# Program to validate an IP address

• Difficulty Level : Medium
• Last Updated : 14 Nov, 2022

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 ` `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)

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` `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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