我写了这样的测试代码:

class MyProgram
{
    int count = 0;
    public static void main(String[] args)
    {
        System.out.println(count);
    }
}

但它给出了以下错误:

Main.java:6: error: non-static variable count cannot be referenced from a static context
        System.out.println(count);
                           ^

我如何让我的方法识别我的类变量?


当前回答

The first thing is to know the difference between an instance of a class, and the class itself. A class models certain properties, and the behaviour of the whole in the context of those properties. An instance will define specific values for those properties. Anything bound to the static keyword is available in the context of the class rather than in the context of an instance of the class As a corollary to the above variables within a method can not be static static fields, and methods must be invoked using the class-name e.g. MyProgram7.main(...) The lifetime of a static field/method is equivalent to the lifetime of your application

如。 例如,汽车具有属性颜色,并表现出行为“运动”。 以25公里每小时的速度行驶的红色大众甲壳虫为例。

现在,汽车的静态属性将是道路上的车轮数量(4),这将适用于所有汽车。

HTH

其他回答

现在您可以在方法中添加/使用实例with

public class Myprogram7 {

  Scanner scan;
  int compareCount = 0;
  int low = 0;
  int high = 0;
  int mid = 0;  
  int key = 0;  
  Scanner temp;  
  int[]list;  
  String menu, outputString;  
  int option = 1;  
  boolean found = false;  

  private void readLine() {

  }

  private void findkey() {

  }

  private void printCount() {

  }
  public static void main(String[] args){

    Myprogram7 myprg=new Myprogram7();
    myprg.readLine();
    myprg.findkey();
    myprg.printCount();
  }
}

在这个程序中,你想使用count,所以将count方法声明为静态方法

class MyProgram<br>
{
    int count = 0;
    public static void main(String[] args)
    {
        System.out.println(count);
    }
}

您可以将此方法声明为公共私有并受保护。如果使用此方法,则可以创建安全应用程序。

class MyProgram
{
    static int count = 0;
    public static void main(String[] args)
    {
        System.out.println(count);
    }
}

静态字段和方法连接到类本身,而不是它的实例。如果你有一个类a,一个“正常”(通常称为实例)方法b和一个静态方法c,并且你为你的类a创建了一个实例a,那么对A.c()和a.b()的调用是有效的。方法c()不知道连接的是哪个实例,因此它不能使用非静态字段。

解决方案是要么使字段静态,要么使方法非静态。你的主要内容可以是这样的:

class Programm {

    public static void main(String[] args) {
        Programm programm = new Programm();
        programm.start();
    }

    public void start() {
        // can now access non-static fields
    }
}

让我们先分析一下你的程序。 在你的程序中,你的第一个方法是main(),记住它是静态方法…然后声明该方法的局部变量(comparate、low、high等)。这个变量的作用域只是声明的方法,不管它是静态方法还是非静态方法。你不能在这个方法之外使用这些变量。这是你犯的基本错误。

Then we come to next point. You told static is killing you. (It may be killing you but it only gives life to your program!!) First you must understand the basic thing. *Static method calls only the static method and use only the static variable. *Static variable or static method are not dependent on any instance of that class. (i.e. If you change any state of the static variable it will reflect in all objects of the class) *Because of this you call it as a class variable or a class method. And a lot more is there about the "static" keyword. I hope now you get the idea. First change the scope of the variable and declare it as a static (to be able to use it in static methods).

给你们的建议是:你们误解了变量作用域和静态函数的概念。对此要有清晰的认识。

I will try to explain the static thing to you. First of all static variables do not belong to any particular instance of the class. They are recognized with the name of the class. Static methods again do not belong again to any particular instance. They can access only static variables. Imagine you call MyClass.myMethod() and myMethod is a static method. If you use non-static variables inside the method, how the hell on earth would it know which variables to use? That's why you can use from static methods only static variables. I repeat again they do NOT belong to any particular instance.