The assignment operator of Array1D < Array1D <T> >

[chucthanh]

Hi all,
I wrote a class Array1D which is of the form:

template <class T>
class Array1D
{
	int n;		//Length of the array
	T* pointer;	//Pointer to the data
public:
-Some methods here...
}

The copy constructor is:

	Array1D (Array1D <T> const & A)
	{
		n = A.length();
		if (n != 0)
		{
			pointer = new T [n];
		} else
		{
			pointer = 0;
		};
		
		for (int k = 0; k < n; k++)
		{
			(pointer[k]) = A.at(k); //(**)
		};
	};

(the method at just returns the pointer[k])

When I used the class in the form Array1D < Array1D <double> > A (B) , I checked that the program did not invoke the assignment operator at (**), which was written as

	void operator = (Array1D <T> const & A)
	{
		cout << "copying" << endl;
		if (n != 0)
		{
			delete [] pointer;
		};
		n = A.length();
		if (n != 0)
		{
			pointer = new T [n];
		} else
		{
			pointer = 0;
		};
		for (int k = 0; k < n; k++)
		{
			pointer[k] = A.at(k);
		};
	};

As a result, all went wrong.
Can you please give me a hint to get around this?
Thank you very much!

[D80DSL3A]

The problem description "all went wrong" is too vague. What specifically went wrong? Error codes?

The line pointer[k] = A.at(k); would invoke the = operator for the T object, not for the Array1D object.
You could check for a problem here by making the assignment directly as pointer[k] = A.pointer[k]; instead.

Are you sure about this Array1D < Array1D <double> > A (B); ?
It seems like it should be Array1D<double> A(B); instead.

[chucthanh]

Hi fun2code, thank you. It is my fault, I found the error, which is in different place.
Yes, I meant to use Array1D < Array1D <double> , the array1D of array1D. So the assignment evoked was of (Array1D <double>) =.
Again, thank you and sorry for bothering you with a wrong question.

[D80DSL3A]

You're welcome. No problem. I see that Array1D < Array1D <double>> is sensible now. It is an array of arrays (a 2D array in essence). I'm not sure why I didn't see it before.
Glad you solved your own problem (this is the best way).