英文:
Tricky but not hard endless loop exercise
问题
我在Java测试中看到了这个练习，最终我不得不寻求想法，因为有一些解决方案我尝试过，但显然无法解决问题，这是由于数学条件造成的。
所以这个练习：
定义变量S和T，使得在下面的无限循环中成立！
while (s <= t && s >= t && s != t) {
}
英文:
I saw this exercise on a Java test, which finally led me to ask for ideas because there are some solutions I tried, but obviously won't solve the problem, due to Math conditions.
So the exercise:
> Define S and T variables as it results an endless while loop below!
while(s <= t && s >= t && s != t)
{
}
答案1
得分: 1
思考面向对象编程，实际上 0！=0 是假的，但是 new Integer(0) != new Integer(0) 是真的，因为它们不是同一引用。
因此 Integer S = new Integer(0); Integer T = new Integer(0);
应该能解决你的问题。
英文:
think about OOP, in fact 0!=0 is false, but new Integer(0) != new Integer(0) is true because it's not the same reference.
so Integer S= new Integer(0);
Integer T = new Integer(0);
should resolve your problem
答案2
得分: 1
这是答案：
这在原始类型中是不可能的。您可以使用封装的整数类型实现：
Integer a = new Integer(1);
Integer b = new Integer(1);
<= 和 >= 比较将使用非封装的值 1，而 != 将比较引用并成功，因为它们是不同的对象。
您可以在此链接中查看更多详细信息：https://stackoverflow.com/questions/19042393/how-can-a-b-b-a-a-b-be-true/19042429
英文:
here is answer :
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.
you can check farther details in this link https://stackoverflow.com/questions/19042393/how-can-a-b-b-a-a-b-be-true/19042429
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