检查IPv4地址是否有效

Question

I am trying to write a basic program in Java that checks if an IP address is valid. I am trying to use no external classes other than the Scanner class and also no regular expressions.

My code, available at this gist takes in 4 integers as input, one for each octet. I also have this code which is a little more readable than the first but is longer.

My question is, is there any other way to implement this idea in significantly fewer lines, and if so, how would I accomplish this?

Answer 1

While this code segment is a bit verbose, it is simple, self-descriptive, and has proven to be tight.

private static boolean isIPAddressValid(String ip) {    
    boolean result = true;
    int i = 0;
    int [] val = new int[4];

    if ((ip == null) || (ip.trim().length() == 0))
    {
        //null ip address entered
        result = false;
    }
    else
    {
        if (!(ip.contains(".")))
        {
            //no '.' found
            result = false;
        }
        else
        {
            String [] parts = ip.split("\\.");
            if (!(parts.length == 4))
            {
                //not 4 quadrants
                result = false;
            }
            else
            {
                for (String s : parts) { 
                    try {
                        val[i] = Integer.parseInt(s);
                        if ((val[i] < 0) || (val[i] > 255))
                        {
                            //this quadrant's value exceeds limits
                            result = false;
                        }
                        i++;
                    } catch (Exception e) {
                        //failed to parse quadrant to an integer");
                        result = false;
                    }
                }
            }
        }
    }
    return result;
}

Answer 2

only some minor enhancements (i think your code looks very good - my opinio so far) it is clearly to read and all work blocks are properly to understand....

boolean isFailed = false;
if (first < 0 || first > 255) {
    System.out.println("Octet 1 is invalid");
    isFailed = true;
}
if (second < 0 || second > 255) {
    System.out.println("Octet 2 is invalid");
    isFailed = true;
}
if (third < 0 || third > 255) {
    System.out.println("Octet 3 is invalid");
    isFailed = true;
}
if (fourth < 0 || fourth > 255) {
    System.out.println("Octet 4 is invalid");
    isFailed = true;
}

if (!isFailed){
    System.out.println("IP Address: " + first + "." + second + "." + third + "." + fourth);
}

so i simply invert the order of printing - that saves you only that big check before...

Answer 3

your approach was to can check each octet...

you can simply do this 4 times or write a method for it:

private static boolean check(int octet, int index){
    if (0xFF & octet < 256) return true;
    System.out.println("Octet "+index+" is invalid";
    return false;
}

and use this method in your main method

if (check(first,0) && check (second, 2) && check (third, 3) && check(fourth, 4) ){
    System.out.println("your ip is valid");
}

note - this only reveals the first invalid octet - if you want to check all you need another boolean

boolean result = check(first,0) && 
    check (second, 2) && 
    check (third, 3) && 
    check(fourth, 4); //reveals all errors

a totally different approach would be to use http://docs.oracle.com/javase/7/docs/api/java/net/InetAddress.html#getByName%28java.lang.String%29

try{
    /*InetAdress adress =*/ InetAdress.
        getByName(""+first+"."+second+"."+third+"."+forth)
    System.out.println("your ip is valid");
}catch (UnknownHostException e){
    //TODO represent that error message into a nice expression
    System.out.println("your ip is invalid");
}

but this as well doesn't provide information about that octet which is invalid...

(by the way - what is wrong with your code? it's fine, really!)

Answer 4

I was just bored and wrote this regexp

public static boolean isValid(String ip) {
    boolean isvalid;

    isvalid = ip.matches(
            "(([0-9]|[0-9]{0,2}|1[0-9]*{0,2}|2[0-5][0-5]|0{0,3}).){3}" +
             "([0-9]|[0-9]{0,2}|1[0-9]*{0,2}|2[0-5][0-5]|0{0,3})"
    );

    return isvalid;
}

And it was tested on the following dataset:

String[] ips = {
        "0.0.0.0",
        "0.111.222.0",
        "0.0.0.000",
        "0.00.0.000",
        "1.1.1.1",
        "2.2.2.2",
        "12.13.14.15",
        "29.29.29.29",
        "99.99.000.1",
        "111.102.144.190",
        "255.255.199.199",
        "266.255.255.255", //inv
        "255.265.255.255", //inv 
        "255.255.258.255", //inv 
        "255.255.255.259", //inv 
        "299.100.110.255" //inv
    };
    for (String s : ips) {
        if (isValid(s) == false) {
            System.err.println(s + " is invalid");
        }
    }