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Question by Fran-Martin · Apr 24, 2015 at 07:03 AM · array

Code error with dynamic arrays

Hello everyone. How I declare a bidimensional structure (matrix) in which each cells contains a undefined-size array of int?

     void CheckSubMeshesLocation()
     {
         _subMeshesLocation = new int[(int)_gridN,(int)_gridM];
         for(int i=0;i<_gridN;i++)
         {
             for(int j=0;j<_gridM;j++)
             {
                 for(int k=0;k<_subMeshArray.Length;k++)
                 {
                     _subMeshesLocation[i,j] = new int[_subMeshArray.Length];
                 }
             }
         }
     }
 
 
     int[,][] _subMeshesLocation;
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Answer by maccabbe · Apr 24, 2015 at 07:21 AM

The line

 _subMeshesLocation = new int[(int)_gridN,(int)_gridM];

should be

 _subMeshesLocation = new int[(int)_gridN,(int)_gridM][];

Also you are declaring i*j*k arrays when you only need i*j arrays of variable length.

Here is an example of working code.

 using UnityEngine;
 
 public class NewBehaviourScript : MonoBehaviour {
      int[,][] _subMeshesLocation;
      int iSize=3;
      int jSize=3;
 
      void Start(){
          _subMeshesLocation = new int[iSize, jSize][];
          for(int i=0; i<iSize; i++){
              for(int j=0; j<jSize; j++){
                  // for each (i, j) position creates an array of random size
                  int[] temp=new int[Random.Range(0, 10)];
                  // and then fills it with 0, 1, ... array.Length-1
                  for(int k=0; k<temp.Length; k++){
                      temp[k]=k;
                  }
                  _subMeshesLocation[i, j]=temp;
              }
          }

          // prints back all the values
          for(int i=0; i<_subMeshesLocation.GetLength(0); i++) {
              for(int j=0; j<_subMeshesLocation.GetLength(1); j++) {
                  int[] temp=_subMeshesLocation[i, j];
                  for(int k=0; k<temp.Length; k++) {
                      Debug.Log(temp[k]);
                  }
              }
          }
      }
 }



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avatar image Fran-Martin · Apr 24, 2015 at 07:25 AM 0
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Thank you very much. Pretty clear

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