為什么我不能為對象的數組設置屬性？

Question

例如，我想為Gabriel設置a[1]名稱，它總是給我錯誤：

Exception in thread "main" java.lang.NullPointerException 
at Register.main(Register.java:4)

我的代碼是：

public class Register {
    public static void main(String[] args) {
        Employee a[] = new Employee[2];
        a[0].setName("Douglas");
    }
}

Answer 1

當您創建一個非原始數組時，您會創建一個指定大小的null數組。 在訪問對象的屬性之前，您需要創建它們（使用new ）：

Employee a[] = new Employee[2];
a[0] = new Employee(); // Here!
a[0].setName("Douglas");

Answer 2

我們可以假設您的Employee class 如下：

class Employee {
    private String name;

    public Employee(String s) {
        name = s;
    }

    public void setName(String s) {
        name = s;
    }

    public String getName() {
        return name;
    }
}

public class Main {
    public static void main(String[] args) {
        Employee a[] = null;

        if (a == null)
            System.out.println("Employee array is null !!!");

        a = new Employee[10]; // You are init your aray

        a[0] = new Employee("Test Name 1"); // it's init first item of array
        a[1] = new Employee("Test Name 2"); // it's init second item of array
        a[2] = new Employee("Test Name 3"); // it's init third item of array

        for (Employee employee : a) {
            if (employee == null) {
                System.out.println("Employee  is null !!!");
            } else
                System.out.println("Employee  is not null : "
                        + employee.getName());
        }
    }
}

正如您在代碼"Employee a[] = new Employee[2];"中看到的只為員工數量分配空間。 但它是用null設置的。 因為它保留了您的null員工參考資料，所以您應該一一初始化它們。 由於這個原因"a[0].setName("Douglas");" 試圖到達 null 地址並拋出異常。

如果運行它，你可以看到這個 output：

Employee array is null !!!
Employee  is not null : Test Name 1
Employee  is not null : Test Name 2
Employee  is not null : Test Name 3
Employee  is null !!!
Employee  is null !!!
Employee  is null !!!
Employee  is null !!!
Employee  is null !!!
Employee  is null !!!
Employee  is null !!!

Answer 3

您可以通過lombok使用構建器模式，並且可以使用非空對象初始化列表。

import lombok.Builder;
import org.assertj.core.util.Arrays;

import java.util.List;

public static class Main {

    @Builder
    public class Employee {
        private String name;
    }

    public static void main(String[] args) {
        List<Main.Employee> employees = List.of(
                Employee.builder().name("Douglas").build(),
                Employee.builder().name("Maik").build(),
                Employee.builder().name("Alex").build()
        );
    }
}

Answer 4

您應該向Employee class 添加一個構造函數。 然后可以按如下方式創建數組：

public static void main(String... args) {
    Employee[] arr = {
            new Employee("Douglas"),
            new Employee("Douglas2")};

    System.out.println(Arrays.toString(arr));
    // [Douglas, Douglas2]
}

public static class Employee {
    String name;

    public Employee(String name) {
        this.name = name;
    }

    @Override
    public String toString() {
        return name;
    }
}