Does list::pop_front call destructor

[srkumar]

Hi,
Can any one give some idea whether the call to list::pop_front invokes destructor or not. As per definition given for list::pop_front on cplusplus.com site

call to list::pop_front() removes the first element in the list container, effectively reducing the list size by one.This calls the removed element's destructor. (http://www.cplusplus.com/reference/stl/list/pop_front.html)

But when I tried that, it was not invoking the destructor. Will I have to explicitely invoke the destructor

Thanks
Shiv

[Bazzy]

Why do you say that it doesn't call the destructor?
Try this code:

struct S
{
    ~S() { cout << "Destructor Called"; }
}aValue;
list<S> li(2,aValue);
li.pop_front();

It should display "Destructor Called"

[jsmith]

If you have a list of pointers-to-something then keep in mind that what is being destroyed is the pointer, which is not the same thing as destroying the object.

So to give a counterexample using Bazzy's code,

struct S
{
    ~S() { cout << "Destructor Called"; }
};

list<S*> li;
li.push_back( new S );
li.pop_front();

In this case ~S is NOT called, because what is being destroyed is an S*, not an S.