更​​改为：

$.when(first, second).done(logthree);

正在执行 logthree（）并通过返回的结果为 .done（）。你需要传递一个函数引用 .done（）这仅仅是 logthree 没有括号。当您添加括号，即指示JS跨preTER立即执行。这是一个常见的​​错误。

You are executing logthree() and passing the return result to .done(). You need to pass a function reference to .done() which is just logthree without the parens. When you add parens, that instructs the JS interpreter to execute it immediately. This is a common mistake.