import java.io.BufferedReader; import java.io.InputStreamReader;
//完成下面的模板
公共课问题3_1 {
//complete this class, called Problem3_1, with the following items:
//1. Declare four attributes, name, age, height, and weight of types String and int-s.
//Write a constructor for this class that initializes ONLY the name, age, and height to three incoming arguments,
//and sets the weight to always be -1 (the latter is not an incoming argument).
String name;
int age;
int height;
int weight;
Address address;
public Problem3_1(String name, int age, int height) {
this.name = name;
this.age = age;
this.height = height;
weight = -1;
}
void setAddress(int number, String street) {
address = new Address(number, street);
}
//2. Imagine there is a class called Address that you have access to (it's below).
//Its constructor takes an integer street number and a String street. Add an attribute called address to
//the Problem3_1 class, and create a method called setAddress that sets the attribute to the incoming argument.
public static class Address{
int number;
String street;
public Address(int number, String street){
this.number = number;
this.street = street;
}
}
}
那么为什么代码说我在错误的位置有这个无效信息?它的位置在哪里?或者为什么不能“应用”?
PS:我得到的错误:
问题3_1.java:82:问题3_1中的setAddress(int,java.lang.String) 无法应用于(Problem3_1.Address) p.setAddress(一);
测试用例:
public static void main(String[] args){
/*Below are tests that will check if you completed the code above correctly; if
your code doesn't compile, you'll need to fix those errors first.
DO NOT WRITE CODE BELOW THIS POINT
*/
int failed = 0;
Problem3_1 p = new Problem3_1("Jane", 22, 65);
if (p.name.compareTo("Jane") == 0 && p.age == 22 && p.height == 65 && p.weight == -1)
System.out.println("Test 1 passed!");
else{
failed ++;
System.out.println("Please check your code for question 1! ");
}
Address a = new Address(12, "Fairfax Dr");
p.setAddress(a);
if (p.address == a)
System.out.println("Test 2 passed!");
else{
failed ++;
System.out.println("Please check your code for question 2! ");
}
if (failed > 0)
System.out.println(systemCall("Failed"));
else{
System.out.println("GREAT WORK! EVERYTHING PASSED!");
System.out.println(systemCall("Nice"));
}
}
答案 0 :(得分:0)
类似的东西:
void setAddress(Address a) {
this.address = a;
}
目前,您没有将任何值传递到setAddress
,因此setAddress
无法做任何事情。
答案 1 :(得分:0)
public class Problem3_1{
String name;
int age;
int height;
int weight;
static Address address;
public Problem3_1(String name, int age, int height,int weight) {
this.name = name;
this.age = age;
this.height = height; //cm
this.weight = weight; //kg
}
void setAddress(int number,String street) {
address = new Address(number,street);
}
public static void main(String [] args){
Problem3_1 problem=new Problem3_1("Tom",28,180,78);
problem.setAddress(4460,"new Delhi");
System.out.println(address.getAddressNumber() +" number of street : "+address.getAddressString());
}
public static class Address{
private int number;
private String street;
public Address(int number, String street){
this.number = number;
this.street = street ;
}
int getAddressNumber(){
return number;
}
String getAddressString(){
return street;
}
}
}
答案 2 :(得分:0)
首先,将Address类作为参数传递。
void setAddress(Address s)
由于静态嵌套类中的Address,但您已在Problem3_1类文件中将Address声明为String
。因此,您需要手动连接值,如
void setAddress(Address s) {
//address = s;
address = s.number + " , " + s.street;
}
如果您正确地创建和传递值,这应该可行。
示例:
Problem3_1 a = new Problem3_1("name1",20,5);
Problem3_1.Address a_address= new Problem3_1.Address(20,"1st street");
a.setAddress(a_address);
System.out.println(a.address);
如果你在外部类和内部类中重写了toString()方法,那么应该打印你想要的输出。
<强>更新强>
http://ideone.com/rS20uZ - 点击此处查看完整代码和输出。