【发布时间】：2014-11-25 02:49:15
【问题描述】：
这个问题在这里有详细解答：Overloading operator<<: cannot bind lvalue to ‘std::basic_ostream<char>&&’
我试图重载一个嵌套子类，并花了一个小时试图重载operator<<。在这里研究了一下，但仍然无法解决。有什么帮助吗？ :)
每当我尝试使用 g++ -std=c++11 -lm -ggdb -g -O0 -Wall p_7_1.cpp -o p_7_1 编译它时，它都会给我一个错误：
Undefined symbols for architecture x86_64:
"operator<<(std::__1::basic_ostream<char, std::__1::char_traits<char> >&, LinkedBinaryTree<int>::Position const&)", referenced from:
std::__1::basic_ostream<char, std::__1::char_traits<char> >& operator<<<int>(std::__1::basic_ostream<char, std::__1::char_traits<char> >&, LinkedBinaryTree<int> const&) in p_7_1-9fc2c2.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)
p_7_1.cpp：
#include "tree.hpp"
typedef LinkedBinaryTree<int> Tree;
#include <iostream>
using namespace std;
int main() {
// Test if tree works:
Tree lbt;
cout << lbt.empty() << endl;
lbt.addRoot();
lbt.addRoot();
cout << lbt.empty() << endl;
cout << lbt.size() << endl;
lbt.expandExternal(lbt.root());
cout << lbt.empty() << endl;
cout << lbt.size() << endl;
*(lbt.root()) = 12;
cout << lbt;
}
tree.hpp:
#ifndef LINKED_BINARY_TREE_HPP
#define LINKED_BINARY_TREE_HPP
#include <list>
template <typename T> class LinkedBinaryTree;
template <typename T> std::ostream& operator<<(std::ostream& os, const LinkedBinaryTree<T>& lbt);
template <typename T> std::ostream& operator<<(std::ostream& os, const typename LinkedBinaryTree<T>::Position& p);
template <typename T>
class LinkedBinaryTree {
protected:
struct Node { // a node of the tree
T elt; // element value
Node* par; // parent
Node* left; // left child
Node* right; // right child
Node() : elt(), par(NULL), left(NULL), right(NULL) { } // constructor
};
public:
class Position { // position in the tree
private: //
Node* v; // pointer to the node
public:
Position(Node* _v = NULL) : v(_v) { } // constructor
T& operator*() // get element
{ return v->elt; } //
Position left() const // get left child
{ return Position(v->left); } //
Position right() const // get right child
{ return Position(v->right); } //
Position parent() const // get parent
{ return Position(v->par); } //
bool isRoot() const // root of the tree?
{ return v->par == NULL; } //
bool isExternal() const // an external node?
{ return v->left == NULL && v->right == NULL; } //
friend class LinkedBinaryTree; // give tree access
public:
friend std::ostream& operator<<(std::ostream& os, const typename LinkedBinaryTree<T>::Position& p);
friend std::ostream& operator<< <T>(std::ostream& os, const LinkedBinaryTree<T>& lbt);
};
typedef std::list<Position> PositionList; // list of positions
public: //
LinkedBinaryTree(); // constructor
int size() const; // number of nodes
bool empty() const; // is tree empty?
Position root() const; // get the root
PositionList positions() const; // list of nodes
void addRoot(); // add root to empty tree
void expandExternal(const Position& p); // expand external node
Position removeAboveExternal(const Position& p); // remove p and parent
// housekeeping functions omitted...
protected: // local utilities
void preorder(Node* v, PositionList& pl) const; // preorder utility
public:
friend std::ostream& operator<< <T>(std::ostream& os, const LinkedBinaryTree<T>& lbt);
private: //
Node* _root; // pointer to the root
int n; // number of nodes
}; //
template <typename T>
LinkedBinaryTree<T>::LinkedBinaryTree() // constructor
: _root(NULL), n(0) { }
template <typename T>
int LinkedBinaryTree<T>::size() const // number of nodes
{ return n; }
template <typename T>
bool LinkedBinaryTree<T>::empty() const // is tree empty?
{ return size() == 0; }
template <typename T>
typename LinkedBinaryTree<T>::Position LinkedBinaryTree<T>::root() const // get the root
{ return Position(_root); }
template <typename T>
typename LinkedBinaryTree<T>::PositionList LinkedBinaryTree<T>::positions() const {
PositionList pl;
preorder(_root, pl); // preorder traversal
return PositionList(pl); // return resulting list
}
template <typename T>
void LinkedBinaryTree<T>::addRoot() // add root to empty tree
{ _root = new Node; n = 1; }
template <typename T>
void LinkedBinaryTree<T>::expandExternal(const Position& p) {
Node* v = p.v; // p's node
v->left = new Node; // add a new left child
v->left->par = v; // v is its parent
v->right = new Node; // and a new right child
v->right->par = v; // v is its parent
n += 2; // two more nodes
}
template <typename T>
typename LinkedBinaryTree<T>::Position // remove p and parent
LinkedBinaryTree<T>::removeAboveExternal(const Position& p) {
Node* w = p.v; Node* v = w->par; // get p's node and parent
Node* sib = (w == v->left ? v->right : v->left);
if (v == _root) { // child of root?
_root = sib; // ...make sibling root
sib->par = NULL;
}
else {
Node* gpar = v->par; // w's grandparent
if (v == gpar->left) gpar->left = sib; // replace parent by sib
else gpar->right = sib;
sib->par = gpar;
}
delete w; delete v; // delete removed nodes
n -= 2; // two fewer nodes
return Position(sib);
}
// preorder traversal
template <typename T>
void LinkedBinaryTree<T>::preorder(Node* v, PositionList& pl) const {
pl.push_back(Position(v)); // add this node
if (v->left != NULL) // traverse left subtree
preorder(v->left, pl);
if (v->right != NULL) // traverse right subtree
preorder(v->right, pl);
}
template <typename T>
std::ostream& operator<<(std::ostream& os, const LinkedBinaryTree<T>& lbt){
os << lbt.root();
// os << *(lbt.root());
return os;
}
template <typename T>
std::ostream& operator<<(std::ostream& os, const typename LinkedBinaryTree<T>::Position& p) {
os << *p; // Other func stuff will be here later
return os;
}
#endif
更新：
在嵌套类中有一个关于运算符重载的explanation by david-rodríguez-dribeas。他建议最好声明operator<< inline。
【问题讨论】：
-
如果我在类中声明朋友运算符，我可以让它工作：
class A { class B { friend std::ostream& operator<<(...) {...} }; };。如果我把定义从课堂上拿出来，我仍然无法让它发挥作用 -
使用 -Wall 编译并实际阅读您的警告
-
如果您阅读问题的开头，您会看到我确实使用 -Wall 编译它，并且我还提供了错误。而且没有警告。你实际上应该阅读这个问题。
-
如果您再次阅读我的评论，您会发现您需要阅读您的警告，而不仅仅是启用它们。您是否提到您在问题中有警告？如果没有，为什么不呢？
标签： c++ c++11 operator-overloading binary-tree nested-class