Linux C判断两个IPv6地址是否相等的方法

IPv6地址用冒号和16进制数表示，其中遇到连续的0可以作省略处理，如2001:0:0:0:0:0:0:1可以写成2001::1，这样对于书写很方便，但是带来一个额外的问题：两个地址比较的时候不能像IPv4那样调用字符串比较函数进行比较。本文通过比较两个IPv6地址的网络字节序来判断是否相等。

#include <stdio.h>
#include <arpa/inet.h>

int ipv6_equal(char *addr1, char *addr2)
{
	int ret = -1;
	int i = 0;
	unsigned char n_addr1[16] = {-1};
	unsigned char n_addr2[16] = {-1};

	if (!addr1) {
		printf("addr1 is NULL\n");
		return -1;
	}

	if (!addr2) {
		printf("addr2 is NULL\n");
		return -1;
	}

	ret = inet_pton(AF_INET6, addr1, &(n_addr1));
	if (ret <= 0 ) {
		if (ret == 0) {
			printf("addr1: Invalid IPv6 address\n");
		}

		return -1;
	}
	ret = inet_pton(AF_INET6, addr2, &(n_addr2));
	if (ret <=0 ) {
		if (ret == 0) {
			printf("addr2: Invalid IPv6 address\n");
		}

		return -1;
	}

	for (i = 0; i < 16; i++) {
		//printf("i: %d, addr1: %u, addr2: %u\n", i, n_addr1[i], n_addr2[i]);
		if (n_addr1[i] != n_addr2[i]) {
			return 1;
		}	
	}
	return 0;
}

int main(void)
{
	if (ipv6_equal("2001::1", "2001::1") == 0) {
		printf("test: 2001::1 equal 2001::1\n");
	}

	if (ipv6_equal("2001:0:0:0:0:0:0:1", "2001::1") == 0) {
		printf("test: 2001:0:0:0:0:0:0:1 equal 2001::1\n");
	}

	if (ipv6_equal("2001::1", "2001::4") == 1) {
		printf("test: 2001::1 not equal 2001::4\n");
	}

	if (ipv6_equal("2001:::1", "2001::4") == -1) {
		printf("test: Invalid address\n");
	}
	
	return 0;
}

结果：

test: 2001::1 equal 2001::1
test: 2001:0:0:0:0:0:0:1 equal 2001::1
test: 2001::1 not equal 2001::4
addr1: Invalid IPv6 address
test: Invalid address