Overloading << C++

Question

我试图超载< <运算符，所以我可以输出链表的内容。

所以，如果我与值1，2，3一链表和我称：

cout << list << endl; 

其中列表的类型是链表的。我应该得到这样的东西：

1 2 3或1 | 2 | 3在终端。

我应该使用循环遍历列表并将每个节点的数据添加到输出？想实现，这并不需要看到节点类的方法...

ostream& operator <<(ostream& outs, const LinkedList& source) 
{ 

    //use a loop to cycle through, adding each data to outs, seperated by a space or a | 

    return outs; 
} 

谨以此制定出一个结构好吗？：

LinkedList::report(ostream& outs) //thanks beta :) 
{ 
    Node *currentPtr = headPtr; 

    while(currentPtr != NULL) 
    { 
     outs << currentPtr->getData() << " "; //thanks Rob :) 

     currentPtr = currentPtr->getNextPtr(); 
    } 

    return outs; 
} 

Answer 1

我建议这样的：

ostream& operator <<(ostream& outs, const LinkedList& source) 
{ 
    source.report(outs); 
    return outs; 
} 

此方法看不到节点类。你在LinkedList::report(ostream& outs)中保留你的实现 - 你使用循环。

编辑：
一对夫妇更正到您的report方法：

// It must have a return type (e.g. void), and it must be const because your 
// operator takes a const argument, which must call this method 
void LinkedList::report(ostream& outs) const 
{ 
    const Node *currentPtr = headPtr; // might as well make this const too 

    while(currentPtr != NULL) 
    { 
     outs << currentPtr->getData() << " "; // you can use "|" if you want 
     currentPtr = currentPtr->getNextPtr(); 
    } 

    // no need to return anything 
} 

Answer 2

是的，你应该实现一个循环通过您LinkedList类进行迭代。

既然你不显示怎么LinkedList的作品，我不能告诉你，在循环正是去，但一般可能是这样的：

ostream& operator<<(ostream& outs, const LinkedList& source) { 
    while(there_remains_data_to_print(source)) { 
     outs << get_the_next_datum(source) << " "; 
    } 
    return source; 
} 

例如，如果您有LinkedList STL风格的迭代器：

ostream& operator<<(ostream& outs, const LinkedList& source) { 
    LinkedList::iterator it = source.begin(); 
    while(it != source.end()) { 
     outs << *it << " "; 
     ++it; 
    } 
    return source; 
} 

或者，对于传统的C风格的链接列表：

ostream& operator<<(ostream& outs, LinkedList source) { 
    while(source) { 
     outs << source->datum << " "; 
     source = source->next; 
    } 
    return source; 
}