Thread: error when returning template type~

hi all~

i met some problem when coding a template class, the trouble is one of its method(const method) due to grab the data with position specified which return the type the same as template one, but when i compiled the compiler report an error which said: "warning: argument to non-pointer type 'int' from NULL". i wonder what this means and want the solution to my method, here is the code:

Code:

template <class T>
class LinkedList
{
	public:
		LinkedList();
		~LinkedList();
		void clear();
		bool isEmpty() const;
		int getLength() const;
		T getData(int pos) const;
		int find(T t) const;
		bool insert(int pos, T t);
		bool remove(int pos);
		void trace() const;
	private:
		LinkedListNode<T>* __first;
};
template <class T>
T LinkedList<T>::getData(int pos) const
{
	if (pos < 0) return NULL;
	int _len = 0;
	LinkedListNode<T>* _ptNode = __first;
	while (_ptNode)
	{
		if (_len == pos) return _ptNode->data;
		_ptNode = _ptNode->next;
		_len++;
	}
	return NULL;
}


int main()
{
	LinkedList<int> myLinkedList;
	int num = 10;
	for (int i=0; i<num; i++)
	{
		myLinkedList.insert(i, i);
	}
	myLinkedList.remove(9);
	myLinkedList.getData(1);                   // Error here !
	myLinkedList.trace();
	return 0;
}

thanx !

Simply return 0 instead of NULL. As the compiler tells you, the return value is not a pointer, but an int.

Originally Posted by Carlos

Simply return 0 instead of NULL. As the compiler tells you, the return value is not a pointer, but an int.

ie to say, NULL is indeed a pointer itself ??? if then it will points to what if i set something like: int x = NULL ?

Originally Posted by black

ie to say, NULL is indeed a pointer itself ??? if then it will points to what if i set something like: int x = NULL ?

It depends. Check your compiler's definition for NULL.
In case of VC++ 6.0 it's:

Code:

/* Define NULL pointer value */

#ifndef NULL
#ifdef __cplusplus
#define NULL    0
#else
#define NULL    ((void *)0)
#endif
#endif

In this case if you're compiling C++ code (and as you're using templates it should be ) there should be no problem, I guess.
However, there's no reason to prefer NULL above 0 for built-in types.

it works ! thanx man !!! code fixed are like this:

Code:

T LinkedList<T>::getData(int pos) const
{
	if (pos < 0) return 0;
	int _len = 0;
	LinkedListNode<T>* _ptNode = __first;
	while (_ptNode)
	{
		if (_len == pos) return _ptNode->data;
		_ptNode = _ptNode->next;
		_len++;
	}
	return 0;
}

but, it is strange that compiler permit a number return say, 0 to pass, and if the T type are applied into some other data type like another class, will it be error again when return a 0 ?

Normally the conversion should be legal only if you declare a one-parameter constructor which performs implicit conversions. In this case, the parameter should be of type int.

Code:

class ClassA
{
public:
ClassA();
ClassA( int i_in ); // allows implicit conversion - unless declared explicit
};

Implicit conversion are not always welcome, as such conversions are uncontrollable and therefore unwanted. Declaring the constructor with the keyword explicit prohibits this behaviour.