Without the theFlooble reference being volatile, the code in doWork() would be at risk for seeing a partially constructed Flooble as it dereferences the theFlooble reference.

如果 theFlooble 引用不是 volatile 类型，doWork() 中的代码在解除对 theFlooble 的引用时，将会得到一个不完全构造的 Flooble。

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