通过引用或通过值将共享指针作为参数传递给类

问题描述

如果要复制到成员变量中，共享指针应该通过引用还是通过值作为参数传递给类?

Should a shared pointer be passed by reference or by value as a parameter to a class if it is going to be copied to a member variable?

共享指针的复制将增加引用计数，我不想制作任何不必要的副本，因此引用计数会增加.将共享指针作为参考传递会解决这个问题吗?我认为确实如此，但还有其他问题吗?

The copying of the shared pointer will increment the refrence count and I don't want to make any unnecessary copies and thus ref count increments. Will passing the shared pointer as a refrence solve this? I assume it does but are there any other problems with this?

按值传递:

class Boo {
public:
    Boo() { }
};

class Foo {
public:
    Foo(std::shared_ptr<Boo> boo)
        : m_Boo(boo) {}
private:
    std::shared_ptr<Boo> m_Boo;
};

int main() {
    std::shared_ptr<Boo> boo = std::make_shared<Boo>();

    Foo foo(boo);
}

通过引用:

class Boo {
public:
    Boo() { }
};

class Foo {
public:
    Foo(std::shared_ptr<Boo>& boo)
        : m_Boo(boo) {}
private:
    std::shared_ptr<Boo> m_Boo;
};

int main() {
    std::shared_ptr<Boo> boo = std::make_shared<Boo>();

    Foo foo(boo);
}

推荐答案

传值然后移入成员:

class Foo {
public:
    Foo(std::shared_ptr<Boo> boo)
        : m_Boo(std::move(boo)) {}
private:
    std::shared_ptr<Boo> m_Boo;
};

这在所有情况下都是最有效的——如果调用者有一个右值引用，那么就不会有一个 add-ref，如果调用者有一个值，就会有一个 add-ref.

This will be the most efficient in all cases - if the caller has a rvalue-reference then there wont be a single add-ref, if the caller has a value there'll be a single add-ref.

如果您通过了 const&，即使在不必要的情况下，您也会强制添加引用.如果您按值传递然后在没有 std::move 的情况下进行设置，您可能会得到 2 个 add-refs.

If you pass by const& you force an add-ref even in cases where its unnecessary. If you pass by value and then set without a std::move you may get 2 add-refs.

如果您有一个移动比复制便宜得多的类，并且您有一个始终复制传入的实例的函数调用，那么这是一个很好的模式 - 就像在这种情况下一样.您通过按值获取它来强制复制在函数调用边界发生，然后如果调用者有一个右值引用，则复制根本不需要发生 - 它会被移动.

This is a good pattern to use if you've got a class where a move is significantly cheaper than a copy, and you have a function call which will always copy the instance passed in - as in this case. You force the copy to happen at the function call boundary by taking it by value, and then if the caller has a rvalue reference the copy need never happen at all - it will be moved instead.

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