如何在Kotlin中初始化一个线程？ [英] How to initialize a Thread in Kotlin?

问题描述

在Java中它通过接受一个实现runnable的对象来工作：

In Java it works by accepting an object which implements runnable :

Thread myThread = new Thread(new myRunnable())

其中 myRunnable 是一个实现<$的类C $ C>可运行。

但是当我在Kotlin尝试这个时，它似乎不起作用：

But when I tried this in Kotlin, it doesn't seems to work:

var myThread:Thread = myRunnable:Runnable

推荐答案

初始化 Thread 的对象时，只需调用构造函数：

For initialising an object of Thread you simply invoke the constructor:

val t = Thread()

然后，您还可以传递一个可选的带有lambda（SAM转换）的Runnable ，如下所示：

Then, you can also pass an optional Runnable with a lambda (SAM Conversion) like this:

Thread {
    Thread.sleep(1000)
    println("test")
}

越多显式版本传递 Runnable 的匿名实现，如下所示：

The more explicit version is passing an anonymous implementation of Runnable like this:

Thread(Runnable {
    Thread.sleep(1000)
    println("test")
})

请注意，之前显示的示例仅创建 Thread 的实例，但实际上并未启动它。为了实现这一点，你需要明确地调用 start（）。

Note that the previously shown examples do only create an instance of a Thread but don't actually start it. In order to achieve that, you need to invoke start() explicitly.

最后但并非最不重要的，你需要要知道标准库函数 thread ，我建议使用它：

Last but not least, you need to know the standard library function thread, which I'd recommend to use:

public fun thread(start: Boolean = true, isDaemon: Boolean = false, contextClassLoader: ClassLoader? = null, name: String? = null, priority: Int = -1, block: () -> Unit): Thread {

您可以像这样使用它：

thread(start = true) {
    Thread.sleep(1000)
    println("test")
}

它有许多可选参数，例如直接启动线程，如下所示。

It has many optional parameters for e.g. directly starting the thread as shown here.

这篇关于如何在Kotlin中初始化一个线程？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！