Does checking against Boolean.TRUE/Boolean.FALSE avoid boxing/unboxing?

Question

Let's say I have a Map<Integer, Boolean> and I want to filter out all the integers that have a boolean-value of true. Here's some code:

for (Map.Entry<Integer, Boolean> e : map.entrySet()){
    if (e.getValue() == true){ // Unboxing here
        // Do something useful...
    }
}

In this code, the Boolean-object is unboxed every time the if is executed (which might be a problem if the map is really large).

Now, the Boolean-class offers constants (Boolean.TRUE and Boolean.FALSE), which represent the normal boolean values true and false as an already boxed version (right?). So using this code would avoid unboxing completely:

for (Map.Entry<Integer, Boolean> e : map.entrySet()){
    if (e.getValue() == Boolean.TRUE){ // No Unboxing done (?)
        // Do something useful...
    }
}

Am I right about this assumption? Or is it even worse and both values are unboxed to be compared?

Answer 1

It will work if you used the primitive boolean literals to add elements to your map.

When using primitive literals where a reference type is expected, the compiler will use Boolean#valueOf(..) implemented as

public static Boolean valueOf(boolean b) {
    return (b ? TRUE : FALSE);
}

to return a value of type Boolean. So you will get the same two Boolean objects every time.

If you instead added to the map with

new Boolean(true) // or new Boolean(false)

it will be a different reference, so comparison with == will fail.

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This does seem like premature optimization.