C++ 智能指针和数组
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smart pointers and arrays
提问by helloworld922
How do smart pointers handle arrays? For example,
智能指针如何处理数组？例如，
void function(void)
{
std::unique_ptr<int> my_array(new int[5]);
}
When my_arraygoes 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)?
当my_array超出范围并被破坏时，整个整数数组是否会被重新声明？是否只回收数组的第一个元素？或者还有其他事情发生（例如未定义的行为）？
回答by Alok Save
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:
它将调用delete[]，因此整个数组将被回收，但我相信您需要表明您使用的是数组形式unique_ptr：
std::unique_ptr<int[]> my_array(new int[5]);
This is called as Partial Specializationof the unique_ptr.
这被称为局部特殊化的unique_ptr。
回答by Nathan Monteleone
Edit: This answer was wrong, as explained by the comments below. Here's what I originally said:
编辑：正如下面的评论所解释的那样，这个答案是错误的。这是我最初所说的：
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.
我不认为 std::unique_ptr 知道调用 delete[]。它实际上有一个 int* 作为成员——当你删除一个 int* 时，它会删除整个数组，所以在这种情况下你没问题。
delete[] 与普通删除相反的唯一目的是它调用数组中每个元素的析构函数。对于原始类型，这无关紧要。
I'm leaving it here because I learned something -- hope others will too.
我把它留在这里是因为我学到了一些东西——希望其他人也能。
