Problem with for() loops

Processing Beginners
1 
int th = 36;
  int tw = 36;
  
  for(int y = 0; y < th - 1; y++)
  {
    for(int x = 0; x < tw - 1; x++)
  {
    
    PVector p1 = normals[x][y]; 
    //works
    PVector p2 = normals[34][y]; 
    // works
    PVector p3 = normals[35][y]; 
    // returns null object?
2

Your end of loop conditions are incorrect because they do not access the last element in the array. Try this

for(int y = 0; y < th; y++)
{
for(int x = 0; x < tw; x++)
{

Also in Java a 2D array is created as a ‘1D array’ of ‘1D arrays’ so your for-loops initialize an array

normals[y][x];

This is not causing a problem in your code because tb == th a square 2D array if you want

normals[x][y];

then swap the for loops like this

{
for(int x = 0; x < tw; x++)
{
for(int y = 0; y < th; y++)
3

Hello @JALAJ_0106 ,

I suspect that you may have used the same loop to initialize array.

  PVector p3 = normals[35][0];  
  // null There is no element 35!      
  println(p3);

Example:

int th = 36;
int tw = 36;

PVector normals[][] = new PVector[tw][th];  // Create a 2D array with `tw` columns and `th` rows

for(int y = 0; y < th - 1; y++)   // <36-1 is 0 to 34 
  {
  for(int x = 0; x < tw - 1; x++) // <36-1 is 0 to 34
    {
    normals[y][x] = new PVector(0, 0, 0);  // Initialize each element with a new PVector
    }
  }
  
  PVector p2 = normals[34][0];
  println(p2);
  // works
  
  PVector p3 = normals[35][0];  
  // null There is no element 35!      
  println(p3);

Try and replace th-1 with th and see what happens.
print x am y in the loop to examine what you are doing.

:)