最近，类似的问题也出现在了程序员网站上。SE

为什么在Java和c#中使用静态主方法，而不是构造函数? 从主要或次要来源中寻找一个明确的答案，为什么(特别是)Java和c#决定将静态方法作为它们的入口点-而不是通过应用程序类的实例表示应用程序实例，入口点是一个适当的构造函数?

公认的答案是，

In Java, the reason of public static void main(String[] args) is that Gosling wanted the code written by someone experienced in C (not in Java) to be executed by someone used to running PostScript on NeWS   For C#, the reasoning is transitively similar so to speak. Language designers kept the program entry point syntax familiar for programmers coming from Java. As C# architect Anders Hejlsberg puts it, ...our approach with C# has simply been to offer an alternative... to Java programmers... ...

最近，类似的问题也出现在了程序员网站上。SE

为什么在Java和c#中使用静态主方法，而不是构造函数? 从主要或次要来源中寻找一个明确的答案，为什么(特别是)Java和c#决定将静态方法作为它们的入口点-而不是通过应用程序类的实例表示应用程序实例，入口点是一个适当的构造函数?

公认的答案是，

In Java, the reason of public static void main(String[] args) is that Gosling wanted the code written by someone experienced in C (not in Java) to be executed by someone used to running PostScript on NeWS   For C#, the reasoning is transitively similar so to speak. Language designers kept the program entry point syntax familiar for programmers coming from Java. As C# architect Anders Hejlsberg puts it, ...our approach with C# has simply been to offer an alternative... to Java programmers... ...

静态方法不需要任何对象。它直接运行，所以main直接运行。

public static void main(String args[])是什么意思?

public is an access specifier meaning anyone can access/invoke it such as JVM(Java Virtual Machine. static allows main() to be called before an object of the class has been created. This is neccesary because main() is called by the JVM before any objects are made. Since it is static it can be directly invoked via the class. class demo { private int length; private static int breadth; void output(){ length=5; System.out.println(length); } static void staticOutput(){ breadth=10; System.out.println(breadth); } public static void main(String args[]){ demo d1=new demo(); d1.output(); // Note here output() function is not static so here // we need to create object staticOutput(); // Note here staticOutput() function is static so here // we needn't to create object Similar is the case with main /* Although: demo.staticOutput(); Works fine d1.staticOutput(); Works fine */ } } Similarly, we use static sometime for user defined methods so that we need not to make objects. void indicates that the main() method being declared does not return a value. String[] args specifies the only parameter in the main() method. args - a parameter which contains an array of objects of class type String.

我不知道JVM是否在对象实例化之前调用main方法…但是main()方法是静态的还有一个更有力的原因…当JVM调用类的主方法(比如Person)时。它通过"Person.main()"调用它。您可以看到，JVM通过类名调用它。这就是为什么main()方法应该是静态和公共的，以便JVM可以访问它。

希望有帮助。如果是的话，请在评论中告诉我。

原型public static void main(String[])是在JLS中定义的约定:

方法main必须声明为public、static和void。它必须指定一个形式形参(§8.4.1)，其声明类型为String数组。

在JVM规范5.2中。虚拟机启动我们可以读到:

The Java virtual machine starts up by creating an initial class, which is specified in an implementation-dependent manner, using the bootstrap class loader (§5.3.1). The Java virtual machine then links the initial class, initializes it, and invokes the public class method void main(String[]). The invocation of this method drives all further execution. Execution of the Java virtual machine instructions constituting the main method may cause linking (and consequently creation) of additional classes and interfaces, as well as invocation of additional methods.

有趣的是，在JVM规范中并没有提到主方法必须是静态的。 但是规范还说Java虚拟机执行以下两个步骤:

链接初始类(5.4。链接) 初始化它。初始化)

类或接口的初始化包括执行类或接口的初始化方法。

在2.9。特殊方法:

定义一个类或接口初始化方法:

一个类或接口最多有一个类或接口初始化方法，并通过调用该方法进行初始化(§5.5)。类或接口的初始化方法具有特殊名称<clinit>，不带参数，且为空。

类或接口初始化方法不同于定义如下的实例初始化方法:

在Java虚拟机级别，每个用Java编程语言(JLS§8.8)编写的构造函数都作为实例初始化方法出现，具有特殊名称<init>。

因此JVM初始化一个类或接口初始化方法，而不是实例初始化方法，后者实际上是一个构造函数。 因此，他们不需要在JVM规范中提到主方法必须是静态的，因为在调用主方法之前不创建实例这一事实暗示了这一点。