blocks|key|1315918|text|您试过访问各个字节吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1315919|uint8_t+packet[4];
uint32_t+value;
packet[0]+=+(value+>>+24)+&+0xFF;
packet[1]+=+(value+>>+16)+&+0xFF;
packet[2]+=+(value+>>++8)+&+0xFF;
packet[3]+=+value+&+0xFF;|code-block|syntax|javascript|1315920|可能比调用函数更有效。:-)|1315921|注:以上方法与平台无关。不需要任何关于如何存储value的知识。|offset|length|style|CODE|1315922|解释：|1315923|让packet成为需要以大端格式存储到缓冲区中的uint32_t的缓冲区或内存目的地(首先是最重要的字节格式)。|1315924|表达式(value+>>+24)将最重要的字节移动到最小的(字节)位置。表达式"&+0xff“截断、删除任何无关的值，从而产生一个无符号的8位值。然后，将该值存储在第一位置的缓冲区中的最重要位置。|1315925|对于剩余的字节，也是如此。|1315926|entityMap^0|0|0|0|N|5|0|0|1|6|O|8|0|4|B|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|11|8|@$K|12|L|13|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|15|8|@$K|16|L|17|M|N]|$K|18|L|19|M|N]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1A|8|@$K|1B|L|1C|M|N]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1D|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|X|$]]

Have you tried accessing the individual bytes? 

<pre><code>uint8_t packet[4];
uint32_t value;
packet[0] = (value &gt;&gt; 24) &amp; 0xFF;
packet[1] = (value &gt;&gt; 16) &amp; 0xFF;
packet[2] = (value &gt;&gt; 8) &amp; 0xFF;
packet[3] = value &amp; 0xFF;
</code></pre>

Probably more efficient than calling a function. :-)

Note: the above method is platform independent. Doesn't require any knowledge how the <code>value</code> is stored.

Explanation: 
Let <code>packet</code> be a buffer or memory destination for a <code>uint32_t</code> that needs to be stored into the buffer in Big Endian format (Most Significant Byte First). 

The expression (<code>value &gt;&gt; 24</code>) shifts the most significant byte to the least significant (byte) position. The expression "&amp; 0xff" truncates, chops off, any extraneous values, resulting in an unsigned 8-bit value. The value is then stored in the Most Significant position in the buffer at the first location. 

Likewise for the remaining bytes.

blocks|key|2069510|text|推荐的方法是使用函数htonl+(+16位整数的htons)将主机字节顺序转换为网络字节顺序，使用ntohl+(+16位整数的ntohs)将网络字节顺序转换为主机字节顺序。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2069511|鉴于以下结构：|2069512|struct+UUID_FIELDS
{
++++uint32_t+time_low;++++++++++++++++
++++uint16_t+time_mid;+++++++++++++++++
++++uint16_t+time_hi_and_version;++++++
++++uint8_t+clock_seq_hi_and_reserved;
++++uint8_t+clock_seq_low;+++++++++++++
++++uint8_t+node[6];+++++++++++++++++++
};|code-block|syntax|javascript|2069513|序列化和反序列化的代码可能如下：|2069514|std::string+serialize(UUID_FIELDS+fields)+{
++++std::string+buffer(sizeof(fields),+'\0');

++++//+convert+all+fields+with+size+>+1+byte+to+big+endian+(network+byte+order)
++++fields.time_low+=+htonl(fields.time_low);
++++fields.time_mid+=+htons(fields.time_mid);
++++fields.time_hi_and_version+=+htons(fields.time_hi_and_version);

++++memcpy(&buffer[0],+&fields,+sizeof(fields));

++++return+buffer;
}

UUID_FIELDS+deserialize(const+std::string&+buffer)+{
++++UUID_FIELDS+fields;

++++assert(buffer.size()+==+sizeof(fields));

++++memcpy(&fields,+&buffer[0],+sizeof(fields));

++++//+convert+all+fields+with+size+>+1+byte+to+little+endian+(maybe)+(host+byte+order)
++++fields.time_low+=+ntohl(fields.time_low);
++++fields.time_mid+=+ntohs(fields.time_mid);
++++fields.time_hi_and_version+=+ntohs(fields.time_hi_and_version);

++++return+fields;
}|2069515|请注意，您必须与远程端点中的接收方/发送方达成一致，即在大端点发送/接收数字。|2069516|entityMap|0|LINK|mutability|MUTABLE|url|http://linux.die.net/man/3/ntohl|1^0|A|5|O|5|1D|5|1R|5|A|5|0|1D|5|1|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@$9|1A|A|1B|1|1C]|$9|1D|A|1E|1|1F]]|E|$]]|$1|F|3|G|5|6|7|1G|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1H|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1I|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1J|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1K|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1L|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]|10|$5|V|W|X|E|$Y|Z]]]]

The recommended way is to use the function <a href="http://linux.die.net/man/3/ntohl" rel="nofollow"><code>htonl</code></a> (<code>htons</code> for 16 bit integers) to convert from host byte order to network byte order, and <a href="http://linux.die.net/man/3/ntohl" rel="nofollow"><code>ntohl</code></a> (<code>ntohs</code> for 16 bit integers) to convert from network byte order to host byte order

Given the following structure:

<pre><code>struct UUID_FIELDS
{
 uint32_t time_low; 
 uint16_t time_mid; 
 uint16_t time_hi_and_version; 
 uint8_t clock_seq_hi_and_reserved;
 uint8_t clock_seq_low; 
 uint8_t node[6]; 
};
</code></pre>

The code for serialization and deserialization might be as follows:

<pre><code>std::string serialize(UUID_FIELDS fields) {
 std::string buffer(sizeof(fields), '\0');

 // convert all fields with size &gt; 1 byte to big endian (network byte order)
 fields.time_low = htonl(fields.time_low);
 fields.time_mid = htons(fields.time_mid);
 fields.time_hi_and_version = htons(fields.time_hi_and_version);

 memcpy(&amp;buffer[0], &amp;fields, sizeof(fields));

 return buffer;
}

UUID_FIELDS deserialize(const std::string&amp; buffer) {
 UUID_FIELDS fields;

 assert(buffer.size() == sizeof(fields));

 memcpy(&amp;fields, &amp;buffer[0], sizeof(fields));

 // convert all fields with size &gt; 1 byte to little endian (maybe) (host byte order)
 fields.time_low = ntohl(fields.time_low);
 fields.time_mid = ntohs(fields.time_mid);
 fields.time_hi_and_version = ntohs(fields.time_hi_and_version);

 return fields;
}
</code></pre>

Note that you have to agree with the receiver/sender in the remote endpoint that you are both sending/receiving numbers in big endian.

blocks|key|1315969|text|回复：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1315970|更新:在小终端启动uuid：
446831D3-FBA1-477B-BB07-CB067B00E86B
正确交换必要字段的结果：
FBA1477B-31D3-4468-6BE8-007B06CB07BB|blockquote|1315971|这看起来太不对劲了。我们可以推断，如果掉期是正确的，并且只影响32位和16位字段，那么在不交换所需字段的情况下，将是：|1315972|Original:+446831D3-FBA1-477B-BB07-CB067B00E86B
+++++++++++%7C+++++++_______/+++%7C+++++++_______/
++++++++++++\____+/_______+++++\____+/______
++++++++++++_____/++++++++\++++_____/+++++++\
+++++++++++/+++++++++++++++%7C++/++++++++++++++%7C
Received:+7B47A1FB-D331-6844-6BE8-007B06CB07BB+++#+"de-swapped".|code-block|syntax|javascript|1315973|在这里，您所看到的似乎是在64位单元内的字节交换。甚至字节数组也被颠倒了，这意味着它可能作为64位加载的一部分加载到某个位置，而该加载受swap64操作的限制。|1315974|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Q|8|@]|9|@]|A|$]]|$1|E|3|F|5|6|7|R|8|@]|9|@]|A|$]]|$1|G|3|H|5|I|7|S|8|@]|9|@]|A|$J|K]]|$1|L|3|M|5|6|7|T|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|U|8|@]|9|@]|A|$]]]|O|$]]

Re:

<blockquote>
 Update:
 Starting uuid in little endian:

<pre><code>446831D3-FBA1-477B-BB07-CB067B00E86B
</code></pre>
 
 Result from correctly swapping necessary fields:

<pre><code>FBA1477B-31D3-4468-6BE8-007B06CB07BB
</code></pre>
</blockquote>

This looks horribly wrong. We can deduce that if the swaps are correct and affect only the 32 and 16 bit fields, then without swapping the necessary fields, it would be this:

<pre><code>Original: 446831D3-FBA1-477B-BB07-CB067B00E86B
 | _______/ | _______/
 \____ /_______ \____ /______
 _____/ \ _____/ \
 / | / |
Received: 7B47A1FB-D331-6844-6BE8-007B06CB07BB # "de-swapped".
</code></pre>

What you seem to have have going on here is byte swapping within 64 bit units somewhere in your pipeline. Even the array of bytes is being reversed, suggesting that it might be loaded somewhere as part of a 64 bit load, which is subject to a swap64 operation.

My UUID is structured like this:

<pre><code>struct UUID_FIELDS
{
 uint32_t time_low; 
 uint16_t time_mid; 
 uint16_t time_hi_and_version; 
 uint8_t clock_seq_hi_and_reserved;
 uint8_t clock_seq_low; 
 uint8_t node[6]; 

};
</code></pre>

I have functions that <code>swap</code> along 32 bit boundaries and a <code>swap function</code> that swaps along 16 bit boundaries. I have tried calling <code>swap32()</code> on <code>time_low</code> and <code>swap16()</code> on <code>time_mid</code> and <code>time_hi_and_version</code>. I do not believe I need to swap bytes for the rest of the fields because two of them are 8 bit fields and I have read that the node part of a uuid does not change. <a href="http://howtowriteaprogram.blogspot.com/2009/03/uuid-and-byte-order.html" rel="nofollow">Here is a link for that reference.</a>

The problem is when I complete the swaps the uuid that is printed does not match the one that was in little endian before being converted.

What is the correct way to convert a uuid following RFC-4122 standard from <code>little endian</code> to <code>big endian</code>. And when converted should the uuids match?

How to convert UUID to big endian?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我的UUID的结构如下：struct UUID_FIELDS{    uint32_t time_low;                    uint16_t time_mid;                     uint16_t time_hi_and_version;          uint8_t c...

问如何将UUID转换为大端？
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