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    <p>So basically an IPV4-address and netmask are 32 bit unsigned integer displayed in pairs of 8 bit with a dot between them. </p> <p>So if you have "a.b.c.d" you could get the integer value by doing </p> <pre><code>(a&lt;&lt;24) + (b&lt;&lt;16) + (c&lt;&lt;8) + d </code></pre> <p>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></p> <p>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:</p> <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> <p>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.</p> <p>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 (::)..</p> <p>eg. </p> <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> <p>I guess that on the basis of this you can make a function that does what you want. </p>
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    1. POhow to convert ipaddress/mask to CIDR using c++?
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    1. USAlexis Wilke
    1. USSylwester
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    1. COThanks. I found a similiar approach on the net also. However i was looking for some standard c++ libraries which perform a similiar task? I guess there are not hence will have to device my own function for this. Will post the code once done.
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