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//ts实现数组的concat功能 type Concat<T extends any[], U extends any[]> = [...T, ...U]; type Result = Concat<[1], [2]>;
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个人觉得将any换成unknown会更严谨一些
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类型体操里这么换有啥作用?能举个例子说明下吗?
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The text was updated successfully, but these errors were encountered: