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I don't get what following statement would do (specially second line)?
auto buff = std::make_unique<int[]>(128);
buff = std::make_unique<int[]>(512);
Will the second call to make_unique followed by assignment operator will de-allocate memory allocated by first call, or will there be memory leak? Must I have to use buff.reset(new int[512]); ?
I've debugged it, but didn't find any operator= being called, nor any destructor be invoked (by unique_ptr).
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unique_ptr. An unique_ptr has exclusive ownership of the object it points to and will destroy the object when the pointer goes out of scope.
A unique_ptr does not share its pointer. It cannot be copied to another unique_ptr , passed by value to a function, or used in any C++ Standard Library algorithm that requires copies to be made. A unique_ptr can only be moved.
An explicit delete for a unique_ptr would be reset() . But do remember that unique_ptr are there so that you don't have to manage directly the memory they hold. That is, you should know that a unique_ptr will safely delete its underlying raw pointer once it goes out of scope.
Move assignment operator is called, which does
if (this != &_Right)
{ // different, do the swap
reset(_Right.release());
this->get_deleter() = _STD move(_Right.get_deleter());
}
No leak here, as it does reset, which will deallocate.
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