Inaccessible variable declared in synchronized block - Java

Question

This piece of code will not compile:

    synchronized( obj ) {
        Object a = new Object()
    }

    System.out.println( a.toString() );

Yet I don't know why.. My understanding was that a synchronized block was always eventually executed, so I would expect code following the synchronized block to be aware of any new declared variables. Where am I wrong?

score 6 · Accepted Answer · answered Feb 26 '12 at 01:32

6

It's not the synchronization, it's the {} symbols. They define a scope, no matter whether there's an if, for, synchronized, or even nothing at the beginning of them. So the a goes out of scope once the block finishes, because it was declared within it. (Also there's a missing semicolon at the end of the Object a declaration but I suspect you just forgot to copy that.)

answered Feb 26 '12 at 01:32

sparc_spread

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Thanks! Staring blindly at the synchronized made me completely overlook that fact. – Antiz Feb 26 '12 at 02:11

score 1 · Answer 2 · answered Feb 26 '12 at 01:33

Variable scope is not determined by what gets executed. Variables are visible only inside the block they are declared in. You probably want to do something like this:

Object a;
synchronized( obj ) {
    a = new Object()
}

System.out.println( a.toString() );

score 0 · Answer 3 · answered Feb 26 '12 at 01:35

Variables defined inside a block don't live outside the block.

From the Java Language Specification:

The scope of a local variable declaration in a block (§14.2) is the rest of the block in which the declaration appears, starting with its own initializer (§14.4) and including any further declarators to the right in the local variable declaration statement.

Inaccessible variable declared in synchronized block - Java

3 Answers3