c++ - 用cerr修复的链表分割错误

我有以下代码。该代码的作用是合并两个排序的链表，然后删除两个链表并返回一个新列表：

#include <iostream>

using namespace std;

class node {
    public:
        node(int, node*);
        node* get_next() const;
        void set_next(node*);
        int get_item() const;
        void set_item(int);
    private:
        int item;
        node* next;
};

node* clone_and_destroy(node* a)
{
    node* new_list;
    node* prev;
    node* ret_val;
    while ( a )
    {
        new_list = new node(a->get_item(),NULL);
        if ( prev ){
            prev->set_next(new_list);
            prev = prev->get_next();
        }
        else{
        prev = new_list;
        ret_val = prev;
        }
        node* temp = a;
        a = a->get_next();
        delete temp;
    }
    return ret_val;
}

node* merge(node*& a, node*& b){
    if ( !a ){// cerr<<"\r";
        return clone_and_destroy(b);}
    if ( !b )
        return clone_and_destroy(a);

    node* smaller_node, *bigger_node;
    if ( a->get_item() < b->get_item() ){
        smaller_node = a;
        bigger_node = b;
    }
    else {
        smaller_node = b;
        bigger_node = a;
    }
    node* next_node = smaller_node->get_next();
    node* merged_list = new node(smaller_node->get_item(),smaller_node->get_next());
    delete smaller_node;
    merged_list->set_next(merge(next_node,bigger_node));
    a = NULL ;
    b = NULL ;
    return merged_list;
}

/*node* merge(node*& a, node*& b){
    if ( !a )
        return b;
    if ( !b )
        return a;
    if ( a->get_item() < b->get_item() ){
        node* a_next = a->get_next();
        node* merged_list = a;
        merged_list->set_next(merge(a_next,b));
        delete a;
        return merged_list;
    }
    else {
        node* b_next = b->get_next();
        node* merged_list = b;
        merged_list->set_next(merge(a,b_next));
        delete b;
        return merged_list;
    }
}*/

void print(node* a)
{
    while ( a )
    {
        cout << a->get_item() << " ";
        a = a->get_next();
    }
    cout << endl;
}

node::node(int _item, node* _next)
{
    item = _item;
    next = _next;
}
node* node::get_next() const
{
    return next;
}
void node::set_next(node* new_next)
{
    next = new_next;
}
int node::get_item() const { return item; }
void node::set_item(int _item)
{
    item = _item;
}

int main() {
    node* l1 = NULL, *l2 = NULL;
    cout << "Before Merging: " << endl;
    for (int i = 5; i > 0; i--) {
        l1 = new node(i, l1);
        l2 = new node(2*i-3, l2);
    }
    cout << "List 1 is: \t\t";
    print(l1);
    cout << endl;
    cout << "List 2 is: \t\t";
    print(l2);
    cout << endl << "After Merging:" << endl;
    node* m = merge(l1, l2);
    cout << "List 1 is: \t\t";
    print(l1);
    cout << "Should be: \t\t[ ]" << endl;
    cout << endl;
    cout << "List 2 is: \t\t";
    print(l2);
    cout << "Should be: \t\t[ ]" << endl;
    cout << endl;
    cout << "Merged List is: \t";
    print(m);
    cout << "Should be: \t\t[ -1 1 1 2 3 3 4 5 5 7 ]" << endl;


    for (node* h = m; h != NULL; h = m) {
        m = m->get_next();
        delete h;
    }
    return 0;
}

代码的重要部分是合并功能，其余只是用于实现该功能的工具。现在发生了一件有趣的事！当我运行它时，这段代码遇到了段错误，但是当我尝试调试代码时，我在合并函数的第一行使用了cerr（它被注释），突然工作正常！有人可以向我解释一下吗？这个cerr在做什么？以及如何在没有此错误的情况下并且最小限度地更改代码来修复我的代码！现在我知道cerr阻止了cout的缓冲，但是我认为情况并非如此！

最佳答案

* prev没有任何初始化，您正在尝试访问它。因此段。故障。

node* clone_and_destroy(node* a) { .. ..

     if ( prev ){

编辑：初始化它将无效修复段错误。但是您要检查其背后的逻辑。

node* prev = NULL;

Edit2：为什么cerr修复分段错误

Code crashes unless I put a printf statement in it

摘自“这个问题问“为什么printf（）声明'解决'问题”。答案是因为它解决了问题。您有一个Heisenbug，因为滥用了分配的内存，并且存在printf（）设法稍微改变了代码的行为。”