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
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 codeint 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;} |
C
#include <stdbool.h>#include <stdio.h>#include <string.h>#define DELIM "."/* function to check whether the string passed is valid or not */bool valid_part(const 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') return false; int x = atoi(s); // 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 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 codeint 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) ? printf("Valid\n") : printf("Not valid\n"); is_valid_ip(ip2) ? printf("Valid\n") : printf("Not valid\n"); is_valid_ip(ip3) ? printf("Valid\n") : printf("Not valid\n"); is_valid_ip(ip4) ? printf("Valid\n") : printf("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 every split has leading zero or not. if len(n)>1: if n[0] == "0": return True return Falsedef 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. |
Java
public class GFG { public static boolean inRange(int n) { // check if every split is in range 0-255 if (n >= 0 && n <= 255) { return true; } return false; } public static boolean hasLeadingZero(String n) { // check if every split has leading zero or not. if (n.length() > 1) { if (n.charAt(0) == '0') { return true; } } return false; } public static int isValid(String s) { String[] parts = s.split("\\."); if (parts.length != 4) { // if number of splitting element is not // 4 it is not a valid IP address return 0; } for (String part : parts) { if (hasLeadingZero(part)) { return 0; } if (part.length() == 0) { return 0; } try { int num = Integer.parseInt(part); if (!inRange(num)) { return 0; } } catch (NumberFormatException e) { return 0; } } return 1; } public static void main(String[] args) { String ip1 = "222.111.111.111"; String ip2 = "5555..555"; String ip3 = "0000.0000.0000.0000"; String ip4 = "1.1.1.1"; System.out.println(isValid(ip1)); System.out.println(isValid(ip2)); System.out.println(isValid(ip3)); System.out.println(isValid(ip4)); }} |
Javascript
function inRange(n) { // check if every split is in range 0-255 if (n >= 0 && n <= 255) { return true; } return false;}function hasLeadingZero(n) { // check if every split has leading zero or not. if (n.length > 1) { if (n.charAt(0) === '0') { return true; } } return false;}function isValid(s) { let parts = s.split('.'); if (parts.length !== 4) { // if number of splitting element is not 4 it is not a valid IP address return 0; } for (let i = 0; i < parts.length; i++) { let part = parts[i]; if (hasLeadingZero(part)) { return 0; } if (part.length === 0) { return 0; } try { let num = parseInt(part, 10); if (!inRange(num)) { return 0; } } catch (e) { return 0; } } return 1;}let ip1 = "222.111.111.111";let ip2 = "5555..555";let ip3 = "0000.0000.0000.0000";let ip4 = "1.1.1.1";console.log(isValid(ip1));console.log(isValid(ip2));console.log(isValid(ip3));console.log(isValid(ip4)); |
Output
1 0 0 1
Time complexity : O(n)
Auxiliary Space : O(1)
Method 3 – Using String stream and vector
Approach
- Using string stream to separate all the string from ‘.’ and push back into vector like for ex – 222.111.111.111 vector is v = [“222” , “111” , “111” , “111”]
- If the vector size != 4 return false, like 222.111.111.111 v = [“222” , “111” , “111” , “111”].
- Iterating over the generated vector of string
- for leading zero , test case like 222.0.0.10 this is valid but this is not 222.00.100.100 , we check for the size of the i th string if temp.size() > 1 and if(temp[0] == ‘0’) return false;
- For test case like a.b.c.d , checking the alpha values like abcde…… if any present simply return false
- And lastly we are checking if the number is greater than 255 or not
C++
// Program to check if a given// string is valid IPv4 address or not#include <bits/stdc++.h>using namespace std;/* return 1 if IP string isvalid, else return 0 */int is_valid_ip(string s){ // code here int n = s.size(); // for test case like 1...1 or something lesser than 7 if (n < 7) return false; // Using string stream to separate all the string from // '.' and push back into vector like for ex - // 222.111.111.111 vector is v = ["222" , "111" , "111" // , "111"] vector<string> v; stringstream ss(s); while (ss.good()) { string substr; getline(ss, substr, '.'); v.push_back(substr); } // If the vector size != 4 return false, like // 222.111.111.111 v = ["222" , "111" , "111" , // "111"]. if (v.size() != 4) return false; // Iterating over the generated vector of string for (int i = 0; i < v.size(); i++) { // string temp = v[i]; // for leading zero , test case like 222.0.0.10 this // is valid but this is not vaild 222.00.100.100 , we // check for the size of the i th string if // temp.size() > 1 and if(temp[0] == '0') return // false; if (temp.size() > 1) { if (temp[0] == '0') return false; } // For test case like a.b.c.d , checking the alpha // values like abcde...... if any present simply // return false for (int j = 0; j < temp.size(); j++) { if (isalpha(temp[j])) return false; } // And lastly we are checking if the number is // greater than 255 or not if (stoi(temp) > 255) return false; } return true;}// Driver codeint main(){ string s1 = "128.0.0.1"; string s2 = "125.16.100.1"; string s3 = "125.512.100.1"; string s4 = "125.512.100.abc"; is_valid_ip(s1) ? cout << "Valid\n" : cout << "Not valid\n"; is_valid_ip(s2) ? cout << "Valid\n" : cout << "Not valid\n"; is_valid_ip(s3) ? cout << "Valid\n" : cout << "Not valid\n"; is_valid_ip(s4) ? cout << "Valid\n" : cout << "Not valid\n"; return 0;} |
Output
Valid Valid Not valid Not valid
Time complexity : O(n)
Auxiliary Space : O(1)
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