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Introduction to the usage of main() method in Java

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2019-02-23 16:52:023678browse

This article brings you an introduction to the usage of the main() method in Java. It has certain reference value. Friends in need can refer to it. I hope it will be helpful to you.

The entry point of the Java program---the signature of the main() method is: public static void main(String[] args) {...}, where,

public modifier: Java classes are called by the JVM. In order to allow the JVM to freely call the main() method, the public modifier is used Expose this method.

static modifier: When the JVM calls this main method, it will not create the main class first object, and then call the main method through the object. The JVM calls the main method directly through this class, so use static to modify the main method.

♦ void return value: Because the main method is called by the JVM, the return value of this method will be returned to the JVM, which does not make any sense , so the main() method has no return value.

The above method also includes a string array formal parameter String[] args. According to the rules of method calling: whoever calls the method is responsible for assigning values ​​to the formal parameters. . In other words, the main() method is called by the JVM, that is, the args parameter should be assigned by the JVM. But how does the JVM know to assign a value to the args array? First look at the following program:

public class ArgsTest{
  public static void main(String[] args){
    //输出args数组的长度
    System.out.println(args.length);
    //遍历args数组的每一个元素
    for(String arg : args){
      System.out.println(arg);
    }
  }
}

The above program is the simplest "hello world" program, but this program has added Output the length of the args array and the code for traversing the elements of the arg array. Use the java ArgsTest command to run the above program and see that the program only outputs a 0, which indicates that the args value is an array of length 0. This is reasonable because the computer has no thinking ability and can only faithfully execute user interactions. Given its task, since the program does not set parameter values ​​for the args array, the JVM does not know the elements of the args array, so the JVM sets the args array to an array with a length of 0.

Change the command to run the above program:

java ArgsTest Java Spring

The printed result is:

2
Java
Spring

It can be concluded that if the class name is followed by one or more strings (multiple strings separated by spaces) when running a java program, the JVM will Strings are assigned to args array elements once. The relationship between the parameters and the args array when running a Java program is: the first parameter corresponds to the first array element, the second parameter corresponds to the second array element,..., and so on.

If a parameter itself contains a space, the parameter should be enclosed in double quotes "", otherwise the JVM will treat the space as a parameter separator instead of the parameter itself. For example, use the following command to run the above program:

java ArgsTest "Java Spring"

See that the length of args is 1, there is only one array element, and its value is Java Spring.

#Question: How to output "hello world" before the main() method is executed?

As we all know, in the Java language, the main() method is the entry method of the program. When the program is running, the main() method is loaded first. But does this mean that the main() method is the first module to be executed when the program is running?

the answer is negative. In the Java language, since the static block will be called when the class is loaded, you can use the static block to realize the function of outputting "hello world" before the main() method is executed. The example is as follows:

public class Test{
  static{
    System.out.println("hello world");
  }
  public static void main(String[] args){
    System.out.println("hello my world");
  }
}

The program running result is:

hello world
hello my world

Due to static Regardless of the order, the code blocks will be executed before the main() method is executed. Therefore, the following code will have the same output as the above code:

public class Test{
  public static void main(String[] args){
    System.out.println("hello my world");    
    static{
      System.out.println("hello world");
    }
  }
}

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