Hazelcast map synchronization

问题

I am trying to implement distributed cache using Hazelcast in my application. I am using Hazelcast’s IMap. The problem I have is every time I get a value from a map and update the value, I need to do a put(key, value) again. If my value object has 10 properties and I have to update all 10, then I have to call put(key, value) 10 times. Something like -

IMap<Integer, Employee> mapEmployees = hz.getMap("employees");
Employee emp1 = mapEmployees.get(100);
emp1.setAge(30);
mapEmployees.put(100, emp1);
emp1.setSex(“F”);
mapEmployees.put(100, emp1);
emp1.setSalary(5000);
mapEmployees.put(100, emp1);

If I don’t do this way, some other node which operates on the same Employee object will update it and the final result is that the employee object is not synchronized. Is there any solution to avoid calling put explicitly multiple times? In a ConcurrentHashMap, I don’t need to do this because if I change the object, the map also gets updated.

回答1:

As of version 3.3 you'll want to use an EntryProcessor:

What you really want to do here is build an EntryProcessor<Integer, Employee> and call it using mapEmployees.executeOnKey( 100, new EmployeeUpdateEntryProcessor( new ObjectContainingUpdatedFields( 30, "F", 5000 ) );

This way, Hazelcast handles locking the map on the key for that Employee object and allows you to run whatever code is in the EntryProcessor's process() method atomically including updating values in the map.

So you'd implement EntryProcessor with a custom constructor that takes an object that contains all of the properties you want to update, then in process() you construct the final Employee object that will end up in the map and do an entry.setValue(). Don't forget to create a new StreamSerializer for the EmployeeUpdateEntryProcessor that can serialize Employee objects so that you don't get stuck with java.io serialization.

Source: http://docs.hazelcast.org/docs/3.5/manual/html/entryprocessor.html

回答2:

Probably a transaction is what you need. Or you may want to take a look at distributed lock.

Note that in your solution if this code is ran by two threads changes made by one of them will be overwriten.

回答3:

This may interest you.

You could do something like this for your Employee class (simplified code with one instance variable only):

public final class Employee
    implements Frozen<Builder>
{
    private final int salary;

    private Employee(Builder builder)
    {
        salary = builder.salary;
    }

    public static Builder newBuilder()
    {
        return new Builder();
    }

    @Override
    public Builder thaw()
    {
        return new Builder(this);
    }

    public static final class Builder
        implements Thawed<Employee>
    {
        private int salary;

        private Builder()
        {
        }

        private Builder(Employee employee)
        {
            salary = employee.salary;
        }

        public Builder withSalary(int salary)
        {
            this.salary = salary;
            return this;
        }

        @Override
        public Employee freeze()
        {
            return new Employee(this);
        }
    }
}

This way, to modify your cache, you would:

Employee victim = map.get(100);
map.put(100, victim.thaw().withSalary(whatever).freeze());

This is a completely atomic operation.

回答4:

If there is possibility that another node can update data that your node is working with then using put() will overwrite changes made by another node. Usually it is unwanted behavior, cause it leads to data loss and inconsistent data state.

Take a look at IMap.replace() method and other ConcurrentMap related methods. If replace() is failed then you've faced changes collision. In this case you should give it another attempt:

re-read entry from hazelcast
update it's fields
save to hazelcast with replace

After some failed attempts you can throw StorageException to the upper level.

来源：https://stackoverflow.com/questions/17182831/hazelcast-map-synchronization

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java

hazelcast