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!
This question is related to
c
string
ip-address
From string to int and back
const char * s_ip = "192.168.0.5";
unsigned int ip;
unsigned char * c_ip = (unsigned char *)&ip;
sscanf(s_ip, "%hhu.%hhu.%hhu.%hhu", &c_ip[3], &c_ip[2], &c_ip[1], &c_ip[0]);
printf("%u.%u.%u.%u", ((ip & 0xff000000) >> 24), ((ip & 0x00ff0000) >> 16), ((ip & 0x0000ff00) >> 8), (ip & 0x000000ff));
%hhu instructs sscanf to read into unsigned char pointer; (Reading small int with scanf)
inet_ntoa from glibc
char *
inet_ntoa (struct in_addr in)
{
unsigned char *bytes = (unsigned char *) ∈
__snprintf (buffer, sizeof (buffer), "%d.%d.%d.%d",
bytes[0], bytes[1], bytes[2], bytes[3]);
return buffer;
}
This is what I would do if passed a string buffer to fill and I knew the buffer was big enough (ie at least 16 characters long):
sprintf(buffer, "%d.%d.%d.%d",
(ip >> 24) & 0xFF,
(ip >> 16) & 0xFF,
(ip >> 8) & 0xFF,
(ip ) & 0xFF);
This would be slightly faster than creating a byte array first, and I think it is more readable. I would normally use snprintf, but IP addresses can't be more than 16 characters long including the terminating null.
Alternatively if I was asked for a function returning a char*:
char* IPAddressToString(int ip)
{
char[] result = new char[16];
sprintf(result, "%d.%d.%d.%d",
(ip >> 24) & 0xFF,
(ip >> 16) & 0xFF,
(ip >> 8) & 0xFF,
(ip ) & 0xFF);
return result;
}
void ul2chardec(char*pcIP, unsigned long ulIPN){
int i; int k=0; char c0, c1;
for (i = 0; i<4; i++){
c0 = ((((ulIPN & (0xff << ((3 - i) * 8))) >> ((3 - i) * 8))) / 100) + 0x30;
if (c0 != '0'){ *(pcIP + k) = c0; k++; }
c1 = (((((ulIPN & (0xff << ((3 - i) * 8))) >> ((3 - i) * 8))) % 100) / 10) + 0x30;
if (!(c1 =='0' && c0=='0')){ *(pcIP + k) = c1; k++; }
*(pcIP +k) = (((((ulIPN & (0xff << ((3 - i) * 8)))) >> ((3 - i) * 8))) % 10) + 0x30;
k++;
if (i<3){ *(pcIP + k) = '.'; k++;}
}
*(pcIP + k) = 0; // pcIP should be x10 bytes
}
You actually can use an inet function. Observe.
main.c:
#include <arpa/inet.h>
main() {
uint32_t ip = 2110443574;
struct in_addr ip_addr;
ip_addr.s_addr = ip;
printf("The IP address is %s\n", inet_ntoa(ip_addr));
}
The results of gcc main.c -ansi; ./a.out
is
The IP address is 54.208.202.125
Note that a commenter said this does not work on Windows.
Hint: break up the 32-bit integer to 4 8-bit integers, and print them out.
Something along the lines of this (not compiled, YMMV):
int i = 0xDEADBEEF; // some 32-bit integer
printf("%i.%i.%i.%i",
(i >> 24) & 0xFF,
(i >> 16) & 0xFF,
(i >> 8) & 0xFF,
i & 0xFF);
My alternative solution with subtraction :)
void convert( unsigned int addr )
{
unsigned int num[OCTET],
next_addr[OCTET];
int bits = 8;
unsigned int shift_bits;
int i;
next_addr[0] = addr;
shift_bits -= bits;
num[0] = next_addr[0] >> shift_bits;
for ( i = 0; i < OCTET-1; i ++ )
{
next_addr[i + 1] = next_addr[i] - ( num[i] << shift_bits ); // next subaddr
shift_bits -= bits; // next shift
num[i + 1] = next_addr[i + 1] >> shift_bits; // octet
}
printf( "%d.%d.%d.%d\n", num[0], num[1], num[2], num[3] );
}
Another approach:
union IP {
unsigned int ip;
struct {
unsigned char d;
unsigned char c;
unsigned char b;
unsigned char a;
} ip2;
};
...
char ips[20];
IP ip;
ip.ip = 0xAABBCCDD;
sprintf(ips, "%x.%x.%x.%x", ip.ip2.a, ip.ip2.b, ip.ip2.c, ip.ip2.d);
printf("%s\n", ips);
#include "stdio.h"
void print_ip(int ip) {
unsigned char bytes[4];
int i;
for(i=0; i<4; i++) {
bytes[i] = (ip >> i*8) & 0xFF;
}
printf("%d.%d.%d.%d\n", bytes[3], bytes[2], bytes[1], bytes[0]);
}
int main() {
int ip = 0xDEADBEEF;
print_ip(ip);
}
Source: Stackoverflow.com