Seg Fault只發生在每次運行

Question

這是我發佈的第一個問題，我確保檢查其他線程的主題標題和問題相同，但是迄今爲止，沒有任何解決方案對我有幫助。

問題：當經由一個生成文件彙編包括以下

OBJ = hw1.o 

hw1: $(OBJ) 
    g++ -g -o hw1 $(OBJ) 

hw1.o: 

./PHONE: 
    clean 

clean: 
    rm hw1.o 

我接收有關每3個運行段故障。我在這個網站上看到的解決方案包括在主程序之外移動linked_list*對象，但是當我嘗試這個時，它在每次運行中給我提供了分段錯誤。

下面是我的代碼：

#include<iostream> 
#include<iomanip> 
#include<fstream> 
#include<vector> 

struct linked_list 
{ 
    int data; 

    void set_data(int d) { data = d; } 
    int get_data(void) {return this->data; } 
    void set_next(linked_list* ll) { this->next = ll; } 
    linked_list* get_next(void) { return this->next; } 

    linked_list* next; 
}; 

//linked_list* head; this is where I moved the objects out of main 
//linked_list* tail; which cause seg faults every run 
//linked_list* temp; 

void print_linked_list(linked_list*, int); 

int main() 
{ 
    linked_list* head; 
    linked_list* tail; 
    linked_list* temp; 
    std::ifstream data1, data2; 
    int element = 0;   //elements from list 
    int size_ll = 0; 

    //Open data1.txt and check for errors 
    //data1.txt contains elements to be added in to both data structures 
    data1.open("data1.txt"); 
    if(data1.fail()) 
    { 
     std::cout << "Error: Couldn't open data1.txt\nCheck that file exists...\n"; 

     return 0; 
    } 

    //Open data2.txt and check for errors 
    //data2.txt contains elements which need to be deleted in both data structures 
    data2.open("data2.txt"); 
    if(data2.fail()) 
    { 
     std::cout << "Error Couldn't open data2.txt\nCheck that file exists...\n"; 

     return 0; 
    } 

    head->set_next(NULL); 
    tail->set_next(NULL); 

    for(int i = 0; i <= 150; i ++) 
    { 
     data1 >> element; 

     temp = new linked_list; 
     temp->set_data(element); 
     temp->set_next(NULL); 

     if((size_ll % 50) == 0 && size_ll != 0) 
     { 
      std::cout << "\n\nPrinting current linked list (left to right, top to bottom):"; 
      std::cout << "\nCurrent size: " << size_ll; 
      std::cout << "\n\n"; 
      print_linked_list(head, 0); 
      std::cout << "\n"; 
      std::cout << "End of current list.\n"; 
     } 

     //handling linked list 
     if(size_ll == 0) 
     { 
      head = tail = temp; 

      size_ll++; 
     } 
     else 
     { 
      tail->set_next(temp); 
      tail = temp; 

      size_ll++; 
     } 
    } 

    data1.close(); 
    data2.close(); 
} 

//pass the next element of the linked list 
void print_linked_list(linked_list* current, int lol) 
{ 
    if(current == NULL) 
    { 
     return; 
    } 

    std::cout << std::setw(4) << current->get_data() << " "; 

    if(lol == 9) 
    { 
     std::cout << std::endl; 
     lol = -1; 
    } 

    print_linked_list(current->get_next(), lol+1); 
} 

Answer 1

看你所寫的內容：

linked_list* head; 
linked_list* tail; 

，然後你做：

head->set_next(NULL); 
tail->set_next(NULL); 

沒有初始化head也不tail。這是一個UB

Answer 2

您沒有初始化鏈接列表。如果您將它們作爲靜態變量，則它們將被初始化爲零，並且行head->set_next(NULL)將因嘗試解除NULL引用而崩潰。

在main()中定義它們時，它們具有隨機堆棧值。如果堆棧值是無效地址，則會在同一行崩潰，但如果它恰好包含指向有效內存的地址，則您將開始覆蓋任意內存。