operator = on指針可能導致mac上的段錯誤

Question

我想寫一個類設置鏈接列表來存儲整數。編譯和我的Mac的終端上運行後，這是輸出：

[] 
[10] 
[10, 20] 
Segmentation fault: 11 

但我期待看到以下的輸出：

[] 
[10] 
[10, 20] 
[10, 20] 
[10, 20, 30] 

我想知道，如果它是我的操作有問題=函數，還是我不能用指針的operator =函數？如果是這樣，我該如何糾正這個問題，以便程序按照我的預期輸出？我真的很感激你的幫助。提前致謝！

#include <iostream> 
using namespace std; 

class Node { 
    public: 
    int value; 
    Node* next; 
    Node(int n, Node* ptr = NULL) : value(n), next(ptr) {} 
}; 

class Set { 
    Node* head; 
    friend ostream& operator<<(ostream&, const Set&); 
    public: 
    Set() : head(NULL) {} 
    Set(const Set& another){ *this = another; } 
    ~Set(); 
    Set& operator+=(const int&); 
    Set& operator=(const Set&); 
}; 

int main() { 
    int num1 = 10; 
    int num2 = 20; 
    int num3 = 30; 
    Set set1; 
    cout << set1; 
    Set* set2; 
    set1 += num1; 
    cout << set1; 
    set1 += num2; 
    cout << set1; 
    set2 = new Set(set1); 
    cout << *set2; 
    *set2 += num3; 
    cout << *set2; 
    delete set2; 
    return 0; 
} 

Set::~Set() { 
    Node* current = head; 
    while (current != NULL) { 
     Node* temp = current; 
     current = current->next; 
     delete temp; 
    } 
} 

Set& Set::operator+=(const int& aNum) { 
    if (head == NULL) { 
     head = new Node(aNum); 
     return *this; 
    } 
    Node* previous = head; 
    Node* current = head->next; 
    while (current != NULL) { 
     previous = current; 
     current = current->next; 
    } 
    previous->next = new Node(aNum); 
    return *this; 
} 

Set& Set::operator=(const Set& another) { 
    if (this != &another) { 
     Node* current = head; 
     while (current != NULL) { 
      Node* temp = current; 
      current = current->next; 
      delete temp; 
     } 
     Node* anotherCurrent = another.head; 
     while (anotherCurrent != NULL) { 
      *this += anotherCurrent->value; 
      anotherCurrent = anotherCurrent->next; 
     } 
    } 
    return *this; 
} 

ostream& operator<<(ostream& os, const Set& s) { 
    os << "["; 
    for (Node* p = s.head; p != NULL; p = p->next) { 
     os << p->value; 
     if (p->next != NULL) 
      os << ", "; 
    } 
    os << "]" << endl; 
    return os; 
} 

Answer 1

你必須複製之前刪除以前的列表時，否則+=運營商將使用head設置head到NULL，它是當前未分配，但不NULL

Set& Set::operator=(const Set& another) { 
    if (this != &another) { 
     Node* current = head; 
     while (current != NULL) { 
      Node* temp = current; 
      current = current->next; 
      delete temp; 
     } 
     head = NULL; // <============== code to add 
     Node* anotherCurrent = another.head; 
     while (anotherCurrent != NULL) { 
      *this += anotherCurrent->value; 
      anotherCurrent = anotherCurrent->next; 
     } 
    } 
    return *this; 

BTW一個很有趣設計模式，必須閱讀：copy-and-swap idiom