Replacing all items in a multidimensional array with random values

Question

for a single dimensional array normally I would use a for loop like the one below but I can't think of a way to do this that doesn't involve lots of loops.

for (int i = 0; i < myArray.length; ++i)
{
    myArray[i] = rnd.Next(1, 500);
}

Are you saying you don't know how many dimensions the array has? — juharr, Dec 07 '16 at 12:59
Try this: http://stackoverflow.com/questions/359881/how-to-set-values-in-a-multidimensional-array-using-a-linear-index — Robin Mackenzie, Dec 07 '16 at 13:03

Dmitry Bychenko · Accepted Answer · 2016-12-07T13:25:13.507

5

You can try low level Buffer.BlockCopy to conceal the loop(s):

  // N-D array (whatever dimensions)
  int[,,] array = new int[3, 5, 11]; 

  Buffer.BlockCopy(
    Enumerable
      .Range(0, array.Length)
      .Select(x => rand.Next(0, 500))
      .ToArray(),
    0,
    array,
    0,
    array.Length * sizeof(int)); // sizeof(int) : we copy bytes...

we create 1-D array

    Enumerable
      .Range(0, array.Length)
      .Select(x => rand.Next(0, 500))
      .ToArray()

with the total length of N-D one (array.Length == array.Length(0) * ... * array.GetLEngth(N)) and copy it into N-D one.

edited Dec 07 '16 at 13:25

answered Dec 07 '16 at 13:02

Dmitry Bychenko

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1

This won't work because it copies the result of the `Enumerable` and doesn't manipulate the source array. You basically manipulate a temporary array and do nothing with it. – Tamir Vered Dec 07 '16 at 13:08
1

@Tamir Vered: Thank you! You're quite right: I put the routine prematurely: wrong argument order (source and target swapped) and wrong *bytes* size (should have been multiplied by `sizeof(int)`) – Dmitry Bychenko Dec 07 '16 at 13:19

score 0 · Answer 2 · answered Dec 07 '16 at 12:58

0

Use more than one loop:

for (int i = 0; i < myArray.Length; ++i)
{
    for (int j = 0; j < myArray[i].Length; ++j)
    {
        myArray[i][j] = rand.Next(0, 500);
    }
}

answered Dec 07 '16 at 12:58

PC Luddite

5,883
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++i vs i++ shouldn't be? – asdf_enel_hak Dec 07 '16 at 12:59
3

He says "I can't think of a way to do this that doesn't involve lots of loops.". – Tamir Vered Dec 07 '16 at 12:59
@asdf_enel_hak In this usage it doesn't matter because the return value of the increment is not being used. – juharr Dec 07 '16 at 13:01
@asdf_enel_hak http://stackoverflow.com/questions/484462/difference-between-i-and-i-in-a-loop – Tamir Vered Dec 07 '16 at 13:01
@asdf_enel_hak I got into the habit of using `++i` instead of `i++` after learning C++ where you should use `++it` on iterators. – PC Luddite Dec 07 '16 at 13:02
1

That's not a multidimensional but a jagged array (although the global principle still hold) – Sehnsucht Dec 07 '16 at 13:02
@Sehnsucht Actually it makes the question quite different since the memory is allocated inconsecutively so it isn't possible to use `Buffer.BlockCopy` for example, nor other "look on this array as a single dimension block of memory" solutions. – Tamir Vered Dec 07 '16 at 13:41
@TamirVered I was talking about the principle of that answer not the question – Sehnsucht Dec 07 '16 at 13:45

Replacing all items in a multidimensional array with random values

2 Answers2