C实现判断两个IP是否在同一个子网

在进行网络程序的时候避免不了对给定IP是否跨子网段进行判断。相关原理倒是简单, 贴出相关代码:

 1 #include <stdio.h>
 2 #include <math.h>
 3 #include <string.h>
 4 #include <stdlib.h>
 5 #include <unistd.h>
 6  
 7 #define ALL_BIT 32 /* ip address have 32 bits */
 8  
 9 /*
10 ** description: juge the ip_a and ip_b is in the same subnet or not
11 ** this functio is used IPv4 only, and did not check the input parameter's
12 ** correctness, so make sure the value of parameter before call it
13 ** para: sub_mask is a decimal subnet mask , not dotted decimal
14 ** ip_a/ip_b: the ip address string, in dotted decimal
15 ** return: 0 for the same, -1 for the not
16 */
17 int same_subnet (int sub_mask, char *ip_a, char *ip_b)
18 {
19   int mask = 0xFFFFFFFF;
20   mask = mask << (ALL_BIT - sub_mask);
21  
22   double cnt_a = 0.0;
23   double cnt_b = 0.0;
24   double tmp = 0.0;
25  
26   char *token = NULL;
27  
28  int i = 3;
29  
30   for (token = strtok (ip_a, "."); token != NULL; token = strtok (NULL, ".")) {
31     tmp = atoi (token);
32     tmp = tmp * pow (256, i--);  /* what dose this mean? do you understand */
33     cnt_a += tmp;
34   }
35  
36   i = 3; /* reset i */
37   for (token = strtok (ip_b, "."); token != NULL; token = strtok (NULL, ".")) {
38     tmp = atoi (token);
39     tmp = tmp * pow (256, i--);
40     cnt_b += tmp;
41   }
42  
43   unsigned long mask_a = (unsigned long)cnt_a;
44   unsigned long mask_b = (unsigned long)cnt_b;
45  
46   //printf ("mask_a %u\tmask_b %u\n", mask_a, mask_b);
47  
48   mask_a &= mask;   /* get the ip's netmask for compare */
49   mask_b &= mask;
50   //printf ("mask_a %u\tmask_b %u\n", mask_a, mask_b);
51  
52   if (mask_a == mask_b)
53   return 0;
54   else
55     return -1;
56 }
57  
58 /*
59 ** test code
60 */
61 int main (int argc, char **argv)
62 {
63   int i = same_subnet (atoi (argv[1]), argv[2], argv[3]);
64   printf ("%s subnet\n", i == 0? "same":"not same");
65   return 0;
66 }
67  
68  
69  
70 代码在gcc version 4.7.2 20121109 (Red Hat 4.7.2-8) (GCC)编译通过。

 

posted @ 2014-05-21 23:25  Arnold0Kang  阅读(571)  评论(0编辑  收藏  举报