是否有一个库函数来确定IP地址（IPv4和IPv6）是C / C ++中的私有/本地的？ [英] Is there a library function to determine if an IP address (IPv4 and IPv6) is private/local in C/C++?

问题描述

1，给定一个32位整数值，如何精确确定它是否为私有IPv4地址。

1, Given a 32-bit integer value, how to exactly determine if it is private IPv4 address.

2，给定一个128位整数值，以确切地确定它是否是私有IPv6地址。

2, Given a 128-bit integer value, how to exactly determine if it is private IPv6 address.

考虑不同平台上的IP地址的字节顺序，它是容易错误的写每个时间。所以我认为应该有一个库函数为这个，是有？

Consider the byte order of the IP address on different platforms, it is error-prone to write such a common little function every time. So I think there should be a library function for this, is there?

推荐答案

这将让你开始。我没有打扰，包括链接本地地址范围，但这是一个练习，你可以通过修改下面的代码来完成。

This will get you started. I didn't bother including the "link local" address range, but that's an exercise left for you to complete by modifying the code below.

IPV6略有不同。和你的问题是略有畸形，因为大多数系统没有本地的128位类型。 IPv6地址通常包含在嵌入在sockaddr_in6结构中的16字节数组。

IPV6 is slightly different. And your question is slightly malformed since most systems don't have a native 128-bit type. IPv6 addresses are usually contained as an array of 16 bytes embedded in a sockaddr_in6 struct.

完成此示例所需的一切都在此链接。

Everything you need to know to finish this example is at this link here.

// assumes ip is in HOST order.  Use ntohl() to convert as approrpriate

bool IsPrivateAddress(uint32_t ip)
{
    uint8_t b1, b2, b3, b4;
    b1 = (uint8_t)(ip >> 24);
    b2 = (uint8_t)((ip >> 16) & 0x0ff);
    b3 = (uint8_t)((ip >> 8) & 0x0ff);
    b3 = (uint8_t)(ip & 0x0ff);

    // 10.x.y.z
    if (b1 == 10)
        return true;

    // 172.16.0.0 - 172.31.255.255
    if ((b1 == 172) && (b2 >= 16) && (b2 <= 31))
        return true;

    // 192.168.0.0 - 192.168.255.255
    if ((b1 == 192) && (b2 == 168))
        return true;

    return false;
}

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