java.util.AbstractList 類的 get() 方法用於返回此列表中指定位置的元素。
用法:
public abstract E get(int index)
參數:該方法將元素的索引作為參數,即要返回的元素。
Returns Value:此方法返回此列表中指定位置的元素。
異常:如果索引超出範圍(索引 = size()),此方法將引發 IndexOutOfBoundsException。
以下是說明 get() 方法的示例。
範例1:
// Java program to demonstrate
// get() method
// for Integer value
import java.util.*;
public class GFG1 {
public static void main(String[] argv)
throws Exception
{
try {
// Creating object of AbstractList<Integer>
AbstractList<Integer>
arrlist1 = new ArrayList<Integer>();
// Populating arrlist1
arrlist1.add(10);
arrlist1.add(20);
arrlist1.add(30);
arrlist1.add(40);
arrlist1.add(50);
// print arrlist1
System.out.println("ArrayListlist:"
+ arrlist1);
// getting the value at the index 3
// using get() method
int value = arrlist1.get(3);
// print the value
System.out.println("Element at index 3:"
+ value);
}
catch (IndexOutOfBoundsException e) {
System.out.println("Exception thrown:" + e);
}
}
}
輸出:
ArrayListlist:[10, 20, 30, 40, 50] Element at index 3:40
範例2:
// Java program to demonstrate
// get() method
// for IndexOutOfBoundsException
import java.util.*;
public class GFG1 {
public static void main(String[] argv)
throws Exception
{
try {
// Creating object of AbstractList<Integer>
AbstractList<Integer>
arrlist1 = new ArrayList<Integer>();
// Populating arrlist1
arrlist1.add(10);
arrlist1.add(20);
arrlist1.add(30);
arrlist1.add(40);
arrlist1.add(50);
// print arrlist1
System.out.println("ArrayListlist:"
+ arrlist1);
// getting the value at the index 7
// using get() method
System.out.println("\nTrying to get "
+ "the element from out"
+ " of range index ");
int value = arrlist1.get(7);
// print the value
System.out.println("Element at index 7:"
+ value);
}
catch (IndexOutOfBoundsException e) {
System.out.println("Exception thrown:" + e);
}
}
}
輸出:
ArrayListlist:[10, 20, 30, 40, 50] Trying to get the element from out of range index Exception thrown:java.lang.IndexOutOfBoundsException:Index:7, Size:5
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注:本文由純淨天空篩選整理自RohitPrasad3大神的英文原創作品 AbstractList get() method in Java with Examples。非經特殊聲明,原始代碼版權歸原作者所有,本譯文未經允許或授權,請勿轉載或複製。