本文介绍了非静态方法无法从静态上下文中引用的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

首先是一些代码:

import java.util.*;
//...

class TicTacToe
{
//...

public static void main (String[]arg)
{

    Random Random = new Random() ;
    toerunner () ; // this leads to a path of
                   // methods that eventualy gets us to the rest of the code
}
//...

public void CompTurn (int type, boolean debug)
{
//...

        boolean done = true ;
        int a = 0 ;
        while (!done)
        {
            a = Random.nextInt(10) ;
            if (debug) { int i = 0 ; while (i<20) { System.out.print (a+", ") ; i++; }}
            if (possibles[a]==1) done = true ;
        }
        this.board[a] = 2 ;


}
//...

} //to close the class

以下是错误消息:

TicTacToe.java:85: non-static method nextInt(int) cannot be referenced from a static context
            a = Random.nextInt(10) ;
                      ^

究竟出了什么问题?该错误消息非静态方法无法从静态上下文引用是什么意思?

What exactly went wrong? What does that error message "non static method cannot be referenced from a static context" mean?

推荐答案

您使用 nextInt > Random.nextInt 。

You are calling nextInt statically by using Random.nextInt.

相反,创建一个变量, Random r = new Random(); 然后拨打 r.nextInt(10)

Instead, create a variable, Random r = new Random(); and then call r.nextInt(10).

检查时绝对值得out:

It would be definitely worth while to check out:



  • What is the reason behind "non staticmethod cannot be referenced from a static context"?

你真的应该替换这一行,

You really should replace this line,

Random Random = new Random();

有这样的东西,

Random r = new Random();

如果你使用变量名作为类名,你会遇到很多问题。此外,作为Java约定,使用变量的小写名称。这可能有助于避免一些混淆。

If you use variable names as class names you'll run into a boat load of problems. Also as a Java convention, use lowercase names for variables. That might help avoid some confusion.

这篇关于非静态方法无法从静态上下文中引用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!

07-22 21:32
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