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What is the immutable object in java high concurrency? I believe many inexperienced people don't know what to do about it. Therefore, this paper summarizes the causes and solutions of the problem. Through this article, I hope you can solve this problem.

Once an object is released, it is a secure object, which is an immutable object.

The conditions that immutable objects need to meet:

The state of an object cannot be modified after it is created

All fields of the object are of type final

The object was created correctly (the this reference did not escape during object creation)

Final keywords: classes, methods, variables

Modifier class: cannot be inherited. The member variable in the final class can be set to final as needed. Note that all member methods in the final class are implicitly specified as final methods.

Modification method: 1. The locking method is not modified by the inherited class; 2. Efficiency

Modifier variables: basic data type variables cannot be modified after initialization, and reference type variables cannot point to another object after initialization

Here is an example of a final decorated variable:

@ Slf4j@NotThreadSafepublic class ImmutableExample1 {private final static Integer a = 1; private final static String b = "2"; private final static Map map = new HashMap (); static {map.put (1,2); map.put (2,3);} public static void main (String [] args) {/ / a = 2 String / b = "3"; / / map = new HashMap () Map.put (1,3); log.info ("{}", map.get (1));}}

The map reference variable cannot specify a new reference, but the value in it can be modified.

This can lead to thread safety problems.

In addition to final defining immutable objects, are there any other means to define immutable objects? Yes, of course

Collections.unmodifiableXX: Collection 、 List 、 Set 、 Map.

Guava:ImmutableXXX:Collection 、 List 、 Set 、 Map

@ Slf4j@ThreadSafepublic class ImmutableExample2 {private static Map map = new HashMap (); static {map.put (1,2); map.put (2,3); map = Collections.unmodifiableMap (map);} public static void main (String [] args) {map.put (1,3); map.put (3prime4); log.info ("{}", map.get (3);}}

In this way, the operation will report an error:

Exception in thread "main" java.lang.UnsupportedOperationException at java.util.Collections$UnmodifiableMap.put (Collections.java:1457) at com.vincent.example.immutable.ImmutableExample2.main (ImmutableExample2.java:24)

In other words, if you declare a variable with Collections.unmodifiableMap, its contents cannot be modified. The data will not be contaminated.

@ ThreadSafepublic class ImmutableExample3 {private final static ImmutableList list = ImmutableList.of; private final static ImmutableSet set = ImmutableSet.copyOf (list); private final static ImmutableMap map = ImmutableMap.of; private final static ImmutableMap map2 = ImmutableMap.builder (). Put (1meme 2). Put (3p4). Build (); public static void main (String [] args) {map2.put (4);}}

According to the actual situation of the variable, it is best to become an immutable object, if you can make the object immutable as far as possible, so that there will be no thread safety problems in the case of multithreading.

After reading the above, have you mastered the method of what immutable objects are in java high concurrency? If you want to learn more skills or want to know more about it, you are welcome to follow the industry information channel, thank you for reading!

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