Java中的 HashMap putIfAbsent(key, value) 方法及示例
HashMap类的putIfAbsent(K key, V value)方法用于将指定的键与指定的值进行映射,仅当此 HashMap 实例中不存在(或映射为空)该键时。
句法:
public V putIfAbsent(K key, V value)
参数:此方法接受两个参数:
- 键:这是必须映射提供的值的键。
- value:这是必须与提供的键相关联的值,如果不存在的话。
返回值:
此方法返回null (如果之前没有与提供的键的映射,或者它已映射到空值)或与提供的键关联的当前值。
例外:
此方法可能会抛出以下异常:
- NullPointerException:如果指定的键或值为空,并且此映射不支持空值。
- IllegalArgumentException:如果指定的键或值阻止它存储在映射中。
- UnsupportedOperationException – 如果此映射不支持 put 操作(可选)
- ClassCastException – 如果键或值的类型不适合此映射(可选)
方案一:
Java
// Java program to demonstrate
// putIfAbsent(Key, value) method.
import java.util.HashMap;
public class TestClass {
// Main method
public static void main(String[] args)
{
// create a HashMap and add some values
HashMap map = new HashMap<>();
map.put("a", 10000);
map.put("b", 55000);
map.put("c", 44300);
map.put("e", 53200);
// print original map
System.out.println("HashMap:\n " + map.toString());
// put a new value which is not mapped
// before in map
map.putIfAbsent("d", 77633);
System.out.println("New HashMap:\n " + map);
// try to put a new value with existing key
// before in map
map.putIfAbsent("d", 55555);
// print newly mapped map
System.out.println("Unchanged HashMap:\n " + map);
}
}
Java
// Java program to demonstrate
// putIfAbsent(Key, value) method.
import java.util.*;
public class GFG {
// Main method
public static void main(String[] args)
{
// create a HashMap and add some values
HashMap map = new HashMap<>();
map.put("a", 10000);
map.put("b", 55000);
map.put("c", 44300);
map.put("e", null);
// print original map
System.out.println("HashMap:\n " + map.toString());
// put a new value which is not mapped
// before in map and store the returned
// value in r1
Integer r1 = map.putIfAbsent("d", 77633);
// put a new value for key 'e' which is mapped
// with a null value, and store the returned
// value in r2
Integer r2 = map.putIfAbsent("e", 77633);
// print the value of r1
System.out.println("Value of r1:\n " + r1);
// print the value of r2
System.out.println("Value of r2:\n " + r2);
// print newly mapped map
System.out.println("New HashMap:\n " + map);
}
}
输出
HashMap:
{a=10000, b=55000, c=44300, e=53200}
New HashMap:
{a=10000, b=55000, c=44300, d=77633, e=53200}
Unchanged HashMap:
{a=10000, b=55000, c=44300, d=77633, e=53200}
方案二:
Java
// Java program to demonstrate
// putIfAbsent(Key, value) method.
import java.util.*;
public class GFG {
// Main method
public static void main(String[] args)
{
// create a HashMap and add some values
HashMap map = new HashMap<>();
map.put("a", 10000);
map.put("b", 55000);
map.put("c", 44300);
map.put("e", null);
// print original map
System.out.println("HashMap:\n " + map.toString());
// put a new value which is not mapped
// before in map and store the returned
// value in r1
Integer r1 = map.putIfAbsent("d", 77633);
// put a new value for key 'e' which is mapped
// with a null value, and store the returned
// value in r2
Integer r2 = map.putIfAbsent("e", 77633);
// print the value of r1
System.out.println("Value of r1:\n " + r1);
// print the value of r2
System.out.println("Value of r2:\n " + r2);
// print newly mapped map
System.out.println("New HashMap:\n " + map);
}
}
输出
HashMap:
{a=10000, b=55000, c=44300, e=null}
Value of r1:
null
Value of r2:
null
New HashMap:
{a=10000, b=55000, c=44300, d=77633, e=77633}
参考: https: Java/util/HashMap.html#putIfAbsent-KV-