Which is more efficient, a for-each loop, or an iterator?

Question

Which is the most efficient way to traverse a collection?

List  a = new ArrayList();
for (Integer integer : a) {
  integer.toSt         


        

Answer 1

If you are just wandering over the collection to read all of the values, then there is no difference between using an iterator or the new for loop syntax, as the new syntax just uses the iterator underwater.

If however, you mean by loop the old "c-style" loop:

for(int i=0; i

Then the new for loop, or iterator, can be a lot more efficient, depending on the underlying data structure. The reason for this is that for some data structures, get(i) is an O(n) operation, which makes the loop an O(n²) operation. A traditional linked list is an example of such a data structure. All iterators have as a fundamental requirement that next() should be an O(1) operation, making the loop O(n).

To verify that the iterator is used underwater by the new for loop syntax, compare the generated bytecodes from the following two Java snippets. First the for loop:

List  a = new ArrayList();
for (Integer integer : a)
{
  integer.toString();
}
// Byte code
 ALOAD 1
 INVOKEINTERFACE java/util/List.iterator()Ljava/util/Iterator;
 ASTORE 3
 GOTO L2
L3
 ALOAD 3
 INVOKEINTERFACE java/util/Iterator.next()Ljava/lang/Object;
 CHECKCAST java/lang/Integer
 ASTORE 2 
 ALOAD 2
 INVOKEVIRTUAL java/lang/Integer.toString()Ljava/lang/String;
 POP
L2
 ALOAD 3
 INVOKEINTERFACE java/util/Iterator.hasNext()Z
 IFNE L3

And second, the iterator:

List  a = new ArrayList();
for (Iterator iterator = a.iterator(); iterator.hasNext();)
{
  Integer integer = (Integer) iterator.next();
  integer.toString();
}
// Bytecode:
 ALOAD 1
 INVOKEINTERFACE java/util/List.iterator()Ljava/util/Iterator;
 ASTORE 2
 GOTO L7
L8
 ALOAD 2
 INVOKEINTERFACE java/util/Iterator.next()Ljava/lang/Object;
 CHECKCAST java/lang/Integer
 ASTORE 3
 ALOAD 3
 INVOKEVIRTUAL java/lang/Integer.toString()Ljava/lang/String;
 POP
L7
 ALOAD 2
 INVOKEINTERFACE java/util/Iterator.hasNext()Z
 IFNE L8

As you can see, the generated byte code is effectively identical, so there is no performance penalty to using either form. Therefore, you should choose the form of loop that is most aesthetically appealing to you, for most people that will be the for-each loop, as that has less boilerplate code.