整数IP地址 - Ç [英] Integer to IP Address - C

问题描述

我是一个小测验preparing，我有一个强烈的怀疑我可以实现这样一个函数来负责。基本上，鉴于网络符号格式的IP地址，我们怎样才能从一个32位整数到这一个字符串的点分十进制形式（类似于155.247.182.83）...？显然，我们不能使用任何类型的INET函数要么...我难倒！

I'm preparing for a quiz, and I have a strong suspicion I may be tasked with implementing such a function. Basically, given an IP address in network notation, how can we get that from a 32 bit integer into a string in it's dotted decimal notation (something like 155.247.182.83)...? Obviously we can't be using any type of inet functions either...I'm stumped!

推荐答案

下面是一个简单的方法来做到这一点：在（IP＆GT;＆GT; 8），（IP＆GT;＆GT; 16）和（IP＆GT;＆GT; 24）移动第二，第三和第四个字节为低位字节，而＆安培; 0xFF的在每一步隔离至少显著字节。

Here's a simple method to do it: The (ip >> 8), (ip >> 16) and (ip >> 24) moves the 2nd, 3rd and 4th bytes into the lower order byte, while the & 0xFF isolates the least significant byte at each step.

void print_ip(int ip)
{
    unsigned char bytes[4];
    bytes[0] = ip & 0xFF;
    bytes[1] = (ip >> 8) & 0xFF;
    bytes[2] = (ip >> 16) & 0xFF;
    bytes[3] = (ip >> 24) & 0xFF;	
    printf("%d.%d.%d.%d\n", bytes[3], bytes[2], bytes[1], bytes[0]);        
}

有一个隐含的字节[0] =（IP＆GT;＆GT; 0）及为0xFF; 在第一步

使用的snprintf（）将其打印为一个字符串。

Use snprintf() to print it to a string.

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