Different output for array.clone() for 1- and multidimensional arrays

Question

Case 1, one-dimensional:

double[]vals = new double[4]; vals[0] = 1; vals[1] = 123; vals[2] = -1; vals[3] = 2; double[] vals4 = vals.clone(); double newval = 55; vals[0] += newval; for (int k=0;k<vals.length;k++){ System.out.println("k= " + vals[k]); System.out.println("k= " + vals4[k]); }

The output is

k= 56.0 k= 1.0 k= 123.0 k= 123.0 k= -1.0 k= -1.0 k= 2.0 k= 2.0

Case 2, 2-dimensional:

double[][] vals3 = new double[][]{ {1,2}, {3,4} }; double[][] vals2 = vals3.clone(); vals3[0][0] += 112; for (int i=0;i<vals3.length;i++){ for (int j=0;j<vals3[0].length;j++){ System.out.println(vals2[i][j]); System.out.println(vals3[i][j]); } }

The output is

113.0 113.0 2.0 2.0 3.0 3.0 4.0 4.0

I use the same function, array.clone(), but why do the array elements change in the 2-dim case, when I change the array's inputs, but don't in the 1-dim case?

Peter Lawrey · Accepted Answer · 2014-11-16 19:28:29Z

4

Java only has single dimensional arrays. A double[][] is an array of references to double[] so when you clone an array of references, you get another array of references to the same underlying arrays.

To do a deep clone, you can to create a new double[][] of the same size and clone the individual double[]s

e.g.

 public static double[][] deepClone(double[][] d) { double[][] ret = new double[d.length][]; for(int i = 0; i < ret.length; i++) ret[i] = d[i].clone(); return ret; }

answered Nov 16, 2014 at 19:28

Peter Lawrey

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7 Comments

Alex Over a year ago

would a System.arraycopy() be a better option for 2-dim case?

Peter Lawrey Over a year ago

@Alex Using System.arraycopy doesn't alter the fact that you would be copying an array of references to arrays.

Alex Over a year ago

OK then, in the case when I change a value in a 2-dim array, how should I copy it to keep the values in the other one unchanged? With two for loops?

Peter Lawrey Over a year ago

@Alex effectively yes, but the simplest solution is to use clone() in the manner in my answer.

Alex Over a year ago

Thanks. Are they the same in terms of number of function evaluations/memory use/runtime?

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Maksym · Accepted Answer · 2014-11-16 19:28:08Z

Yep, it's because array.clone() copies references, in other words array clone just does "shallow" copy.

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