blocks|key|2309800|text|为了引导您朝着正确的方向前进，请考虑IPv4地址是什么(32位整数)。现在，考虑什么是掩码(在按位操作中使用的位字段)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2309801|在大端系统上取地址127.0.0.1。用十六进制来说，那是0x7f000001。24位掩码是0xffffff00+(+1位24位，0位8位，总共32位).按位的地址和掩码0x7f000001+&+0xffffff00+=+0x7f000000是CIDR格式。|offset|length|style|CODE|2309802|我让您决定如何最好地解析地址，将IPv4地址转换为其整数形式(和back)，并从路由前缀掩码创建一个位掩码，不过我将告诉您，至少有一些用于地址操作的标准函数。|2309803|entityMap^0|0|T|A|1A|A|2D|10|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]|$D|P|E|Q|F|G]|$D|R|E|S|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|T|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|U|8|@]|9|@]|A|$]]]|K|$]]

Just to lead you in the right direction, consider what an IPv4 address is (a 32-bit integer). Now, consider what a mask is (a bit field used in bitwise operations).

Take the address 127.0.0.1 on a big-endian system. In hex, that's <code>0x7f000001</code>. A 24-bit mask is <code>0xffffff00</code> (24 bits of 1, 8 bits of 0, 32 bits total). The bitwise and of the address and the mask <code>0x7f000001 &amp; 0xffffff00 = 0x7f000000</code> is the CIDR format.

I leave it to you to determine how best to parse the address, convert the IPv4 address to its integer form (and back), and create a bit mask from a routing prefix mask, though I will tell you that there are standard functions for at least the address manipulation.

blocks|key|2279829|text|因此，IPV4+4地址和网络掩码基本上是32位无符号整数，它们之间有一个点，以对8位的形式显示。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2279830|因此，如果有"a.b.c.d“，则可以通过以下操作获得整数值|2279831|(a<<24)+%2B+(b<<16)+%2B+(c<<8)+%2B+d+|code-block|syntax|javascript|2279832|如果您不习惯<<，它会将数字中的位移到左边，将次数作为正确的参数。因此，a<<b与a*2%5Eb相同|offset|length|style|CODE|2279833|现在，在CIDR中，斜杠后面的数字告诉您斜杠之前的数字的网络部分是多少位(32位)。例如：10.10.10.1/28有28位网络，32-28=4位主机部分。这意味着，如果您接受ip并将其转换为一个数字(=0x0A0A0A01)，并使用&+(=+0xFFFFFFFF0)执行逻辑0xFFFFFFFF<<4，则会得到0x0A0A0A00。如果您使用以下内容将其转换回ip：|2279834|sprintf("%25d.%25d.%25d.%25d",+(n>>24)&0xff,+(n>>16)&0xff,+(n>>8)&0xff,+n&0xff)+|2279835|你会得到10.10.0。同样，如果使用相同的算法将0xFFFFF0转换为ip，则会得到255.255.255.248，这是对应于掩码长度28的网络掩码。|2279836|IPv6是完全相同的，除了它在8+16位集群中划分的128位，在它们之间以十六进制显示，它们之间有逗号，对于0(：：)的多次运行有一些奇特的快捷方式。|2279837|例如：|2279838|fe80::52e5:49ff:ffc9:1889/64+==+

;;convert+the+mask+
0xffffffffffffffffffffffffffffffff+<<+(128-64)+=
0xffffffffffffffff0000000000000000

;;convert+the+ip:
fe80::52e5:49ff:ffc9:1889+==
fe80:0000:0000:0000:52e5:49ff:ffc9:1889+==
0xfe8000000000000052e549ffffc91889+

++0xfe8000000000000052e549ffffc91889
&+0xffffffffffffffff0000000000000000
=+0xfe800000000000000000000000000000

;;convert+first+address+in+range+back+to+string:
fe80:0000:0000:0000:0000:0000:0000:0000+=+
fe80::|2279839|我想，在此基础上，你可以做一个你想做的函数。|2279840|entityMap^0|0|0|0|10|4|15|5|0|39|1|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|17|8|@$K|18|L|19|M|N]|$K|1A|L|1B|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1C|8|@$K|1D|L|1E|M|N]]|9|@]|A|$]]|$1|Q|3|R|5|F|7|1F|8|@]|9|@]|A|$G|H]]|$1|S|3|T|5|6|7|1G|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1H|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1I|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|F|7|1J|8|@]|9|@]|A|$G|H]]|$1|10|3|11|5|6|7|1K|8|@]|9|@]|A|$]]|$1|12|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]]|13|$]]

So basically an IPV4-address and netmask are 32 bit unsigned integer displayed in pairs of 8 bit with a dot between them. 

So if you have "a.b.c.d" you could get the integer value by doing 

<pre><code>(a&lt;&lt;24) + (b&lt;&lt;16) + (c&lt;&lt;8) + d 
</code></pre>

If you're not used to the &lt;&lt; it's shifts the bits in the number to the left the number of times as the right argument.. Thus <code>a&lt;&lt;b</code> is the same as <code>a*2^b</code>

Now in a CIDR the number after the slash tells you how many bits (of 32) that are the network part of the number before the slash. eg. 10.10.10.1/28 has 28 bits network and 32-28=4 bits host part. That means that if you take the ip and translate it to a number (=0x0A0A0A01) and do a logical <code>&amp;</code> with 0xFFFFFFFF&lt;&lt;4 (= 0xFFFFFFF0) you get 0x0A0A0A00. If you translate that back to a ip with the following:

<pre><code>sprintf("%d.%d.%d.%d", (n&gt;&gt;24)&amp;0xff, (n&gt;&gt;16)&amp;0xff, (n&gt;&gt;8)&amp;0xff, n&amp;0xff) 
</code></pre>

You get 10.10.10.0. Also if you translate 0xFFFFFFF0 to a ip with the same algorithm you get 255.255.255.248 which is the netmask that corresponds to mask length 28.

IPv6 is exactly the same, except it's 128 bits divided in 8 16 bit clusters displayed in hex with commas between them with some fancy shortcut for multiple runs of 0 (::)..

eg. 

<pre><code>fe80::52e5:49ff:ffc9:1889/64 == 

;;convert the mask 
0xffffffffffffffffffffffffffffffff &lt;&lt; (128-64) =
0xffffffffffffffff0000000000000000

;;convert the ip:
fe80::52e5:49ff:ffc9:1889 ==
fe80:0000:0000:0000:52e5:49ff:ffc9:1889 ==
0xfe8000000000000052e549ffffc91889 

 0xfe8000000000000052e549ffffc91889
&amp; 0xffffffffffffffff0000000000000000
= 0xfe800000000000000000000000000000

;;convert first address in range back to string:
fe80:0000:0000:0000:0000:0000:0000:0000 = 
fe80::
</code></pre>

I guess that on the basis of this you can make a function that does what you want.

I have a list of ipaddress/mask which needs to be converted to CIDR notation.

for e.g. 12.174.36.240/24 needs to be converted to 12.174.36.0/24
or something like what <a href="http://www.subnet-calculator.com/cidr.php" rel="nofollow">http://www.subnet-calculator.com/cidr.php</a> does

How can this be acheived?

PS: the mask value is not always 24.

how to convert ipaddress/mask to CIDR using c++?

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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我有一个ipaddress/掩码列表，需要转换为CIDR符号。例如，需要将12.174.36.240/24转换为12.174.36.0/24或类似于所做的事情这怎么能做到呢？PS:掩码值并不总是24。

问如何使用c++将ipaddress/掩码转换为CIDR？
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问如何使用c++将ipaddress/掩码转换为CIDR？EN