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目前我这样做:
useActor<SomeInterpreter>((state as any).children.SomeChild);
由于我不是 TypeScript 中“任何”的忠实粉丝,还有其他方法吗?
David Khourshid 表示,在发表此评论时,这是不可能的。他们将来会添加更好的方法。