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Home/ Questions/Q 6126711
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T16:24:36+00:00 2026-05-23T16:24:36+00:00

I know that an initialization block runs after the call to ‘super()’ in a

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I know that an initialization block runs after the call to ‘super()’ in a constructor. However, when looking through some code this morning, I found the following:

public class SimpleListLocalsAnalysis extends BackwardFlowAnalysis
    FlowSet emptySet;

    public SimpleLiveLocalsAnalysis(UnitGraph graph) {
        super(graph);

        {
            Chain locals = g.getBody().getLocals();
            FlowUniverse localUniverse = new FlowUniverse(locals.toArray());
            emptySet = new ArrayPackedSet(localUniverse);
        }

        doAnalysis();
    }
...
}

The above code shows some initialisation going on within an initialisation block just after ‘super(graph)’ invocation. What’s the purpose of placing the code in an initialisation block within a constructor, as surely it runs anyway after the call to super. Am I missing something here?

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  1. Editorial Team
    Editorial Team
    2026-05-23T16:24:37+00:00Added an answer on May 23, 2026 at 4:24 pm

    Its not initilization block, its simple block

    just like

    public void foo(){
    
      {
          //some code
      }
    }
    

    Purpose:

    You can have restricted scope

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