Class Cast exception which seems to be a valid statement

Question

System.out.println((long)(new Object()!=null ? 0 : new Object()));

On execution this gives a class cast exception as follows

Exception in thread "main" java.lang.ClassCastException: java.lang.Integer cannot be cast to java.lang.Long

This maybe a lame question but I'm not able to understand as to why this is happening even after we are explicitly casting the 0 returned to long.

you need to write `0L`, otherwise it is automaticly boxed as an `int`. Every number without a specification is therefore handelt to be an `int` (despite something like `0.0` which would be a `double`) — SomeJavaGuy, Oct 30 '15 at 10:01
Additionally, as we are talking about objects, Integer can't be typecasted to Long — SacJn, Oct 30 '15 at 10:02
@KevinEsche you should create an answer based on your comment. — Dominik Sandjaja, Oct 30 '15 at 10:13
@DaDaDom i am pretty sure this is a duplicate, but anyway you are right. — SomeJavaGuy, Oct 30 '15 at 10:24

SomeJavaGuy · Accepted Answer · 2015-10-30T10:30:52.060

Your problem is that you are trying to assign an int to a long which results in a ClassCastException.

To get the correct result you will need to define the value 0 that you are trying to cast to a long as an actuall long value.

In that case every non floating number that you are trying to assign is actually handelt to be an int if you don´t explicity define the type of the value.

This can easily be seen if you just write long i = 11111111111;. Your compiler should tell you that the range for this number is out of the scope for an int. To tell the compiler that this value should be handelt as a long you simply need to add an L after the number: long i = 11111111111L;.

In your case your statement should be

System.out.println((long)(new Object()!=null ? 0L : new Object()));

Some additional information:

A similiar scenario happens with the double aswell. float f = 0.; wont work because you would need to define it as a float by adding a F after the value. float f = 0.F;

EDIT:

Your value 0 should get wrapped into Integer because it is possible that it could return an Object. if you write it as System.out.println((long)(new Object()!=null ? 0 : 1)); this would compile again.

Correct me if I'm wrong but I am explicitly typecasting the int to long and AFAIK this is possible. — Sachin Malhotra, Oct 30 '15 at 10:26
@SachinMalhotra you can´t cast an `Integer` to `Long`, aswell as you can´t cast a `Double` to be a `Float`. — SomeJavaGuy, Oct 30 '15 at 10:28

score 1 · Answer 2 · answered Oct 30 '15 at 10:23

1

Why don't you try using;

System.out.println((long)(new Object()!=null ? 0L : new Object()));

You are getting Exception because the ternary returns Integer 0 for true and that can't be cast to Long.

answered Oct 30 '15 at 10:23

Shivam

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score 0 · Answer 3 · edited May 23 '17 at 11:44

Just adding on to what already has been told. According to java specifications for ternary operator,

If one of the second and third operands is of primitive type T, and the type of the other is the result of applying boxing conversion (§5.1.7) to T, then the type of the conditional expression is T.

Also a similar question to this one which is due to yet another rule for ternary operators.

Class Cast exception which seems to be a valid statement

3 Answers3