Question

Josh Bloch mentions using a local variable in double check idiom for possible performance improvement (EJ, 2nd ed. p284) and says the example code runs about 25% faster on his machine compared to that without a local variable. So the question is, why is it faster? In case of local variable version, it accesses the instance variable 3 times and in normal case it accesses the instance variable 4 times. Is this the reason behind ‘25%’ increase or are there any other reasons?

EDIT: 3 or 4 times access is only when the instance is first created. Thereafter, it’s 1 or 2 times.

EDIT2: check the accepted answer for this question to see the example code. I think this is for Java 6. How to solve the "Double-Checked Locking is Broken" Declaration in Java?

Answer 1

The basic thing is that accessing the volatile variable is slower, than accessing the local one. When you declare a local variable, you are basically caching the value of a volatile variable inside the method.

In the normal (without a local variable) case you are accessing the volatile variable:

1. In the first if before the synchronized clause
2. In the second if inside the synchronized clause
3. Inside the second if where you assign a value to it
4. In the return statement

Now, if you introduce a local variable, you only access the volatile variable three times:

1. When assigning the local variable for the first if before the synchronized clause
2. When assigning the local variable for the second if inside the synchronized clause
3. Inside the second if where you assign a value to it

You do not access it in the return statement, by returning the local variable, thus giving you a speed boost.