g++:Mac OS X 上的链接器问题 - 体系结构的未定义符号 x86_64

g++: linker issue on Mac OS X - Undefined symbols for architecture x86_64

我之前问过这个问题 here, but got no answer, just a "detour". Now, I am trying to find an actual solution to the problem (stated below). Before anybody says that this question was asked before, I want to say that I tried solutions provided here, here, , and here - 没有任何帮助:(

问题是链接器说 Undefined symbols for architecture x86_64 没有任何 other 警告或错误。调用、完整错误消息和正在编译的代码如下所示。

注意:如果我定义operator<<内联,问题就消失了,但这不是真正的解决方案,而是绕了个弯路:)

提前谢谢你:)

调用和环境

环境:

运行 个参数:

g++ -std=c++11 -lm -stdlib=libc++ tstLinkedList1.cpp -o tstLinkedList1

g++ -std=c++11 -lm -stdlib=libstdc++ tstLinkedList1.cpp -o tstLinkedList1

我也尝试在两种情况下添加 -lc++ - 同样的事情:(

错误

编辑: operator<<重载发生错误,定义在下面LinkedList.hpp文件的最后

使用 -stdlib=libc++:

Undefined symbols for architecture x86_64:
  "operator<<(std::__1::basic_ostream<char, std::__1::char_traits<char> >&, LinkedList<int> const&)", referenced from:
      _main in tstLinkedList1-66598f.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)

使用 -stdlib=libstdc++:

Undefined symbols for architecture x86_64:
  "operator<<(std::ostream&, LinkedList<int> const&)", referenced from:
      _main in tstLinkedList1-8d9300.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)

代码

LinkedList.hpp:

#pragma once

template <typename T> class LinkedList;
template <typename T>
std::ostream& operator<<(std::ostream& os, const LinkedList<T>& list);


/** Node class
 * 
 * @tparam T template type
 */
template <typename T>
class Node {
private:
  T _elem;                      //!< Stored value
  Node<T>* _next;               //!< Next element

  friend class LinkedList<T>;   //!< Friend class
};


/** Singly Linked List
 *
 * @tparam T template type
 */
template <typename T>
class LinkedList {
public:
  LinkedList();
  ~LinkedList();
  std::size_t size() const;
  bool empty() const;
  const T& front() const;
  void addFront(const T& e);
  void removeFront();
public:
  // Housekeeping
  friend std::ostream& operator<<(std::ostream& os, const LinkedList<T>& list);
private:
  Node<T>* _head;
  std::size_t _size;
};

/** Constructor */
template <typename T>
LinkedList<T>::LinkedList() : _head(nullptr), _size(0) {}

/** Destructor */
template <typename T>
LinkedList<T>::~LinkedList() {
  while (!empty()) removeFront();
}

/** Number of elements in the list
 * 
 * @returns std::size_t Number of elements in the list
 */
template <typename T>
std::size_t LinkedList<T>::size() const {
  return this->_size;
}

/** Empty?
 *
 * @returns bool True if empty
 */
template <typename T>
bool LinkedList<T>::empty() const {
  return _head == nullptr;
}

/** Get front element (read-only)
 *
 * @returns T
 */
template <typename T>
const T& LinkedList<T>::front() const {
  return _head->_elem;
}

/** Add element in the front of the list
 * 
 * @param e Element to be added
 */
template <typename T>
void LinkedList<T>::addFront(const T& e) {
  Node<T>* v = new Node<T>;
  v->_elem = e;
  v->_next = _head;
  _head = v;
  _size++;
}

/** Remove the first element */
template <typename T>
void LinkedList<T>::removeFront() {
  if (empty()) return;
  Node<T>* old = _head;
  _head = old->_next;
  _size--;
  delete old;
}

/** Operator<< for the linked list
 *
 * @returns std::ostream
 * @param LHS->std::ostream
 * @param RHS->LinkedList<T>
 */
template <typename T>
std::ostream& operator<<(std::ostream& os, const LinkedList<T>& list) {
  os << "TEST";
  return os;
}

tstLinkedList1.cpp:

#include <iostream>
#include "LinkedList.hpp"

using namespace std;

int main() {
  LinkedList<int> ll1;
  ll1.removeFront();
  ll1.addFront(1);

  std::cout << ll1 << std::endl;
}

好友声明与您认为的不符。它声明了一个非模板函数,而不是您之前声明的模板。您需要的是:

// within LinkedList:
template<typename U>
friend std::ostream& operator<<(std::ostream&, const LinkedList<U>&);

匹配您声明的模板并使那个成为朋友。嗯,整个模板。或者,您也可以使用

// within LinkedList:
friend std::ostream& operator<<<>(std::ostream&, const LinkedList&);
// NOTE:                       ^^ here you need to add <>

仅将 T 的朋友加为好友(您可以使用 LinkedList 而不是 LinkedList<T> - 在 class 中没有区别)。

如果你只使用

// within LinkedList:
friend std::ostream& operator<<(std::ostream&, const LinkedList<T>&);

你需要

// within the global namespace
// NOTE: not a template!
std::ostream& operator<<(std::ostream&, const LinkedList<int>&);

使您的示例有效。或者你可以定义 friend operator<< inline:

// within LinkedList:
friend std::ostream& operator<<(std::ostream& os, const LinkedList<T>& list)
{
    // implement me!
    return os;
}

并完全删除前向声明。