Time complexity of System.arraycopy(…)?

Question

System.arraycopy(Object src, int srcPos, Object dest, int destPos, int length) is a native method.

What is the time complexity for this method?

Answer 1

Here's some relevant source code from OpenJDK 8 (openjdk-8-src-b132-03_mar_2014). I found it with help from Java native method source code (note: instructions there are confusing; I just searched the source for relevant identifiers). I think Captain Ford's comment is correct; i.e., there are (many) cases where iterating each element isn't necessary. Note that not iterating each element doesn't necessarily mean O(1), it just means "faster". I think that, regardless, an array copy must be fundamentally O(x), even if x isn't the number of items in the array; i.e., however you do it, copying gets more expensive with more elements in the array, and if you have a Very Big array, it will take a linearly Long Time. Caveat: I don't know for certain that this is the actual source code for the Java you are running; only that this is the only implementation I could find in the OpenJDK 8 source. I think that this is a cross-platform implementation, but I might be wrong -- I definitely haven't figured out how to build this code. See also: Differences between Oracle JDK and Open JDK. The following comes from: /openjdk/hotspot/src/share/vm/oops/objArrayKlass.cpp

// Either oop or narrowOop depending on UseCompressedOops.
template <class T> void ObjArrayKlass::do_copy(arrayOop s, T* src,
                               arrayOop d, T* dst, int length, TRAPS) {

  BarrierSet* bs = Universe::heap()->barrier_set();
  // For performance reasons, we assume we are that the write barrier we
  // are using has optimized modes for arrays of references.  At least one
  // of the asserts below will fail if this is not the case.
  assert(bs->has_write_ref_array_opt(), "Barrier set must have ref array opt");
  assert(bs->has_write_ref_array_pre_opt(), "For pre-barrier as well.");

  if (s == d) {
    // since source and destination are equal we do not need conversion checks.
    assert(length > 0, "sanity check");
    bs->write_ref_array_pre(dst, length);
    Copy::conjoint_oops_atomic(src, dst, length);
  } else {
    // We have to make sure all elements conform to the destination array
    Klass* bound = ObjArrayKlass::cast(d->klass())->element_klass();
    Klass* stype = ObjArrayKlass::cast(s->klass())->element_klass();
    if (stype == bound || stype->is_subtype_of(bound)) {
      // elements are guaranteed to be subtypes, so no check necessary
      bs->write_ref_array_pre(dst, length);
      Copy::conjoint_oops_atomic(src, dst, length);
    } else {
      // slow case: need individual subtype checks
      // note: don't use obj_at_put below because it includes a redundant store check
      T* from = src;
      T* end = from + length;
      for (T* p = dst; from < end; from++, p++) {
        // XXX this is going to be slow.
        T element = *from;
        // even slower now
        bool element_is_null = oopDesc::is_null(element);
        oop new_val = element_is_null ? oop(NULL)
                                      : oopDesc::decode_heap_oop_not_null(element);
        if (element_is_null ||
            (new_val->klass())->is_subtype_of(bound)) {
          bs->write_ref_field_pre(p, new_val);
          *p = *from;
        } else {
          // We must do a barrier to cover the partial copy.
          const size_t pd = pointer_delta(p, dst, (size_t)heapOopSize);
          // pointer delta is scaled to number of elements (length field in
          // objArrayOop) which we assume is 32 bit.
          assert(pd == (size_t)(int)pd, "length field overflow");
          bs->write_ref_array((HeapWord*)dst, pd);
          THROW(vmSymbols::java_lang_ArrayStoreException());
          return;
        }
      }
    }
  }
  bs->write_ref_array((HeapWord*)dst, length);
}

Answer 2

I did some investigation and later decided to write a test code, here is what I have.

My testing code is given below:

import org.junit.Test;

public class ArrayCopyTest {

  @Test
  public void testCopy() {
    for (int count = 0; count < 3; count++) {
      int size = 0x00ffffff;
      long start, end;
      Integer[] integers = new Integer[size];
      Integer[] loopCopy = new Integer[size];
      Integer[] systemCopy = new Integer[size];

      for (int i = 0; i < size; i++) {
        integers[i] = i;
      }

      start = System.currentTimeMillis();
      for (int i = 0; i < size; i++) {
        loopCopy[i] = integers[i];
      }
      end = System.currentTimeMillis();
      System.out.println("for loop: " + (end - start));

      start = System.currentTimeMillis();
      System.arraycopy(integers, 0, systemCopy, 0, size);
      end = System.currentTimeMillis();
      System.out.println("System.arrayCopy: " + (end - start));
    }
  }

}

It produces result shown below

for loop: 47
System.arrayCopy: 24

for loop: 31
System.arrayCopy: 22

for loop: 36
System.arrayCopy: 22

So, Bragboy is correct.

Answer 3

I don't understand how Kowser's answer answers his own question. I thought to check the time complexity of an algorithm You have to compare its running time for inputs of different sizes, like this:

import org.junit.Test;

public class ArrayCopyTest {

  @Test
  public void testCopy() {
    int size = 5000000;
    for (int count = 0; count < 5; count++) {
      size = size * 2;
      long start, end;
      Integer[] integers = new Integer[size];
      Integer[] systemCopy = new Integer[size];

      start = System.currentTimeMillis();
      System.arraycopy(integers, 0, systemCopy, 0, size);
      end = System.currentTimeMillis();
      System.out.println(end - start);
    }
  }

}

Output:

10
22
42
87
147

Answer 4

It will have to go through all the elements in the array to do this. Array is a unique data structure where you have to specify a size when you initialize it. Order would be the source array's size or in Big O terms its O(length).

Infact this happens internally in an ArrayList. ArrayList wraps an array. Although ArrayList looks like a dynamically growing collection, internally it does an arrycopy when it has to expand.

Answer 5

Just to sum up relevant comments on another question (flagged as a duplicate of this one).

Surely, it's just adding to the new array with all the entries of the other? Which would be O(n) where n is the number of values to be added.

bragboy's answer agrees of course, but then I had thought that the only way to get a sure answer was to find the source code to get a canonical answer, but that isn't possible. Here is the declaration for System.arraycopy();

public static native void arraycopy(Object src, int src_position,  
                                    Object dst, int dst_position,  
                                    int length);

It's native, written in the language of the operating system, which means that the implementation of arraycopy() is platform dependant.

So, in conclusion it's likely O(n), but maybe not.