首页 \ 问答 \ 将int值转换为3字节数组（反之亦然）(Converting int value into 3 byte array (and vice versa))

将int值转换为3字节数组（反之亦然）(Converting int value into 3 byte array (and vice versa))

 我正在研究一个C / C ++ WinForms应用程序，它可以从硬件设备读取/写入数据。 我的应用程序有一个多选列表框，其中包含数字1 - 100000，用户最多可以选择10个数字。 当他们完成选择每个数字时，用户点击一个按钮，我的事件处理程序代码需要使用3个字节来构建固定大小（30字节）的字节数组来表示每个选定的数字，并且如果少于10个数字则填充数组选择。  
 作为一个例子，假设我的用户选择以下值：  
17
99152
3064
52588
65536
 
 我目前使用此代码将每个数字转换为一个字节数组：  
byte[] bytes = BitConverter.GetBytes(selectedNumber);
Array.Reverse(bytes) // because BitConverter.IsLittleEndian() = true
Debug.WriteLine(BitConverter.ToString(bytes));
 
 对于上面列出的数字，这会产生以下结果：  
00-00-00-11
00-01-83-50
00-00-0B-F8
00-00-CD-6C
00-01-00-00
 
 BitConverter给我回4个字节的数组，我只有空间使用3个字节来存储每个数字在最后的字节数组中。 我可以删除每个单独的字节数组的最重要的字节，然后像这样构建我的最终数组：  
00-00-11-01-83-50-00-0B-F8-00-CD-6C-01-00-00-[padding here]
 
 写入设备应该可以工作。 但是从设备读取数组（或类似的数组）会给我带来一些问题。 当我有一个3字节的数组，并尝试使用此代码将其转换为int ...  
int i = BitConverter.ToInt32(bytes, 0);
 
 ...我得到“目标阵列不够长，无法复制集合中的所有项目。” 我想我可以在每三个字节的开始插入一个0x00的最重要字节，然后转换它，但有没有更好的方法来做到这一点？ 

I am working on a C# WinForms application that reads/writes data to/from a hardware device. My application has a multiselect listbox which contains the numbers 1 - 100000 and the user may select up to 10 numbers. When they're done selecting each number, the user clicks a button and my event handler code needs to build a fixed-size (30 bytes) byte array using 3 bytes to represent each selected number and pad the array if less than 10 numbers were selected. 
As an example, suppose my user chooses the following values: 
17
99152
3064
52588
65536
 
I'm currently using this code to convert each number into a byte array: 
byte[] bytes = BitConverter.GetBytes(selectedNumber);
Array.Reverse(bytes) // because BitConverter.IsLittleEndian() = true
Debug.WriteLine(BitConverter.ToString(bytes));
 
For the numbers I listed above, this produces the following: 
00-00-00-11
00-01-83-50
00-00-0B-F8
00-00-CD-6C
00-01-00-00
 
BitConverter is giving me back a 4 byte array where I only have space to use 3 bytes to store each number in the final byte array. I can drop the most significant byte of each individual byte array and then build my final array like this: 
00-00-11-01-83-50-00-0B-F8-00-CD-6C-01-00-00-[padding here]
 
Writing that to the device should work. But reading the array (or a similar array) back from the device causes a bit of a problem for me. When I have a 3 byte array and try to convert that into an int using this code... 
int i = BitConverter.ToInt32(bytes, 0);
 
...I get "Destination array is not long enough to copy all the items in the collection." I suppose I could insert a most significant byte of 0x00 at the beginning of every three bytes and then convert that but is there a better way to do this?

原文：https://stackoverflow.com/questions/14061448

更新时间：2022-02-24 20:02

最满意答案

 你只需要检查你的itlow是否已经是items.begin() 。 如果是，地图中没有这样的元素：  
itlow=items.lower_bound(key);
if(itlow->first == key)
    return itlow->second;
else if(itlow != items.begin())
    itlow--;
    return itlow->second;
else
    throw some_exception();
 
 您可以返回迭代器，而不是抛出异常，然后如果找不到这样的元素，则可以返回items.end() 。  
#include <iostream>
#include <map>

using namespace std;


map<int, int>::const_iterator find(const map<int, int> &items, int value)
{
    auto itlow = items.lower_bound(value);

    if(itlow->first == value)
        return itlow;
    else if(itlow != items.cbegin())
        return --itlow;
    else 
        return items.cend();

}

int main()
{
  map<int, int> items;
          items[2]=0;
  items[6]=10;
  items[15]=18;
  items[20]=22;

  auto i = find(items, 0);
  if(i != items.cend())
  {
     cout << i->second << endl;
  }
  i = find(items, 15);
  if(i != items.cend())
  {
    cout << i->second << endl;
  }
  i = find(items, 9);
  if(i != items.cend())
  {
    cout << i->second << endl;
  }
}

You just need to check if your itlow is already items.begin(). If it is, there's no such element in the map: 
itlow=items.lower_bound(key);
if(itlow->first == key)
    return itlow->second;
else if(itlow != items.begin())
    itlow--;
    return itlow->second;
else
    throw some_exception();
 
Instead of throwing exception, you may return iterator, and then you can return items.end() if no such element is found. 
#include <iostream>
#include <map>

using namespace std;


map<int, int>::const_iterator find(const map<int, int> &items, int value)
{
    auto itlow = items.lower_bound(value);

    if(itlow->first == value)
        return itlow;
    else if(itlow != items.cbegin())
        return --itlow;
    else 
        return items.cend();

}

int main()
{
  map<int, int> items;
          items[2]=0;
  items[6]=10;
  items[15]=18;
  items[20]=22;

  auto i = find(items, 0);
  if(i != items.cend())
  {
     cout << i->second << endl;
  }
  i = find(items, 15);
  if(i != items.cend())
  {
    cout << i->second << endl;
  }
  i = find(items, 9);
  if(i != items.cend())
  {
    cout << i->second << endl;
  }
}

将int值转换为3字节数组（反之亦然）(Converting int value into 3 byte array (and vice versa))

最满意答案

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