问题

I will only provide translations for the code-related portions of your text. Here's the translated code:

第一次在这里提问，请容忍我如果问题不够好。

Visual Studio。
C++。
UDP。

我接收到一个UDP数据包，将其存储在一个char缓冲区中。
我有一个按字节对齐的结构体，用来将缓冲区拆分成结构体的各个元素。

我需要从结构体中获取经度和纬度，并输出到控制台，但结果看起来不正确。

当我使用Wireshark查看UDP数据包时，我可以看到经度和纬度，并知道它们代表的值：
3FA24690119236AC -> 12.41890000（N）
3FCBBE7C9C146695 -> 2.04515000（E）

我正在使用的经度和纬度的数据类型是unsigned long long。

我无法弄清楚如何从3FCBBE7C9C146695转换为12.41890000，考虑到它可能是小端字节序。

是否有人可以友善地解释我应该做什么，以初学者的思维方式，并演示一些代码。

我明白你不会有UDP数据包。

这是我希望看到的理解（即使我是否正确理解过程也不确定）。

```cpp
char buffer[] = { '3F', 'A2', '46', '90', '11', '92', '36', 'AC' };
struct payload { unsigned long long lat; };

payload p;
std::memcpy(&p, buffer, sizeof(p));

.... 转换为十进制度数

printf("Lat: %llu\n", p.lat);

再次说明，我是新手，希望学习！

非常感谢提前的帮助，如果需要改进问题的格式，请告诉我！

```cpp
p.lat = ntohll(p.lat);
p.lat = ntohd(p.lat);

尝试了这些，并将数据类型更改为double、long、int64、uint64，但现在感到困惑。

更多信息....

我想要看到的输出是12.41250000°N 2.43520000°E。

仅有UDP数据包（lat和lon是唯一感兴趣的部分）

39000000000000000000000000030000006500000059ffffffffffffffff41d9205f3f80000000000001c0a801b73fcbbad397da94bd3fa5c2db1fc191f800000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000017ae

文档说明这些（lat、lon）应该是
8字节双精度浮点数（接受的值：-π/2 - π/2（弧度））用于纬度
8字节双精度浮点数（接受的值：-π - π（弧度））用于经度

现在使用unsigned char数组（UDP -> 结构体）

然后

double result;
std::memcpy(&result, p.lat, sizeof(double));
std::cout << "Lat: " << result << std::endl;

得到：Lat: -4.74162e-12

打印出数组：
3fa5c2db1fc191f8
所以似乎数组包含了我在Wireshark中看到的正确十六进制值。

请尽量“简化”它，因为我真的想了解我遇到的问题，而不仅仅是解决它。

再次感谢！

<details>
<summary>英文:</summary>

First ever time asking a question on here so go easy on me if its not up to par.

Visual Studio.
C++.
UDP.

I receive a UDP payload which I store in a char buffer.
I have a struct which is byte-aligned to split the buffer into the elements of the struct.

I need to get the lon, lat from the struct and cout to console but it doesnt look correct.

When i wireshark the UDP payload i can see the lon, lat and i know the values it represents:
3FA24690119236AC -&gt; 12.41890000 (N)
3FCBBE7C9C146695-&gt; 2.04515000  (E)

The data type for the lon, lat im using is unsigned long long.

I cannot work out how to go from 3FCBBE7C9C146695 -&gt; 12.41890000 bearing in mind its probably little-endian.

Could someone be so kind as to explain what I should be doing, in a &lt;1 year juniors mindset and demonstrate with some code. 

I understand you wont have the UPD payload.

This is my understanding of what I would like to see (if I even understand the process correctly).

char buffer[] = { '3F', 'A2', '46', '90', '11', '92', '36', 'AC'};
struct payload { unsigned long long lat; };

payload p;
std::memcpy(&p, buffer, sizeof(p));

.... convert to its dec degree

printf("Lat: %llu\n", p.lat);

Again I am a newbie and wish to learn!

Many thanks in advance, also let me know if I need to format my question better!




`p.lat = ntohll(p.lat);`
`p.lat = ntohd(p.lat);
`

Tried these and changing the data type to double, long, int64, uint64 but I'm chasing my tail now.


More Info....

12.41250000°N 2.43520000°E is the output i want to see.

UDP payload only (**lat**, *lon* are the only parts of interest)

39000000000000000000000000030000006500000059ffffffffffffffff41d9205f3f80000000000001c0a801b7**3fcbbad397da94bd***3fa5c2db1fc191f8*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000017ae

Documentation states these (lat, lon) should be 
8 byte double	(accepted values: -π/2 - &pi/2; (Radians)) for lat
8 byte double	(accepted values: -π - π (Radians) for lon

Now using `unsigned char arrays (UPD -> Struct)` 



then 


    double result;
    std::memcpy(&result, p.lat, sizeof(double));
    std::cout << "Lat: " << result << std::endl;

gives: Lat: -4.74162e-12 

Printing out the array:
3fa5c2db1fc191f8
 so it seems the array hold the correct hex i see in wireshark.

Please do try and 'dumb it down' as I really would like to understand the issue i have and not just to solve it.

Thanks Again

</details>


# 答案1
**得分**: 0

Clue1: 弧度

Clue2: 没有什么能阻止你尝试将角度和弧度的双精度转换成char[]，看看它给出了什么结果。

事实证明，它转换的是弧度，而不是角度版本的角度。至少在我的电脑上，你还必须反转字节的顺序。（但我不知道这是多么普遍的现象。）

```cpp
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const double PI = 4.0 * atan( 1.0 );

double decode( unsigned char *buffer, int n )
{
   reverse( buffer, buffer + n);
   double result;
   memcpy( &result, buffer, n );
   return result * 180.0 / PI;
}

int main()
{
   const int N = sizeof( double );
   unsigned char buffer1[N] = { 0x3f, 0xcb, 0xba, 0xd3, 0x97, 0xda, 0x94, 0xbd };
   unsigned char buffer2[N] = { 0x3f, 0xa5, 0xc2, 0xdb, 0x1f, 0xc1, 0x91, 0xf8 };
   cout << "Latitude:  " << decode( buffer1, N ) << '\n';
   cout << "Longitude: " << decode( buffer2, N ) << '\n';
}

输出：

纬度：12.4125
经度：2.4352

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const double PI = 4.0 * atan( 1.0 );

double decode( unsigned char *buffer, int n )
{
   reverse( buffer, buffer + n);
   double result;
   memcpy( &result, buffer, n );
   return result * 180.0 / PI;
}

int main()
{
   const int N = sizeof( double );
   unsigned char buffer1[N] = { 0x3f, 0xcb, 0xba, 0xd3, 0x97, 0xda, 0x94, 0xbd };
   unsigned char buffer2[N] = { 0x3f, 0xa5, 0xc2, 0xdb, 0x1f, 0xc1, 0x91, 0xf8 };
   cout << "Latitude:  " << decode( buffer1, N ) << '\n';
   cout << "Longitude: " << decode( buffer2, N ) << '\n';
}

Latitude:  12.4125
Longitude: 2.4352