How can "a <= b && b <= a && a != b" be true? [duplicate]

Question

Here is the code i have to figure it out how is it possible. I have a clue but i do not know how to do it. I think it is about negative and positive numbers and maybe the variable modifiers as well. I am a beginner i looked the solution everywhere but i could not find anything usable.

the question is that: You need to declare and initialize the two variables. The if condition must be true.

the code:

if( a <= b && b <= a && a!=b){
       System.out.println("anything...");
}

I appreciate you taking the time.

Answer 1

This is not possible with primitive types. You can achieve it with boxed Integers:

Integer a = new Integer(1);
Integer b = new Integer(1);

The <= and >= comparisons will use the unboxed value 1, while the != will compare the references and will succeed since they are different objects.

Answer 2

This works too:

Integer a = 128, b = 128;

This doesn't:

Integer a = 127, b = 127;

Auto-boxing an int is syntactic sugar for a call to Integer.valueOf(int). This function uses a cache for values less than 128. Thus, the assignment of 128 doesn't have a cache hit; it creates a new Integer instance with each auto-boxing operation, and a != b (reference comparison) is true.

The assignment of 127 has a cache hit, and the resulting Integer objects are really the same instance from the cache. So, the reference comparison a != b is false.