如何返回std :: make_unique< SubClass>工作? [英] How does returning std::make_unique<SubClass> work?

问题描述

我有一个基类及其子类:

I have a base class and its subclass:

class Base {
    public:
    virtual void hi() {
        cout << "hi" << endl;
    } 
};

class Derived : public Base {
    public:
    void hi() override {
        cout << "derived hi" << endl;
    } 
};

试图创建一个辅助函数，该函数创建一个派生对象的唯一指针.

Trying to create a helper function that creates a unique pointer of a Derived object.

1)这个有效:

std::unique_ptr<Base> GetDerived() {
    return std::make_unique<Derived>(); 
}

2)但是，这个编译失败:

2) But, this one fails to compile:

std::unique_ptr<Base> GetDerived2() { 
    auto a = std::make_unique<Derived>(); 
    return a; 
}

3)std :: move works:

3) std::move works:

std::unique_ptr<Base> GetDerived3() {
    auto a = std::make_unique<Derived>();
    return std::move(a); 
}

4)如果我创建一个Base实例，则两者都可以工作:

4) If I create a Base instance, both work:

std::unique_ptr<Base> GetDerived4() {
    auto a = std::make_unique<Base>();
    return a; 
}

std::unique_ptr<Base> GetDerived5() {
    auto a = std::make_unique<Base>();
    return std::move(a); 
}

为什么(2)失败，但是其他人可以工作?

Why (2) fails but others work?

推荐答案

std::unique_ptr是不可复制的，只能移动.您可以从声明为返回std::unique_ptr<Base>的函数中return std::make_unique<Derived>的原因是，存在从一个到另一个的转换.

std::unique_ptr is not copyable, only movable. The reason you can return std::make_unique<Derived> from a function declared to return std::unique_ptr<Base> is that there is a conversion from one to the other.

所以1)等效于:

std::unique_ptr<Base> GetDerived() {
    return std::unique_ptr<Base>(std::made_unique<Derived>());
}

由于从std::make_unique返回的值是一个右值，因此返回值是移动构造的.

Since the value returned from std::make_unique is an rvalue, the return value is move-constructed.

将其与2)进行对比，等效于:

Contrast that to 2), which is equivalent to:

std::unique_ptr<Base> GetDerived2() { 
    std::unique_ptr<Derived> a = std::make_unique<Derived>(); 
    return std::unique_ptr<Base>(a); 
}

由于a是左值，因此返回值必须是复制构造的，并且std::unique_ptr是不可复制的.

since a is an lvalue, the return value must be copy-constructed, and std::unique_ptr is non-copyable.

3)之所以起作用，是因为将左值a转换为右值，并且返回值可以移动构造.

3) works because you cast the lvalue a to an rvalue, and the return value can be move-constructed.

4)和5)之所以起作用，是因为您已经有了一个std::unique_ptr<Base>并且不需要构造一个来返回.

4) and 5) work because you already have a std::unique_ptr<Base> and don't need to construct one to return.

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