C++ Class variable destructor

Question

I have a question about class variables and scopes in C++. Let's say I have the class below:

class TestClass {
public:
    std::vector<int> v;

    void foo()
    {
        v = std::vector<int>(10);
    }
}

Now, say I call the following code:

TestClass c;
c.foo();

When is the destructor of the vector (assigned to v) called? Is it called when foo() returns or when c goes out of scope?

Answer 1

Destructor for a member vector will be called when TestClass object destructor is called. It will happen when TestClass object goes out of scope

{
  TestClass c;
  c.foo();
} // destructor for c calls destructor for v

This is because C++ Standard 12.6.2 § 10:

In a non-delegating constructor, initialization proceeds in the following order:

— First, and only for the constructor of the most derived class (1.8), virtual base classes are initialized in the order they appear on a depth-first left-to-right traversal of the directed acyclic graph of base classes, where “left-to-right” is the order of appearance of the base classes in the derived class base-specifier-list.

— Then, direct base classes are initialized in declaration order as they appear in the base-specifier-list (regardless of the order of the mem-initializers).

— Then, non-static data members are initialized in the order they were declared in the class definition (again regardless of the order of the mem-initializers).

— Finally, the compound-statement of the constructor body is executed. 11 [ Note: The declaration order is mandated to ensure that base and member subobjects are destroyed in the reverse order of initialization. — end note ]

In line v = std::vector<int>(10) there will be also call to destructor for temporary object, because temporary vector std::vector<int>(10) is being created just to initialize v, and then it is destroyed.

Answer 2

v = std::vector<int>(10);

Temporary object is created, and then copied (or moved C++11) into v. After this line, temporary object's destructor is called.

When c goes out of scope, c.v's destructor is called.

Note: you can also do

v.resize(10);

Answer 3

Okay, let's go through this step by step:

• c is constructed, v is a member of c, so v must be constructed too.
• foo is called
• v is assigned a new value. The new value is constructed, old v is destructed, temporary vector is moved onto v. then moved or copied into v.
• foo returns
• c goes out of scope
• c gets destructed, therefore v is also destructed.

So, v is destructed twice when c is destructed. The temporary may will also get destructed if we have no move constructor.

Answer 4

when your variable c goes out of scope destructor gets called automatically for c, which was automatically allocated, the destructor will take care of deleting the vector