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Program to validate an IP address

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  • Difficulty Level : Medium
  • Last Updated : 16 Jun, 2022
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Write a program to Validate an IPv4 Address. 
According to Wikipedia, IPv4 addresses are canonically represented in dot-decimal notation, which consists of four decimal numbers, each ranging from 0 to 255, separated by dots, e.g., 172.16.254.1

Recommended Practice

Following are steps to check whether a given string is a valid IPv4 address or not:

step 1) Parse string with “.” as delimiter using “strtok()” function. 

e.g.ptr = strtok(str, DELIM);

step 2) 
A) If ptr contains any character which is not digit then return 0 
B) Convert “ptr” to decimal number say ‘NUM’ 
C) If NUM is not in range of 0-255 return 0 
D) If NUM is in range of 0-255 and ptr is non-NULL increment “dot_counter” by 1 
E) if ptr is NULL goto step 3 else goto step 1
step 3) if dot_counter != 3 return 0 else return 1

C++




// Program to check if a given
// string is valid IPv4 address or not
#include <bits/stdc++.h>
using namespace std;
#define DELIM "."
 
/* function to check whether the
   string passed is valid or not */
bool valid_part(char* s)
{
    int n = strlen(s);
     
    // if length of passed string is
    // more than 3 then it is not valid
    if (n > 3)
        return false;
     
    // check if the string only contains digits
    // if not then return false
    for (int i = 0; i < n; i++)
        if ((s[i] >= '0' && s[i] <= '9') == false)
            return false;
    string str(s);
     
    // if the string is "00" or "001" or
    // "05" etc then it is not valid
    if (str.find('0') == 0 && n > 1)
        return false;
    stringstream geek(str);
    int x;
    geek >> x;
     
    // the string is valid if the number
    // generated is between 0 to 255
    return (x >= 0 && x <= 255);
}
 
/* return 1 if IP string is
valid, else return 0 */
int is_valid_ip(char* ip_str)
{
    // if empty string then return false
    if (ip_str == NULL)
        return 0;
    int i, num, dots = 0;
    int len = strlen(ip_str);
    int count = 0;
     
    // the number dots in the original
    // string should be 3
    // for it to be valid
    for (int i = 0; i < len; i++)
        if (ip_str[i] == '.')
            count++;
    if (count != 3)
        return false;
     
    // See following link for strtok()
   
    char *ptr = strtok(ip_str, DELIM);
    if (ptr == NULL)
        return 0;
 
    while (ptr) {
 
        /* after parsing string, it must be valid */
        if (valid_part(ptr))
        {
            /* parse remaining string */
            ptr = strtok(NULL, ".");
            if (ptr != NULL)
                ++dots;
        }
        else
            return 0;
    }
 
    /* valid IP string must contain 3 dots */
    // this is for the cases such as 1...1 where
    // originally the no. of dots is three but
    // after iteration of the string we find
    // it is not valid
    if (dots != 3)
        return 0;
    return 1;
}
 
// Driver code
int main()
{
    char ip1[] = "128.0.0.1";
    char ip2[] = "125.16.100.1";
    char ip3[] = "125.512.100.1";
    char ip4[] = "125.512.100.abc";
    is_valid_ip(ip1) ? cout<<"Valid\n" : cout<<"Not valid\n";
    is_valid_ip(ip2) ? cout<<"Valid\n" : cout<<"Not valid\n";
    is_valid_ip(ip3) ? cout<<"Valid\n" : cout<<"Not valid\n";
    is_valid_ip(ip4) ? cout<<"Valid\n" : cout<<"Not valid\n";
    return 0;
}

Output

Valid
Valid
Not valid
Not valid

Time complexity : O(n) 
Auxiliary Space : O(1)

This article is compiled by Narendra Kangralkar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Python Solution :-

Approach:- We will check all the cases where ip address may be invalid

1. First we will split the given input  using split() function then check if it has a length of 4  or not .If length  is not equal to 4 then we will directly return 0.

2. in second step we will check if any split element contains any leading zero or not .if it is then we will return zero.

3.If any of the split does not contain any number then it is not a valid ip address .so we will return 0

4. Then we will check if all the splits are in the range of  0-255 or not .If not we will return 0.

5.Finally if none of the above condition is true we can finally say that it is a valid ip address.And we will return True.

Here is the code for above approach .

Python3




def in_range(n):   #check if every split is in range 0-255
    if n >= 0 and n<=255:
        return True
    return False
     
def has_leading_zero(n): # check if eery split has leading zero or not.
    if len(n)>1:
        if n[0] == "0":
            return True
    return False
def isValid(s):
     
    s = s.split(".")
    if len(s) != 4#if number of splitting element is not 4 it is not a valid ip address
        return 0
    for n in s:
         
        if has_leading_zero(n):
            return 0
        if len(n) == 0:
            return 0
        try#if int(n) is not an integer it raises an error
            n = int(n)
 
            if not in_range(n):
                return 0
        except:
            return 0
    return 1
         
   
if __name__=="__main__":
     
     
    ip1 = "222.111.111.111"
    ip2 = "5555..555"
    ip3 = "0000.0000.0000.0000"
    ip4 = "1.1.1.1"
    print(isValid(ip1))
    print(isValid(ip2))
    print(isValid(ip3))
    print(isValid(ip4))
 
     
 # this code is contributed by Vivek Maddeshiya.

Output

1
0
0
1

Time complexity : O(n) 
Auxiliary Space : O(1)


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