在Java规范中无法理解volatile的例子

class Test {
    static volatile int i = 0,j = 0;
    static void one() { i++; j++; }
    static void two() {
        System.out.println("i=" + i + " j=" + j);
    }
}

This allows method one and method two to be executed concurrently,but
guarantees that accesses to the shared values for i and j occur
exactly as many times,and in exactly the same order,as they appear
to occur during execution of the program text by each thread.
Therefore,the shared value for j is never greater than that for i,
because each update to i must be reflected in the shared value for i
before the update to j occurs. It is possible,however,that any given
invocation of method two might observe a value for j that is much
greater than the value observed for i,because method one might be
executed many times between the moment when method two fetches the
value of i and the moment when method two fetches the value of j.

怎么

j never greater than i

,但同时

any given invocation of method two might observe a value for j that is
much greater than the value observed for i

??

看起来像矛盾.

运行示例程序后,我的j比我大.为什么使用挥发性呢？它给出几乎相同的结果没有挥发性(我也可以大于j,前面的规范之一).为什么这个例子在这里作为同步的替代方法？