CODEWITHSUNDEEP
c++pointers
tambalolo
tambalolo

Reputation: 1065

Pointers on STL container

I'm new to c++ and find it confusing handling pointers on stl containers. How does stl containers handle pointers?

Point *p1 = new Point(10, 10);
std::vector<Point*> points;
points.push_back(p1);

delete p1; // or delete points[0]

std::cout << points[0]->getID() << "\n"; //why does this still display 10, 10 after deleting above?
std::cout << p1->getID(); //ofcourse, this one will output garbage

//getID method displays xy coordinates given as parameters when object is created
//The result displayed
10, 10
-1, 12337

Upvotes: 0

Views: 1223

Answers (3)

Vijay
Vijay

Reputation: 67211

After the delete statement, Its upto the compiler to when to reuse that freed memory. so it is undefined behaviour.

Upvotes: 1

juanchopanza
juanchopanza

Reputation: 227370

Standard library containers do nothing special about pointers. If the pointers point to dynamically allocated objects, these have to be deleted by someone at some point. Who, when and how really depends on the details of the application, but it is usual to automate this process by use of smart pointers.

std::cout << points[0]->getID() << "\n"; //why does this still display 10, 10 after deleting above?

points[0] has been deleted, and dereferencing a deleted pointer is undefined behaviour. That means anything could happen. What has probably happened is that the memory you have released by calling delete has not been reclaimed for anything, so the data are still there. You cannot rely on that behaviour.

Upvotes: 0

justin
justin

Reputation: 104698

the container, if declared as std::vector<Point*> points; will treat it like declaring Point* point(new Point(blah));, meaning you will need some matching delete in this context. of course, you should be using either a) values in that vector std::vector<Point> points;, or b) smart pointers. in most cases, the former (a).

in this case, you would write delete p1;, and the access of the deleted object is undefined behavior -- any access/use of that object after the delete is useless. you're lucky if the program crashes and points out the error to you.

Upvotes: 1

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