问题
How do smart pointers handle arrays? For example,
void function(void)
{
std::unique_ptr<int> my_array(new int[5]);
}
When my_array goes out of scope and gets destructed, does the entire integer array get re-claimed? Is only the first element of the array reclaimed? Or is there something else going on (such as undefined behavior)?
回答1:
It will call delete[] and hence the entire array will be reclaimed but I believe you need to indicate that you are using an array form of unique_ptrby:
std::unique_ptr<int[]> my_array(new int[5]);
This is called as Partial Specialization of the unique_ptr.
回答2:
Edit: This answer was wrong, as explained by the comments below. Here's what I originally said:
I don't think std::unique_ptr knows to call delete[]. It effectively has an int* as a member -- when you delete an int* it's going to delete the entire array, so in this case you're fine.
The only purpose of the delete[] as opposed to a normal delete is that it calls the destructors of each element in the array. For primitive types it doesn't matter.
I'm leaving it here because I learned something -- hope others will too.
来源：https://stackoverflow.com/questions/6713484/smart-pointers-and-arrays