Question

我是Java新手。创建类对象时，传递作为构造函数定义一部分的List变量时遇到问题

.u.ups

Answer 1

您需要一个List<String>而不是一个String，Arrays.asList(T...)可能是最简单的解决方案：

Patient patientobj = new Patient("sean", "john", Arrays.asList("allergy1"));

如果您还有更多过敏反应

Patient patientobj = new Patient("sean", "john", 
        Arrays.asList("allergy1", "allergy2"));

Answer 2

    LinkedList<Integer> linkedList = new LinkedList<>();
    long  v3 = 10000_0000_00L;
    long longmodulo= v3 % 10;
    linkedList.add((int) longmodulo);

    v3 = v3/10;

    while (v3 != 0){
        long longmodulo2= v3 % 10;
        linkedList.add((int) longmodulo2);

        v3 = v3/10;
        System.out.println("value of v3 "+v3);
    }
    linkedList.forEach(System.out::println);

Answer 3

您还有一个解决方案，就是只添加一个类Patient的构造函数。

public Patient (String patientfirstName,String patientLastName,String allergeyList){
this.patientfirstName  = patientfirstName;
this.patientLastName = patientLastName;\
this.allergeyList = new ArrayList<>( Arrays.asList(allergeyList));
}

Answer 4

我认为您可以使用 Varargs 。多亏了varargs，您可以在参数中输入所需的参数

public class Patient {

public String patientfirstName;
public String patientLastName;
public List<String> allergyList;



public Patient(String fName,String lName,String...aList) {
    this.patientfirstName = fName;
    this.patientLastName = lName;
    this.allergyList = Arrays.asList(aList);
}


public static void main(String[] args) {

    Patient firstPatient = new Patient("Foo", "Bar", "First Allergy","Second Allergy");

    Patient secondPatient = new Patient("Foo", "Baz", "First Allergy","Second Allergy","Third Allergy","Fourth Allergy");

    Patient ThirdPatient = new Patient("Foo", "Foo", "First Allergy");
}

参数“ aList” 就像一个数组，因为varargs就像一个没有特定长度的数组，如您所见，输入参数时您选择的长度

过敏列表的类型是可选的。您也可以执行以下操作：

在“患者”属性中：

 public String[] allergyList;

在构造函数中：

public Patient(String fName,String lName,String...aList) {
        this.patientfirstName = fName;
        this.patientLastName = lName;
        this.allergyList = allergyList;
    }

如何将列表变量传递给构造函数/方法

4 个答案: