CODEWITHSUNDEEP
javaatomic
JamesJenkins
JamesJenkins

Reputation: 141

Atomicity of Reads and Writes to Variables

I've just read a question here, and read the most rated answer by @JB Nizet, and I got confused... According to the answer, in the following code,

private int a=0;

public void foo(){
  int temp=35;
  a=28;
  a=temp;
}

a=28; is an atomic operation.

In some other questions and answers that I've read in Stackoverflow, the information was different, saying that a=28; is not an atomic operation, because first a read operation of the right operand should take place, then the write operation takes place, and each of these 2 operations is atomic, but the entire assignment is not (To be honest, this is how I thought it works).

And what about a=temp; ? Is it any different than a=28; in terms of atomicity?

By the way, I know about the need of volatile for double and long to make read/write to them atomic, just confused about what I wrote above.

Can someone plz elaborate on this?

Thanks..

Upvotes: 1

Views: 199

Answers (1)

Malt
Malt

Reputation: 30335

According to the official documentation:

Reads and writes are atomic for reference variables and for most primitive variables (all types except long and double).

Since a=28; is a write into a primitive which isn't long or double, it's atomic.

However a=temp isn't atomic since it consists of two separate operations - the read from temp and the write to int. Each of these is atomic, but not their composition.

Upvotes: 6

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