isInstance not working if the member is in a array

Question

Class X{
    Integer x=new Integer(5);
    Integer y;
    public static void main (String[] args) throws java.lang.Exception
    {
         X i = new X();
         String[] str={"x", "y"};
         System.out.println(Integer.class.isInstance(str[0]));
    }
}

It returns false as str[0] is a instance of Class String. Is there a way it can return true as str[0]=x and variable "x" is instance of Integer Class?

Thanks.

Why don't you use `Integer[] numbers = {x, y};` and convert to String whenever you want? — Maroun, Aug 12 '15 at 07:17
Yes, it is. You should try to parse string value to Integer (like `Integer.parseInt(str[i])`) and then it should return true. But keep in mind if `str[i]` is not a string representing a number, an exception will be thrown when parsing. — peterremec, Aug 12 '15 at 07:18
if it's only member variable (and not local) then you probably could use reflection to make such a check but I really don't see a reasonable use case for doing it — Mateusz Dymczyk, Aug 12 '15 at 07:18
Your code is not clear. In String array your are keeping "x" and "y". That can never be integers. — Amit Bhati, Aug 12 '15 at 07:21
I think OP tries to refer to the *fields* `x` and `y` through string identifiers `"x"` and `"y"`. — aioobe, Aug 12 '15 at 07:24
I am sorry if the question was not clear. But i solved it using reflection : getField and getType and then comparing it isAssignableFrom(). — Asish, Aug 12 '15 at 08:34

Asoub · Answer 1 · 2015-08-12T07:30:37.483

When you're doing String[] str={"x", "y"}; you're not saving the variable x in the array, you're saving the String containing only the character "x". It's not because it's an array or anything that it doesn't work, if you want to get x as an Integer, you have to do this.x or i.x. In the String array, it's just two Strings, not the values you created in i that happens to have the same name.

EDIT: If you want to save x and y from i in an array, you have to do:

Integer[] ints= {i.x, y.x};
System.out.println(Integer.class.isInstance(ints[0]);

If you want to get these value as String:

Integer.parseInt(ints[]);

score 0 · Accepted Answer · answered Aug 12 '15 at 07:34

Your class is equivalent to the below code:-

public  class X {
        public static void main (String[] args) throws java.lang.Exception
        {
             String[] str={"x", "y"};
             System.out.println(Integer.class.isInstance(Integer.parseInt(str[0])));
        }
    }

You are trying to do comparison between an Integer and a random string(x and y in this case) which cannot be parsed to an integer. You cannot parse string to an integer in this case. See below examples, it might make things clear for you:-

public  class X {
        public static void main (String[] args) throws java.lang.Exception
        {
             String[] str={"5", "7"};
              System.out.println(Integer.class.isInstance(str[0]));
        }
    }

Still returns false.

Change it to

 System.out.println(Integer.class.isInstance(Integer.parseInt(str[0])));

will return true.

score 0 · Answer 3 · answered Aug 12 '15 at 08:42

Thanks for the help. I did it in this way.

class A
{
    public Integer x=new Integer(5);
    public Integer y=new Integer(7);
    public static void main (String[] args) throws java.lang.Exception
    {
        A i=new A();
        String[] s = {"allowedFileTypeMap","x","y"};
        Field field = i.getClass().getField(s[1]);
        if(field!=null){
            Object fieldType = field.getType();
            System.out.println(fieldType);
            if(field.getType().isAssignableFrom(Integer.class)){
                System.out.println("Working");
            }
        }       
    }
}

isInstance not working if the member is in a array

3 Answers3