考虑来上海吗，纯正外企，10-6-5, 随时申请 WFH 和无限制年假，顶配 mac ，入职 400 美金 perks ，口语雅思 6 左右（其实没那么严格），react 技术栈

> First, if x and done are ordinary variables, then thread 2's loop may never stop. A common compiler optimization is to load a variable into a register at its first use and then reuse that register for future accesses to the variable, for as long as possible. If thread 2 copies done into a register before thread 1 executes, it may keep using that register for the entire loop, never noticing that thread 1 later modifies done.

rsc 的博客里有提到这种情况