将两个整数合并成一个，然后解码 [英] Combine two integers into one and decode them later

问题描述

使用C ++我需要将两个不同的ID组合成一个16位整数。然后稍后我需要将这个16位整数解码为两个原始ID值。

示例：

  //将两个整数存储到一个
 unsigned short Identifier1 = 12793; //（maximum number 30000）
 unsigned short Identifier1 = 5450; //（maximum number 30000）
 unsigned short CombinedIDs = 34283; //这是例子，我不知道的代码
 
 //解码一个整数为两个
 //此时我只有CombinedIDs值，我需要提取它
 // into the two original IDs 
 
 unsigned short OriginalIdentifier1 = ... CombinedIDs ..获取代码12793 
无符号short OriginalIdentifier2 = ... CombinedIDs ..获取代码5450 
  

解决方案

这是不可能的。

假设你的两个标识符可以在 [0,30000] 的范围内，有30000 x 30000 =〜2 ^ 30可能的对身份标识。然而，只有2 ^ 16个可能的16位数字。因此，您不能将标识符对映射到16位整数，并希望从中恢复标识符。

---

您可以使用32位整数来存储组合，这种情况下编码和解码都很简单：

  unsigned short标识符1 = 12793; 
 unsigned short Identifier2 = 5450; 
 unsigned int CombinedIDs =（Identifier1<< 16）|标识符2; 
  

解码：

 code> unsigned short Identifier1 = CombinedIDs>> 16; 
 unsigned short Identifier2 = CombinedIDs& 0x0000FFFF 
  

请注意，现在不需要限制标识符在[0,30000]

回答您的意见：4和12位是可能的。

编码：

  unsigned short Identifier1; // 4 bits 
 unsigned short Identifier2; // 12 bits 
 unsigned short CombinedIDs =（Identifier1 << 12）|标识符2; 
  

解码：

 code> unsigned short Identifier1 = CombinedIDs>> 12; 
 unsigned short Identifier2 = CombinedIDs& 0x0FFF; 
  

Using C++ I need to combine two distinct IDs into one 16bit integer. Then later on I need to decode this 16bit integer into the two original ID values.

Example:

// Store two integers into one
unsigned short Identifier1 = 12793; //(maximum number 30000)
unsigned short Identifier1 = 5450; //(maximum number 30000)
unsigned short CombinedIDs = 34283; // this is example, I don't know the code for that

// Decode one integer into two
// At this point I only have CombinedIDs value, I need to extract it
// into the two original IDs

unsigned short OriginalIdentifier1 = ...CombinedIDs.. code to get 12793
unsigned short OriginalIdentifier2 = ...CombinedIDs.. code to get 5450

解决方案

This is impossible.

Assuming your two identifiers can be in the range [0, 30000], there are 30000 x 30000 = ~2^30 possible pairs of identifiers. However, there are only 2^16 possible 16-bit numbers. So, you can't possibly map pairs of identifiers to a 16-bit integer and expect to recover the identifiers from it.

---

Instead, you can use a 32-bit integer to store the combination, in which case both encoding and decoding is straightforward:

Encoding:

unsigned short Identifier1 = 12793;
unsigned short Identifier2 = 5450;
unsigned int CombinedIDs = (Identifier1 << 16) | Identifier2;

Decoding:

unsigned short Identifier1 = CombinedIDs >> 16;
unsigned short Identifier2 = CombinedIDs & 0x0000FFFF

Note that now the restriction that the identifiers be in the range [0, 30000] is not necessary - they be any unsigned short value.

---

EDIT Answering your comment: 4 and 12 bits are possible.

Encoding:

unsigned short Identifier1;  // 4 bits
unsigned short Identifier2;  // 12 bits
unsigned short CombinedIDs = (Identifier1 << 12) | Identifier2;

Decoding:

unsigned short Identifier1 = CombinedIDs >> 12;
unsigned short Identifier2 = CombinedIDs & 0x0FFF;

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