4 ms·
Normally, the deallocation of the destroyed object has no visible effect, and you're not expected to do anything with object after it's destroyed. So, in case
by ansgri 15y ago
Normally, the deallocation of the destroyed object has no visible effect, and you're not expected to do anything with object after it's destroyed.
So, in case of C++, object is destroyed when goes out of scope, but the physical act of deallocation may be deferred by the standard library as whatever 'smart memory manager' wishes.