java - 为什么会抛出空点异常？

Question

这个问题已经有答案了:

What is a NullPointerException, and how do I fix it? (12 个回答)

已关闭 3 年前。

出于某种原因，每当我对此进行测试时，IDE 都会在第 46 行(检查数组名称的 for 循环)抛出 NullPointerException。它似乎有效，因为它获取名称并返回它，但随后立即出现异常。谁能解释一下吗？谢谢!

导入java.util.*;电话号码类{

private String name;
private String number;

PhoneNumbers(String n, String numb) {
    this.name = n;
    this.number = numb;
}

public String getName() {
    return name;
}

public String getNumber() {
    return number;
}

public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    boolean repeat = true;
    int count = 11;
    PhoneNumbers[] contacts = new PhoneNumbers[30];

    contacts[0] = new PhoneNumbers("Gina", "(847) 342-0912");
    contacts[1] = new PhoneNumbers("Marcia", "(847) 341-2392");
    contacts[2] = new PhoneNumbers("Rita", "(847) 354-0654");
    contacts[3] = new PhoneNumbers("Jennifer", "(414) 234-0912");
    contacts[4] = new PhoneNumbers("Fred", "(414) 435-0434");
    contacts[5] = new PhoneNumbers("Neil", "(608) 123-0914");
    contacts[6] = new PhoneNumbers("Judy", "(608) 123-0312");
    contacts[7] = new PhoneNumbers("Arlene", "(608) 123-0312");
    contacts[9] = new PhoneNumbers("LaWanda", "(920) 787-9813");
    contacts[10] = new PhoneNumbers("Deepak", "(930) 412-0991");

    while (repeat) {
        boolean found = false;

        System.out.print("Enter your friends name: ");
        String nameInput = sc.nextLine();

         if (!nameInput.equalsIgnoreCase("quit")){

             for(int i = 0; i < count - 1; i++){
                    if (contacts[i].getName().equalsIgnoreCase(nameInput)) {<---Here's where the issue occurs***
                        found = true;
                        System.out.println(contacts[i].getName() + "'s number is " + contacts[i].getNumber());
                }
             }
             if (!found){
                 System.out.println("Enter your friends phone number: ");
                 String num = sc.nextLine();

                 contacts[count + 1] = new PhoneNumbers(nameInput, num);
                 count++;
             }
         }else if (nameInput.equalsIgnoreCase("quit")) {
            repeat = false;
        }else if (count == 30){
             repeat = false;
             }
        }
    }
}

Answer 1

您在初始化中缺少 contacts[8]，因此在尝试访问不存在的联系人 8 的姓名时，您将收到 NullPointerException:

// ...
contacts[7] = new PhoneNumbers("Arlene", "(608) 123-0312");
contacts[9] = new PhoneNumbers("LaWanda", "(920) 787-9813");
// ...