How can I concatenate two arrays in Java?

Question

I need to concatenate two String arrays in Java.

void f(String[] first, String[] second) {
    String[] both = ???
}

What is the easiest way to do this?

Answer 1

I found a one-line solution from the good old Apache Commons Lang library.
ArrayUtils.addAll(T[], T...)

Code:

String[] both = ArrayUtils.addAll(first, second);

Answer 2

Here's a simple method that will concatenate two arrays and return the result:

public <T> T[] concatenate(T[] a, T[] b) {
    int aLen = a.length;
    int bLen = b.length;

    @SuppressWarnings("unchecked")
    T[] c = (T[]) Array.newInstance(a.getClass().getComponentType(), aLen + bLen);
    System.arraycopy(a, 0, c, 0, aLen);
    System.arraycopy(b, 0, c, aLen, bLen);

    return c;
}

Note that it will not work with primitive data types, only with object types.

The following slightly more complicated version works with both object and primitive arrays. It does this by using T instead of T[] as the argument type.

It also makes it possible to concatenate arrays of two different types by picking the most general type as the component type of the result.

public static <T> T concatenate(T a, T b) {
    if (!a.getClass().isArray() || !b.getClass().isArray()) {
        throw new IllegalArgumentException();
    }

    Class<?> resCompType;
    Class<?> aCompType = a.getClass().getComponentType();
    Class<?> bCompType = b.getClass().getComponentType();

    if (aCompType.isAssignableFrom(bCompType)) {
        resCompType = aCompType;
    } else if (bCompType.isAssignableFrom(aCompType)) {
        resCompType = bCompType;
    } else {
        throw new IllegalArgumentException();
    }

    int aLen = Array.getLength(a);
    int bLen = Array.getLength(b);

    @SuppressWarnings("unchecked")
    T result = (T) Array.newInstance(resCompType, aLen + bLen);
    System.arraycopy(a, 0, result, 0, aLen);
    System.arraycopy(b, 0, result, aLen, bLen);        

    return result;
}

Here is an example:

Assert.assertArrayEquals(new int[] { 1, 2, 3 }, concatenate(new int[] { 1, 2 }, new int[] { 3 }));
Assert.assertArrayEquals(new Number[] { 1, 2, 3f }, concatenate(new Integer[] { 1, 2 }, new Number[] { 3f }));

Answer 3

Using Stream in Java 8:

String[] both = Stream.concat(Arrays.stream(a), Arrays.stream(b))
                      .toArray(String[]::new);

Or like this, using flatMap:

String[] both = Stream.of(a, b).flatMap(Stream::of)
                      .toArray(String[]::new);

To do this for a generic type you have to use reflection:

@SuppressWarnings("unchecked")
T[] both = Stream.concat(Arrays.stream(a), Arrays.stream(b)).toArray(
    size -> (T[]) Array.newInstance(a.getClass().getComponentType(), size));

Answer 4

It's possible to write a fully generic version that can even be extended to concatenate any number of arrays. This versions require Java 6, as they use Arrays.copyOf()

Both versions avoid creating any intermediary List objects and use System.arraycopy() to ensure that copying large arrays is as fast as possible.

For two arrays it looks like this:

public static <T> T[] concat(T[] first, T[] second) {
  T[] result = Arrays.copyOf(first, first.length + second.length);
  System.arraycopy(second, 0, result, first.length, second.length);
  return result;
}

And for a arbitrary number of arrays (>= 1) it looks like this:

public static <T> T[] concatAll(T[] first, T[]... rest) {
  int totalLength = first.length;
  for (T[] array : rest) {
    totalLength += array.length;
  }
  T[] result = Arrays.copyOf(first, totalLength);
  int offset = first.length;
  for (T[] array : rest) {
    System.arraycopy(array, 0, result, offset, array.length);
    offset += array.length;
  }
  return result;
}

Answer 5

Or with the beloved Guava:

String[] both = ObjectArrays.concat(first, second, String.class);

Also, there are versions for primitive arrays:

• Booleans.concat(first, second)
• Bytes.concat(first, second)
• Chars.concat(first, second)
• Doubles.concat(first, second)
• Shorts.concat(first, second)
• Ints.concat(first, second)
• Longs.concat(first, second)
• Floats.concat(first, second)

Answer 6

You can append the two arrays in two lines of code.

String[] both = Arrays.copyOf(first, first.length + second.length);
System.arraycopy(second, 0, both, first.length, second.length);

This is a fast and efficient solution and will work for primitive types as well as the two methods involved are overloaded.

You should avoid solutions involving ArrayLists, streams, etc as these will need to allocate temporary memory for no useful purpose.

You should avoid for loops for large arrays as these are not efficient. The built in methods use block-copy functions that are extremely fast.

Answer 7

Using the Java API:

String[] f(String[] first, String[] second) {
    List<String> both = new ArrayList<String>(first.length + second.length);
    Collections.addAll(both, first);
    Collections.addAll(both, second);
    return both.toArray(new String[both.size()]);
}

Answer 8

A solution 100% old java and without System.arraycopy (not available in GWT client for example):

static String[] concat(String[]... arrays) {
    int length = 0;
    for (String[] array : arrays) {
        length += array.length;
    }
    String[] result = new String[length];
    int pos = 0;
    for (String[] array : arrays) {
        for (String element : array) {
            result[pos] = element;
            pos++;
        }
    }
    return result;
}

Answer 9

I've recently fought problems with excessive memory rotation. If a and/or b are known to be commonly empty, here is another adaption of silvertab's code (generified too):

private static <T> T[] concatOrReturnSame(T[] a, T[] b) {
    final int alen = a.length;
    final int blen = b.length;
    if (alen == 0) {
        return b;
    }
    if (blen == 0) {
        return a;
    }
    final T[] result = (T[]) java.lang.reflect.Array.
            newInstance(a.getClass().getComponentType(), alen + blen);
    System.arraycopy(a, 0, result, 0, alen);
    System.arraycopy(b, 0, result, alen, blen);
    return result;
}

Edit: A previous version of this post stated that array re-usage like this shall be clearly documented. As Maarten points out in the comments it would in general be better to just remove the if statements, thus voiding the need for having documentation. But then again, those if statements were the whole point of this particular optimization in the first place. I'll leave this answer here, but be wary!

Answer 10

The Functional Java library has an array wrapper class that equips arrays with handy methods like concatenation.

import static fj.data.Array.array;

...and then

Array<String> both = array(first).append(array(second));

To get the unwrapped array back out, call

String[] s = both.array();

Answer 11

ArrayList<String> both = new ArrayList(Arrays.asList(first));
both.addAll(Arrays.asList(second));

both.toArray(new String[0]);

Answer 12

Another way with Java8 using Stream

  public String[] concatString(String[] a, String[] b){ 
    Stream<String> streamA = Arrays.stream(a);
    Stream<String> streamB = Arrays.stream(b);
    return Stream.concat(streamA, streamB).toArray(String[]::new); 
  }

Answer 13

Here's an adaptation of silvertab's solution, with generics retrofitted:

static <T> T[] concat(T[] a, T[] b) {
    final int alen = a.length;
    final int blen = b.length;
    final T[] result = (T[]) java.lang.reflect.Array.
            newInstance(a.getClass().getComponentType(), alen + blen);
    System.arraycopy(a, 0, result, 0, alen);
    System.arraycopy(b, 0, result, alen, blen);
    return result;
}

NOTE: See Joachim's answer for a Java 6 solution. Not only does it eliminate the warning; it's also shorter, more efficient and easier to read!

Answer 14

You could try converting it into a Arraylist and use the addAll method then convert back to an array.

List list = new ArrayList(Arrays.asList(first));
  list.addAll(Arrays.asList(second));
  String[] both = list.toArray();

Answer 15

If you use this way so you no need to import any third party class.

If you want concatenate String

Sample code for concate two String Array

public static String[] combineString(String[] first, String[] second){
        int length = first.length + second.length;
        String[] result = new String[length];
        System.arraycopy(first, 0, result, 0, first.length);
        System.arraycopy(second, 0, result, first.length, second.length);
        return result;
    }

If you want concatenate Int

Sample code for concate two Integer Array

public static int[] combineInt(int[] a, int[] b){
        int length = a.length + b.length;
        int[] result = new int[length];
        System.arraycopy(a, 0, result, 0, a.length);
        System.arraycopy(b, 0, result, a.length, b.length);
        return result;
    }

Here is Main method

    public static void main(String[] args) {

            String [] first = {"a", "b", "c"};
            String [] second = {"d", "e"};

            String [] joined = combineString(first, second);
            System.out.println("concatenated String array : " + Arrays.toString(joined));

            int[] array1 = {101,102,103,104};
            int[] array2 = {105,106,107,108};
            int[] concatenateInt = combineInt(array1, array2);

            System.out.println("concatenated Int array : " + Arrays.toString(concatenateInt));

        }
    }  

We can use this way also.

Answer 16

Please forgive me for adding yet another version to this already long list. I looked at every answer and decided that I really wanted a version with just one parameter in the signature. I also added some argument checking to benefit from early failure with sensible info in case of unexpected input.

@SuppressWarnings("unchecked")
public static <T> T[] concat(T[]... inputArrays) {
  if(inputArrays.length < 2) {
    throw new IllegalArgumentException("inputArrays must contain at least 2 arrays");
  }

  for(int i = 0; i < inputArrays.length; i++) {
    if(inputArrays[i] == null) {
      throw new IllegalArgumentException("inputArrays[" + i + "] is null");
    }
  }

  int totalLength = 0;

  for(T[] array : inputArrays) {
    totalLength += array.length;
  }

  T[] result = (T[]) Array.newInstance(inputArrays[0].getClass().getComponentType(), totalLength);

  int offset = 0;

  for(T[] array : inputArrays) {
    System.arraycopy(array, 0, result, offset, array.length);

    offset += array.length;
  }

  return result;
}

Answer 17

Using Java 8+ streams you can write the following function:

private static String[] concatArrays(final String[]... arrays) {
    return Arrays.stream(arrays)
         .flatMap(Arrays::stream)
         .toArray(String[]::new);
}

Answer 18

Here a possible implementation in working code of the pseudo code solution written by silvertab.

Thanks silvertab!

public class Array {

   public static <T> T[] concat(T[] a, T[] b, ArrayBuilderI<T> builder) {
      T[] c = builder.build(a.length + b.length);
      System.arraycopy(a, 0, c, 0, a.length);
      System.arraycopy(b, 0, c, a.length, b.length);
      return c;
   }
}

Following next is the builder interface.

Note: A builder is necessary because in java it is not possible to do

new T[size]

due to generic type erasure:

public interface ArrayBuilderI<T> {

   public T[] build(int size);
}

Here a concrete builder implementing the interface, building a Integer array:

public class IntegerArrayBuilder implements ArrayBuilderI<Integer> {

   @Override
   public Integer[] build(int size) {
      return new Integer[size];
   }
}

And finally the application / test:

@Test
public class ArrayTest {

   public void array_concatenation() {
      Integer a[] = new Integer[]{0,1};
      Integer b[] = new Integer[]{2,3};
      Integer c[] = Array.concat(a, b, new IntegerArrayBuilder());
      assertEquals(4, c.length);
      assertEquals(0, (int)c[0]);
      assertEquals(1, (int)c[1]);
      assertEquals(2, (int)c[2]);
      assertEquals(3, (int)c[3]);
   }
}

Answer 19

This should be one-liner.

public String [] concatenate (final String array1[], final String array2[])
{
    return Stream.concat(Stream.of(array1), Stream.of(array2)).toArray(String[]::new);
}

Answer 20

Wow! lot of complex answers here including some simple ones that depend on external dependencies. how about doing it like this:

String [] arg1 = new String{"a","b","c"};
String [] arg2 = new String{"x","y","z"};

ArrayList<String> temp = new ArrayList<String>();
temp.addAll(Arrays.asList(arg1));
temp.addAll(Arrays.asList(arg2));
String [] concatedArgs = temp.toArray(new String[arg1.length+arg2.length]);

Answer 21

This works, but you need to insert your own error checking.

public class StringConcatenate {

    public static void main(String[] args){

        // Create two arrays to concatenate and one array to hold both
        String[] arr1 = new String[]{"s","t","r","i","n","g"};
        String[] arr2 = new String[]{"s","t","r","i","n","g"};
        String[] arrBoth = new String[arr1.length+arr2.length];

        // Copy elements from first array into first part of new array
        for(int i = 0; i < arr1.length; i++){
            arrBoth[i] = arr1[i];
        }

        // Copy elements from second array into last part of new array
        for(int j = arr1.length;j < arrBoth.length;j++){
            arrBoth[j] = arr2[j-arr1.length];
        }

        // Print result
        for(int k = 0; k < arrBoth.length; k++){
            System.out.print(arrBoth[k]);
        }

        // Additional line to make your terminal look better at completion!
        System.out.println();
    }
}

It's probably not the most efficient, but it doesn't rely on anything other than Java's own API.

Answer 22

This is a converted function for a String array:

public String[] mergeArrays(String[] mainArray, String[] addArray) {
    String[] finalArray = new String[mainArray.length + addArray.length];
    System.arraycopy(mainArray, 0, finalArray, 0, mainArray.length);
    System.arraycopy(addArray, 0, finalArray, mainArray.length, addArray.length);

    return finalArray;
}

Answer 23

How about simply

public static class Array {

    public static <T> T[] concat(T[]... arrays) {
        ArrayList<T> al = new ArrayList<T>();
        for (T[] one : arrays)
            Collections.addAll(al, one);
        return (T[]) al.toArray(arrays[0].clone());
    }
}

And just do Array.concat(arr1, arr2). As long as arr1 and arr2 are of the same type, this will give you another array of the same type containing both arrays.

Answer 24

A simple variation allowing the joining of more than one array:

public static String[] join(String[]...arrays) {

    final List<String> output = new ArrayList<String>();

    for(String[] array : arrays) {
        output.addAll(Arrays.asList(array));
    }

    return output.toArray(new String[output.size()]);
}

Answer 25

A generic static version that uses the high performing System.arraycopy without requiring a @SuppressWarnings annotation:

public static <T> T[] arrayConcat(T[] a, T[] b) {
    T[] both = Arrays.copyOf(a, a.length + b.length);
    System.arraycopy(b, 0, both, a.length, b.length);
    return both;
}

Answer 26

Using only Javas own API:

String[] join(String[]... arrays) {
  // calculate size of target array
  int size = 0;
  for (String[] array : arrays) {
    size += array.length;
  }

  // create list of appropriate size
  java.util.List list = new java.util.ArrayList(size);

  // add arrays
  for (String[] array : arrays) {
    list.addAll(java.util.Arrays.asList(array));
  }

  // create and return final array
  return list.toArray(new String[size]);
}

Now, this code ist not the most efficient, but it relies only on standard java classes and is easy to understand. It works for any number of String[] (even zero arrays).

Answer 27

public String[] concat(String[]... arrays)
{
    int length = 0;
    for (String[] array : arrays) {
        length += array.length;
    }
    String[] result = new String[length];
    int destPos = 0;
    for (String[] array : arrays) {
        System.arraycopy(array, 0, result, destPos, array.length);
        destPos += array.length;
    }
    return result;
}

Answer 28

Here's my slightly improved version of Joachim Sauer's concatAll. It can work on Java 5 or 6, using Java 6's System.arraycopy if it's available at runtime. This method (IMHO) is perfect for Android, as it work on Android <9 (which doesn't have System.arraycopy) but will use the faster method if possible.

  public static <T> T[] concatAll(T[] first, T[]... rest) {
    int totalLength = first.length;
    for (T[] array : rest) {
      totalLength += array.length;
    }
    T[] result;
    try {
      Method arraysCopyOf = Arrays.class.getMethod("copyOf", Object[].class, int.class);
      result = (T[]) arraysCopyOf.invoke(null, first, totalLength);
    } catch (Exception e){
      //Java 6 / Android >= 9 way didn't work, so use the "traditional" approach
      result = (T[]) java.lang.reflect.Array.newInstance(first.getClass().getComponentType(), totalLength);
      System.arraycopy(first, 0, result, 0, first.length);
    }
    int offset = first.length;
    for (T[] array : rest) {
      System.arraycopy(array, 0, result, offset, array.length);
      offset += array.length;
    }
    return result;
  }

Answer 29

Another way to think about the question. To concatenate two or more arrays, one have to do is to list all elements of each arrays, and then build a new array. This sounds like create a List<T> and then calls toArray on it. Some other answers uses ArrayList, and that's fine. But how about implement our own? It is not hard:

private static <T> T[] addAll(final T[] f, final T...o){
    return new AbstractList<T>(){

        @Override
        public T get(int i) {
            return i>=f.length ? o[i - f.length] : f[i];
        }

        @Override
        public int size() {
            return f.length + o.length;
        }

    }.toArray(f);
}

I believe the above is equivalent to solutions that uses System.arraycopy. However I think this one has its own beauty.

Answer 30

How about :

public String[] combineArray (String[] ... strings) {
    List<String> tmpList = new ArrayList<String>();
    for (int i = 0; i < strings.length; i++)
        tmpList.addAll(Arrays.asList(strings[i]));
    return tmpList.toArray(new String[tmpList.size()]);
}