我想写一个类设置链接列表来存储整数。编译和我的Mac的终端上运行后,这是输出:operator = on指针可能导致mac上的段错误
[]
[10]
[10, 20]
Segmentation fault: 11
但我期待看到以下的输出:
[]
[10]
[10, 20]
[10, 20]
[10, 20, 30]
我想知道,如果它是我的操作有问题=函数,还是我不能用指针的operator =函数?如果是这样,我该如何纠正这个问题,以便程序按照我的预期输出?我真的很感激你的帮助。提前致谢!
#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;
}
解决此类问题的正确工具是您的调试器。在*堆栈溢出问题之前,您应该逐行执行您的代码。如需更多帮助,请阅读[如何调试小程序(由Eric Lippert撰写)](https://ericlippert.com/2014/03/05/how-to-debug-small-programs/)。至少,您应该\编辑您的问题,以包含一个[最小,完整和可验证](http://stackoverflow.com/help/mcve)示例,该示例再现了您的问题,以及您在调试器。 –
'Set(const Set&another){* this = another; }' - 这不是编写拷贝构造函数的好方法。你应该在赋值操作符的帮助下编写拷贝构造函数(假设赋值操作不存在)。然后,一旦你这样做,写赋值操作符变成了2行函数。 – PaulMcKenzie
@PaulMcKenzie thx寻求帮助。我已经重写了拷贝构造函数,而不使用赋值操作符,它工作。但是,我不知道为什么复制构造函数不能使用赋值运算符?另外,如何使用复制构造函数来编写赋值运算符?非常感谢^ – mlkw