如果两个不同的异步回调启动两个不同的同步方法,会发生什么情况?

我自己也试过了,多亏了“Droidev”。我用过 Anko library 对于异步任务。下面是代码:

  doAsync {
        Log.d("ASYNC", "Async A started")
        foo1()
        uiThread {
            Log.d("ASYNC", "Async A done")
        }
    }

    Thread.sleep(1000)

    doAsync {
        Log.d("ASYNC", "Async B started")
        foo2()
        uiThread {
            Log.d("ASYNC", "Async B done")
        }
    }
}

fun foo1() {
    Log.d("ASYNC", "foo1() called")
    Thread.sleep(3000)
    Log.d("ASYNC", "foo1() done")
}

fun foo2() {
    Log.d("ASYNC", "foo2() called")
    Log.d("ASYNC", "foo2() done")
}

这是我得到的结果/日志:

07-12 11:44:55.195 23990-24054/gemini.com.jetpack D/ASYNC: Async A started
foo1() called
07-12 11:44:56.195 23990-24028/gemini.com.jetpack D/ASYNC: Async B started
foo2() called
07-12 11:44:56.205 23990-24028/gemini.com.jetpack D/ASYNC: foo2() done
07-12 11:44:56.215 23990-23990/gemini.com.jetpack D/ASYNC: Async B done
07-12 11:44:58.195 23990-24054/gemini.com.jetpack D/ASYNC: foo1() done
07-12 11:44:58.205 23990-23990/gemini.com.jetpack D/ASYNC: Async A done

因此,方法foo2()不会等待foo1()完成。