练习(1)(1.2)(1.3)

1.1

package duixiang;

public class lianxi1 {
    //“+”= public

    String name;
    int age;

    /**
     *sex:0表明是女性
     * 1表明是男性
     */
    int sex;

    public void study(){
        System.out.println("studying");
    }

    public void showAge(){
        System.out.println("age:"+age);
    }

    public int addAge(int i){//(形参):不能赋值,调用时才能赋值
        age+=i;
        return age;
    }

}
public class lianxi1Text {
    public static void main(String[] args) {
        lianxi1 t1=new lianxi1();//设置对象

        t1.name="李四";//调用属性
        t1.age=20;
        t1.sex=1;
        //调用方法
        t1.study();
        t1.showAge();

        int newAge= t1.addAge(2);
        System.out.println(t1.name+"的新年龄为:"+newAge);
        System.out.println(t1.age);//22

        lianxi1 t2=new lianxi1();
        System.out.println("t2:"+newAge);
        t2.showAge();
        int Age2=t2.addAge(2);
        System.out.println(Age2);
        t2.showAge();
    }
}

1.2

public class lianxi2 {
    public static void main(String[] args) {
            lianxi2 t1 =new lianxi2();

             t1.method0();

            //方法一
//            int re= t1.method();
//            System.out.println(re);
     //方式二
        System.out.println(t1.method());

            int re3=t1.method2(10,10);
            System.out.println(re3);

    }
    public void method0(){
        for (int i=0;i<10;i++){
            for (int j=0;j<8;j++){
                System.out.print("* ");
            }
            System.out.println();
        }
        System.out.println("+++++++++++++++++");
    }


    public int method(){
        for (int i=0;i<10;i++){
            for (int j=0;j<8;j++){
                System.out.print("* ");
            }
            System.out.println();
        }
        return 10*8;
}

    public int method2(int m,int n){
        for (int i=0;i<m;i++){
            for (int j=0;j<n;j++){
                System.out.print("* ");
            }
            System.out.println();
        }
        return m*n;
    }
}

1.3

public class lianxi3 {

    public static void main(String[] args) {
        //声明student类型数组
        Student[] stus = new Student[20];

        for (int i = 0; i < stus.length; i++) {//输出20个null new对象
            //给数组元素赋值
            stus[i] = new Student(); //数组中每一个都有了空间,(对象):引用型变量的默认初始值是null,所以需要new一个
//            System.out.println(stus[i]);

            //给Student对象赋值
            stus[i].number = (i + 1);
            System.out.println(i);//0,1,2,3,...19
            //年级:【1.6】 随机数 a ~ b 公式:int value = (int)(Math.random() * (b - a + 1) + a)
            stus[i].state = (int) (Math.random() * (6 - 1 + 1) + 1);
            //[0.100]
            stus[i].score = (int) (Math.random() * (100 - 0 + 1) + 0);

        }
        //遍历学生数组
        for (int i = 0; i < stus.length; i++) {
            //1,2,3,..20
//            System.out.println(stus[i].number+","+stus[i].state+","+stus[i].score);
            System.out.println(stus[i].info());
        }
        System.out.println("=============================");
        //输出班级为3的
        for (int i = 0; i < stus.length; i++) {
            if (stus[i].state == 3) {
                System.out.println(stus[i].info());
            }
        }
        System.out.println("=================================");
        //冒泡排序
        for (int i=0;i< stus.length-1;i++){
            for (int j=0;j< stus.length-1-i;j++){
//                System.out.println(stus.length);
                if (stus[j].score >stus[j+1].score){
                    //如果需要换序,交换的是数组元素,Student对象
                    Student temp =stus[j];
                    stus[j] = stus[j+1];
                    stus[j+1]=temp;

                }
            }
        }
        //遍历学生数组
        for (int i=0;i< stus.length;i++){
            System.out.println(stus[i].info());
        }
    }
}

class Student {//默认值还是0
    int number;//学号
    int state;//年级
    int score;//成绩

    //显示学生信息的方法
    public String info() {
        return "学号:" + number + ",年级" + state + ",成绩" + score;
    }
}

 

posted @ 2021-02-20 16:29  谷先生  阅读(71)  评论(0)    收藏  举报