所以我在同一个文件中有两个类。 ArrayLinkedList
和ArrayLinkedListRow
在第一个提到的里面,我有一个方法
template<class T> friend ostream& operator <<(ostream& s, ArrayLinkedList<T>& ll){
//Extra code for giving s content
return s;
}
以及有
template<class T> friend ostream& operator <<(ostream& s, ArrayLinkedListRow<T>& ll){
//Extra code for giving s content
return s;
}
在ArrayLinkedListRow内部。
我收到以下错误
而且让我发疯,不知道如何解决它。我已经完成了研究,但是仍然无法弄清楚该怎么做。我坚信,尽管错误仅指出了一行,但这两个类可能在问题中相关。
额外信息:
对于那些对我的简短解释感到困惑的人来说,这是类ArrayLinkedList头。
template<class DT>
class ArrayLinkedList {
private:
DT* _info[MAX_SIZE]; // store data
int _next[MAX_SIZE]; // next node
int _nextEmpty[MAX_SIZE]; //next empty slot
ArrayClass< ArrayLinkedListRow<DT> >* _rows;
int _head; // head of the list
int _firstEmpty; // first empty slot
int _size;
void copy(const ArrayLinkedList<DT>& ll);//copy from another list
// add a new node with next as it's next node and returns the index of new node
int newNode( DT& newObject, int next);
public:
ArrayLinkedList(); // empty and copy constructors
ArrayLinkedList(const ArrayLinkedList<DT>& ll);
//copy constructors linked list object to an existing object. This is a deep copy.
~ArrayLinkedList(); // destructor
ArrayLinkedList(DT& newObject); // Constructor that create a list with newObject as the head
ArrayLinkedList(int capacity); // Constructor with a give capacity
ArrayLinkedList(DT& newObject,int capacity);// Constructor with newObject as the head and capacity
bool isEmpty(); // is the list empty?
int size(); // return the number of nodes stored
void add(DT& newObject); // add an object to the tail
void insertAt(DT& newObject, int position); // insert an object at the position specified
DT remove(); // remove the head
DT removeAt(int position); // remove an object at the position specified
int find(DT key); // find the object that matches key, index of the object
void operator=(const ArrayLinkedList<DT>& ll); // = operator
// overloading [] operator, return a reference to object at the
// Add a new data element to the start of a linked list.
DT& operator[] (const int position); // position in the linked list
// ostream operator
template<class T> friend ostream& operator <<(ostream& s, ArrayLinkedList<T>& ll){
return s;
}
void displayRaw(); // display raw data of the data members
};
最佳答案
尝试删除template<class T>
部分:
friend ostream& operator <<(ostream& s, ArrayLinkedList& ll){
//Extra code for giving s content
return s;
}
// and analogically with ArrayLinkedListRow
起作用的原因是here:
ArrayLinkedList<int>
,那么并且只有这样,才会使用模板参数operator <<
和T
(未使用)创建DT
。如果您对此进行编译,则一切正常。 ArrayLinkedList<float>
的变量,则将再次定义运算符,这将产生错误。 仅使用
DT
可使其按预期工作。关于c++ - 即时问题,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/9897606/