C++ 如何为 std::vector 创建 shared_ptr？
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How can I create a shared_ptr to a std::vector?
提问by Oscar_Mariani
I need to create a shared_ptr to a std::vector, what is the correct syntax?
我需要为 std::vector 创建一个 shared_ptr，正确的语法是什么？
std::vector<uint8_t> mVector;
shared_ptr<std::vector<uint8_t>> mSharedPtr = &mVector;
The code above does not compile.
上面的代码不能编译。
Thanks.
谢谢。
回答by davidhigh
What you are trying to do is to let a smart pointer manage a stack object. This doesn't work, as the stack object is going to kill itself when it goes out of scope. The smart pointer can't prevent it from doing this.
您要做的是让智能指针管理堆栈对象。这不起作用，因为堆栈对象在超出范围时会自行终止。智能指针无法阻止它这样做。
std::shared_ptr<std::vector<uint8_t> > sp;
{
std::vector<uint8_t> mVector;
sp=std::shared_ptr<std::vector<uint8_t> >(&mVector);
}
sp->empty(); // dangling reference, as mVector is already destroyed
Three alternatives:
三种选择：
(1) Initialize the vector and let it manage by the shared_ptr:
(1) 初始化向量并让它由以下管理shared_ptr：
auto mSharedPtr = std::make_shared<std::vector<uint8_t> >(/* vector constructor arguments*/);
(2) Manage a copy of the vector (by invoking the vector copy constructor):
(2) 管理vector的副本（通过调用vector copy构造函数）：
std::vector<uint8_t> mVector;
auto mSharedPtr = std::make_shared<std::vector<uint8_t> >(mVector);
(3) Move the vector (by invoking the vector move constructor):
(3) 移动向量（通过调用向量移动构造函数）：
std::vector<uint8_t> mVector;
auto mSharedPtr = std::make_shared<std::vector<uint8_t> >(std::move(mVector));
//don't use mVector anymore.
回答by coyotte508
First, what you're doing is very wrong, if you give a pointer to a shared_ptr make sure it's dynamically allocated with new, not on the stack. Otherwise you may just as well use a pointer instead of a shared_ptr.
首先，你在做什么是非常错误的，如果你给一个 shared_ptr 一个指针，确保它是动态分配的new，而不是在堆栈上。否则，您也可以使用指针而不是 shared_ptr。
Your code should be:
你的代码应该是：
std::vector<uint8_t> mVector;
/* Copy the vector in a shared pointer */
std::shared_ptr<std::vector<uint8_t> > mSharedPtr ( new std::vector<uint8_t>(mVector) );
or:
或者：
std::shared_ptr<std::vector<uint8_t> > mSharedPtr ( new std::vector<uint8_t>() );
As for why it doesn't compile, you need to use the constructor, not the =operator.
至于为什么不编译，你需要使用构造函数，而不是=运算符。
As pointed out by @remyabel, make_sharedis more efficient:
正如@remyabel 所指出的，make_shared更有效：
std::vector<uint8_t> vector;
/* Copy the vector in a shared pointer */
auto sharedPtr = std::make_shared<std::vector<uint8_t>> (vector);
回答by M. Khalil
your code doesn't compile because you are assigning a raw pointer '&mVector' to smart pointer 'mSharedPtr' which are two different objects and no casting is allowed.
您的代码无法编译，因为您将原始指针 ' &mVector'分配给智能指针 ' mSharedPtr'，这是两个不同的对象，并且不允许进行强制转换。
you can do that by other approaches
你可以通过其他方法做到这一点
(1) intializing your shared_ptrwith the raw pointer from the begining
(1)shared_ptr从一开始就用原始指针初始化你
std::shared_ptr<std::vector<uint8_t>> sPtr (&mVector);
(2) using reset()method of shared_ptr
(2)使用reset()方法shared_ptr
std::shared_ptr<std::vector<uint8_t>> sPtr;
sPtr.reset(&mVector);
assigning a stack object raw pointer to smart pointer , you should also supply an empty deleter to the smart pointer, so that the smart pointer doesn't delete the object when it is still on the stack.
将堆栈对象原始指针分配给智能指针，您还应该为智能指针提供一个空删除器，以便智能指针在对象仍在堆栈上时不会删除该对象。
std::shared_ptr<std::vector<uint8_t>> sPtr (&mVector,[](std::vector<uint8_t>*){});
