Passing a pointer by reference

[abu]

Hello,

I want to pass a pointer, of type double **x, by reference. I have tried two different option by passing **x and **&x, and both give same result for the test code I shown below. I was wondering which option is the correct. I am trying to implement the concept for my project.

void myfun1 (double **x, int n)
{
 for (int i = 0;i<n;i++)
	 for(int j = 0;j<n;j++)
		 x[i][j] = i*j;
}

void myfun2 (double **&x, int n)
{
 for (int i = 0;i<n;i++)
	 for(int j = 0;j<n;j++)
		 x[i][j] = i*j;
}

int main ()
{
  int n = 5;
  double **y;

  // Allocating memory for y
  y = new double*[n];
  for( int i = 0 ; i < n ; i++ )
	  y[i] = new double [n];

  // test myfun1
  myfun1(y, n);
  // print y
  for(int i = 0;i<n;i++)
  {
	  for(int j = 0;j<n;j++)
	  {	
		  cout <<y[i][j]<<"\t ";
	  }cout<<"\n";
  }cout<<"\n";

   // test myfun2
  myfun2(y, n);
  // print y
  for(int i = 0;i<n;i++)
  {
	  for(int j = 0;j<n;j++)
	  {	
		  cout <<y[i][j]<<"\t ";
	  }cout<<"\n";
  }
  
  system("PAUSE");
  return 0;
}

Thanks

[Cubbi]

The results are the same because the expression x[i][j] accesses the same double whether x is the original pointer to pointer to double (pass-by-reference) or x is a copy of the original pointer to pointer to double (pass by value).

To experience a difference in behavior, try to assign a new value to x itself, from within the function (e.g. allocate a new array of pointers to arrays)

Also, don't forget to delete everything you new'd or use vectors to save yourself the headache (or a third-party matrix library, to make this even simpler)

[vlad from moscow]

There is no any sense to pass the pointer by reference because the pointer itself is not being changed.

[abu]

Ok Thanks so much.