I'm trying to create a doubly linked list where each list has a first node, last node, and num_elements. However, for some reason, when I try to test the code in a UseList.cpp file, I can't get the num_elements to set to zero as default.
Let me show you what I mean:
In List.h:
template <class L>
class List
{
private:
Node<L> *first;
Node<L> *last;
int num_elements;
public:
// constructors and destructors
List();
[...]
}
[...]
template <class L>
List<L>::List() {
first = NULL;
last = NULL;
num_elements = 0;
}
[...]
This is the show method lower down in list.h:
template <class L>
// Print out the data in each node separated by a space.
void List<L>::show() {
cout << num_elements << endl;
Node<L> *current_node = first;
while (current_node != NULL) {
cout << current_node->data << " ";
current_node = current_node->next;
}
cout << endl;
}
Note that there is a cout statement there to print the num_elements.
This is the relevant part of UseList.cpp:
int main (int argc, char *argv[]) {
cout << "-----------------------------------------" << endl;
cout << "----------------LIST ONE-----------------" << endl;
cout << "-----------------------------------------" << endl;
List<int> *list1;
srand(time(NULL));
list1->show();
[...]
When show is called, it prints out "1" and gives me a segmentation fault. Why is num_elements defaulting to "1" instead of "0"?
When I do a cout in List<L>::List() {
, nothing is printed... (this implies that the constructor never runs?)
Thanks for the help!
您正在声明一个指向 List<int>
的指针 ,而不是将其初始化为任何内容。
You have created a pointer to a List<int>
object, but no object. So, currently, your program will segmentation fault because the pointer is "dangling". When you try to dereference it with ->
, you are accessing memory that isn't yours, and it fails. To fix this, simply allocate a new List object:
List<int> *list1 = new List<int>();
Don't forget to free it later:
delete list1;
Your other option is to just not use dynamic memory. You shouldn't use it if you don't have to.
List<int> list1;
list1.show()
List<int> *list1;
Declares list1
to be a pointer.
List<int> *list1 = new List<int>();
Would actually create an instance of List
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.