows DLL。

一个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目列表。

例如，如果我们有以下值：

* open
* assigned
* completed
* closed

...我们为每个分配一个数值：

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

等等，其中每个后续值是前一个值的 2 倍。

编码

当我们需要传递任何这些值的组合时，我们将相应的数值相加。例如，对于“打开，已分配”，它是3，对于“已分配，完成，关闭”，它是14。这涵盖了所有独特的组合。正如我们所见，“编码”部分非常简单。

解码

要解码该值，我能想到的唯一方法是 switch..case 语句，如下所示（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

该算法显然在以下假设下有效：

问题：

  1. 什么是“解码”值而不是非常复杂的 switch..case 语句的更优化方式/算法？
  2. 这个算法有名字吗？

注意：该问题被标记winapi主要是为了便于发现。该算法相当普遍。

标签： algorithmwinapiencodinghashencode

解决方案

您所描述的正式称为位掩码，其中整数中的每个位都被分配了一个含义。位被分配为二进制 2 的幂的数值（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 逻辑位运算符来设置/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，这最好使用 aSet来处理，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_Assigned, State_Completed, State_Closed];
  // whatever combination you need ...
  DoSomething(lState);
  ...
end;

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