问题
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.
回答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.
回答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.
回答3:
Another rare case for class-variables may be that another thread could change the values of a and b while the comparison is executing.
来源：https://stackoverflow.com/questions/19042393/how-can-a-b-b-a-a-b-be-true