Question

Right now I’m trying to implement this interface with an int which represents a binary number.

I know I shouldn’t implementto string(); , but Let’s assume that I need to.

This is what I wrote so far:

public class IPAddressInt implements IPAddress {
private int ipAdress;

public IPAddressInt(int num1,int num2, int num3, int num4){
    String n1=Integer.toBinaryString(num1);
    String n2=Integer.toBinaryString(num2);
    String n3=Integer.toBinaryString(num3);
    String n4=Integer.toBinaryString(num4);
    
    String finalString=n1+n2+n3+n4;
    this.ipAdress=Integer.parseInt(finalString);
}

public String toString() {
    return this.toString();
    
    
}

when I’m tring to return this.ipAdress.toString(); or even ipAdress.toString()
the compiler says that it Cannot invoke toString() on the primitive type int,
and when I write only this.toString(); it works. why? I know that an int can be converted to a string, and how come the this works and the whole statement is not? shouldn’t it be the same? will I get what I want anyway?

Thank you.

Answer 1

Calling this.toString() is just going to blow up – it’s calling the same method recursively, with no exit.

The reason you can’t call toString directly on ipAddress is that int is a primitive type, and you can’t call methods on primitive types.

To convert anything to a string, use String.valueOf(ipAddress). For int in particular, you could use Integer.toString(ipAddress).

It’s not really clear why you’re doing the conversion to a binary string in the first place though… that doesn’t look like a good idea to me. Any reason why you’re not using

ipAddress = (num1 << 24) | (num2 << 16) | (num3 << 8) | num4;

(assuming each value is really in the range 0-255).

I highly doubt that the binary representation is really want you want, particularly given the lack of padding, and that you’re then trying to parse it as a decimal number.