How to check if a string is a valid IP address in Kotlin
How to check if a string is a valid IP address in Kotlin.
Here's a step-by-step tutorial on how to check if a string is a valid IP address in Kotlin.
Step 1: Split the string into octets
The first step is to split the given string into four octets (substrings) using the dot ('.') as a delimiter. Each octet represents a number between 0 and 255. You can use the split() function to split the string into an array of octets.
val ipAddress = "192.168.0.1"
val octets = ipAddress.split(".")
Step 2: Check the number of octets
Next, we need to check if the string contains exactly four octets. If not, it is not a valid IP address. You can use the size property of the array to check the number of octets.
if (octets.size != 4) {
println("Invalid IP address")
return
}
Step 3: Check the validity of each octet
Now, we need to check the validity of each octet. Each octet should be a number between 0 and 255. You can use the toIntOrNull() function to convert each octet to an integer, and then check if it falls within the valid range.
for (octet in octets) {
val value = octet.toIntOrNull()
if (value == null || value !in 0..255) {
println("Invalid IP address")
return
}
}
Step 4: Check for leading zeros
Leading zeros are not allowed in each octet. For example, "192.168.01.1" is an invalid IP address. To check for leading zeros, you can compare the string representation of each octet with its integer value converted back to a string.
for (octet in octets) {
val value = octet.toIntOrNull()
if (value == null || value !in 0..255 || octet != value.toString()) {
println("Invalid IP address")
return
}
}
Step 5: Print the result
If the string passes all the validation checks, it is a valid IP address. You can print a success message in this case.
println("Valid IP address")
And that's it! You now have a step-by-step tutorial on how to check if a string is a valid IP address in Kotlin. You can use this code snippet to validate IP addresses in your Kotlin projects.