Does a lambda expression create an object on the h

Question

When I iterate over a collection using the new syntactic sugar of Java 8, such as

myStream.forEach(item -> {
  // do something useful
});

Isn't this equivalent to the 'old syntax' snippet below?

myStream.forEach(new Consumer<Item>() {
  @Override
  public void accept(Item item) {
    // do something useful
  }
});

Does this mean a new anonymous Consumer object is created on the heap every time I iterate over a collection? How much heap space does this take? What performance implications does it have? Does it mean I should rather use the old style for loops when iterating over large multi-level data structures?

Answer 1

When an instance representing the lambda is created sensitively depends on the exact contents of your lambda's body. Namely, the key factor is what the lambda captures from the lexical environment. If it doesn't capture any state which is variable from creation to creation, then an instance will not be created each time the for-each loop is entered. Instead a synthetic method will be generated at compile time and the lambda use site will just receive a singleton object that delegates to that method.

Further note that this aspect is implementation-dependent and you can expect future refinements and advancements on HotSpot towards greater efficiency. There are general plans to e.g. make a lightweight object without a full corresponding class, which has just enough information to forward to a single method.

Here is a good, accessible in-depth article on the topic:

http://www.infoq.com/articles/Java-8-Lambdas-A-Peek-Under-the-Hood

Answer 2

You are passing a new instance to the forEach method. Every time you do that you create a new object but not one for every loop iteration. Iteration is done inside forEach method using the same 'callback' object instance until it is done with the loop.

So the memory used by the loop does not depend on the size of the collection.

Isn't this equivalent to the 'old syntax' snippet?

Yes. It has slight differences at a very low level but I don't think you should care about them. Lamba expressions use the invokedynamic feature instead of anonymous classes.

Answer 3

It is equivalent but not identical. Simply said, if a lambda expression does not captures values, it will be a singleton that is re-used on every invocation.

The behavior is not exactly specified. The JVM is given a big freedom how to implement it. Currently, Oracle’s JVM creates (at least) one instance per lambda expression (i.e. doesn’t share instance between different identical expressions) but creates singleton for all expression which don’t capture values.

You may read this answer for more details. There, I not only gave a more detailed description but also testing code to observe the current behavior.

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This is covered by the The Java® Language Specification, chapter “15.27.4. Run-time Evaluation of Lambda Expressions”

Summarized:

These rules are meant to offer flexibility to implementations of the Java programming language, in that:

• A new object need not be allocated on every evaluation.

• Objects produced by different lambda expressions need not belong to different classes (if the bodies are identical, for example).

• Every object produced by evaluation need not belong to the same class (captured local variables might be inlined, for example).

• If an "existing instance" is available, it need not have been created at a previous lambda evaluation (it might have been allocated during the enclosing class's initialization, for example).