Question

我正在学习java并尝试了一些东西。这就是我想要做的。

我正在尝试创建一个对象数组的表。例如我正在创建一个名为animal的对象，通过它我可以根据用户的选择添加任意数量的动物及其品种。

package tt;

import java.util.Scanner;

public class animal {

    String aname;
    String abreed;

    public animal() {

        mainprog aa = new mainprog();

        System.out.printf("eneter name of your %s..\n", aa.Animalcat);
        Scanner name = new Scanner(System.in);
        aname = name.nextLine();

        System.out.printf("eneter breed of your %s..\n", aa.Animalcat);
        Scanner breed = new Scanner(System.in);
        abreed = breed.nextLine();

    }

    public String getbreed() {
        return abreed;

    }

    public String getname() {
        return aname;
    }
}

所以主程序询问我想添加多少动物。

package tt;

import java.util.Scanner;

public class mainprog {

    public static String Animalcat;

    public static void main(String[] args) {

        System.out.println("How many animals you want to add..");

        Scanner an = new Scanner(System.in);

        int animalNumbers = an.nextInt();

        animal[][] addAnimal = new animal[animalNumbers][1];
        animaltype[] at = new animaltype[animalNumbers];

        for (int i = 0; i < animalNumbers; i++) {

            at[i] = new animaltype();
            Animalcat = at[i].getAnimalType();

            for (int j = 0; j < 1; j++) {

                addAnimal[i][j] = new animal();
            }
        }

        Display(addAnimal, at);

    }

    public static void Display(animal x[][], animaltype y[]) {
        System.out.println("Your animals are..");
        for (int m = 0; m < x.length; m++) {
            System.out.printf("Following are the name and the breed of %s ",
                    y[m].getAnimalType());
            System.out.println();
            for (int n = 0; n < x[m].length; n++) {

                System.out.printf(" %s", x[m][n].aname);
                System.out.printf(" %s", x[m][n].abreed);
                System.out.println();

            }
        }
    }

}

package tt;

import java.util.Scanner;

public class animaltype {
    String animalType;

    public animaltype() {
        System.out.println("What kind of animal you want to add..");
        Scanner at = new Scanner(System.in);
        animalType = at.nextLine();
    }

    public String getAnimalType(){
        return animalType;
    }

}

我在这里遇到的问题是我可以询问我想要添加多少种动物，但是我无法控制我想添加多少种这类动物。添加动物时，我只能通过声明[1]手动添加1个动物，并且可以通过用户输入来添加。

addAnimal = new animal[animalNumbers][1];

如果我可以通过声明animal[][] addAnimal = null;然后稍后用以下内容初始化它来解决这个问题：

animal[][] addAnimal = new animal[animalNumbers][animaltypenumbers];

但我一直得到NullPointerException。无论如何我可以做到这一点吗？

Answer 1

创建数组时，它将由其包含的元素的默认值填充。由于Animal是Object，因此它将由null值填充，并且您不能使用任何具有null值的变量。由于您只是在当前代码中填写animal[i][0]，因此您不会遇到任何问题。但是当您尝试访问animal[i][1]时会出现。这种情况发生在Display方法：

public static void Display(animal x[][], animaltype y[]) {
    System.out.println("Your animals are..");
    for (int m = 0; m < x.length; m++) {
        System.out.printf("Following are the name and the breed of %s ",
                y[m].getAnimalType());
        System.out.println();
        for (int n = 0; n < x[m].length; n++) {
            //you only filled elements in x[m][0]
            //x[m][n] when n > 0 is null
            //so you will get NullPointerException
            System.out.printf(" %s", x[m][n].aname);
            System.out.printf(" %s", x[m][n].abreed);
            System.out.println();

        }
    }
}

更好的选择：

使用Animal[] addAnimal代替，您不需要它将其作为数组数组：

public static void main(String[] args) {
    System.out.println("How many animals you want to add..");
    Scanner an = new Scanner(System.in);
    int animalNumbers = an.nextInt();
    animal[] addAnimal = new animal[animalNumbers];
    animaltype[] at = new animaltype[animalNumbers];
    for (int i = 0; i < animalNumbers; i++) {
        //at[i] = new animaltype();
        //Animalcat = at[i].getAnimalType();
        Animalcat = new animaltype();
        //for (int j = 0; j < 1; j++) {
        //    addAnimal[i][j] = new animal();
        //}
        addAnimal[i] = Animalcat;
    }
    Display(addAnimal, at);
}
//modify Display method accordingly

更好的选择：

使用List<Animal>支持的ArrayList<Animal>代替Animal[]。 List让您处理动态增长的元素列表。

public static void main(String[] args) {
    System.out.println("How many animals you want to add..");
    Scanner an = new Scanner(System.in);
    int animalNumbers = an.nextInt();
    List<Animal> animals = new ArrayList<Animal>();
    for (int i = 0; i < animalNumbers; i++) {
        Animalcat = new animaltype();
        animals.add(Animalcat);
    }
    Display(animals);
}
//modify Display method accordingly

对象数组的问题

1 个答案: