Question

出于某种原因，我无法使用Strings compareTo方法来工作。我收到此错误消息：

exception in thread main java.lang.NullpointerException
at java.lang.String.compareTo(Unknown Source)
at Sort.main(Sort.java:27)

我该怎么做才能解决这个问题？（此代码要求to arrays数组已经单独排序。）

import java.util.*;

class Sort{
    public static void main(String[] args){
        String[] ord = new String[6];
        ord[0] = "astra";
        ord[1] = "bilum";
        ord[2] = "deliber";
        ord[3] = "kaliber";
        ord[4] = "suppe";
        ord[5] = "vorte";

        String[] ordet = new String[6];
        ord[0] = "akilles";
        ord[1] = "kopper";
        ord[2] = "lipton";
        ord[3] = "mus";
        ord[4] = "orkester";
        ord[5] = "toving";

        String[] flettet = new String[ord.length + ordet.length];

        for(int i = 0; i < ord.length; i++){
            int teller = i; 
            for(int j = 0; j < ordet.length; j++){
                if(ord[i].compareTo(ordet[j]) > 0){ //line 27
                    teller += 1;
                }
            }
            flettet[teller] = ord[i];
        }
    }   
}

Answer 1

你的意思是 - 看起来你正在设置错误的数组：

    String[] ordet = new String[6];
    ordet[0] = "akilles";
    ordet[1] = "kopper";
    ordet[2] = "lipton";
    ordet[3] = "mus";
    ordet[4] = "orkester";
    ordet[5] = "toving";

Answer 2

您永远不会为ordet分配任何值。所以compareTo将字符串与NULL进行比较

    String[] ord = new String[6];
    ord[0] = "astra";
    ord[1] = "bilum";
    ord[2] = "deliber";
    ord[3] = "kaliber";
    ord[4] = "suppe";
    ord[5] = "vorte";

    String[] ordet = new String[6];
    ord[0] = "akilles";                //overwriting ord[0]
    ord[1] = "kopper";                 //overwriting ord[1]
    ord[2] = "lipton";                 //overwriting ord[2]
    ord[3] = "mus";                    //overwriting ord[3]
    ord[4] = "orkester";               //overwriting ord[4]
    ord[5] = "toving";                 //overwriting ord[5]

Answer 3

String[] ordet = new String[6]; ord[0] = "akilles"; ord[1] = "kopper"; ord[2] = "lipton"; ord[3] = "mus"; ord[4] = "orkester"; ord[5] = "toving"; 您从未填充ordet数组，只是覆盖了ord数组。此时，ordet数组中没有元素，只有null s。

因此...

if(ord[i].compareTo(ordet[j]) > 0){ //line 27 是与不存在的元素进行比较，#compareTo抛出NullPointerException，因为您传入的是null，而不是String

无法使用Strings compareTo方法工作

3 个答案: