C++11 中的移动语义

Question

我想完全理解 C++11 中的移动语义。所以我写了几个类来查看何时调用不同的构造函数：

#include <iostream>
using namespace std;

class A {
public:
    A() : a1_(0) {std::cout << "Calling constructor" << std::endl;}
    A(A&& other) {
        std::cout << "Calling move constructor" << std::endl;
        a1_ = other.a1_;
        other.a1_ = 0;
    }

    // Move assignment operator.
    A& operator=(A&& other) {
        std::cout << "Calling move operator" << std::endl;
        if (this != &other) {
            a1_ = other.a1_;
            other.a1_ = 0;
        }
        return *this;
    }

    // Copy constructor.
    A(const A& other) {
        std::cout << "Calling copy constructor" << std::endl;
        a1_ = other.a1_;
    }

    // Copy assignment operator.
    A& operator=(const A& other) {
        std::cout << "Calling copy assignment operator" << std::endl;
        if (this != &other) {
            a1_ = other.a1_;
        }
        return *this;
    }

private:
    int a1_;
};

class B {
    A oA_;

public:
    B() {}
    void setoA(A a) {oA_ = a;}
        A getoA() {return oA_;}
};

A createA() {
    A a1;
    return a1;
}

B createB() {
    B tmpB;
    A tmpA;
    tmpB.setoA(tmpA);
    return tmpB;
}

int main() {
    B b;
    A a;
    b.setoA(a);
    std::cout << "**************************" << std::endl;
    b.setoA(createA());
    std::cout << "**************************" << std::endl;
    b.setoA(std::move(createA()));
    std::cout << "**************************" << std::endl;
    B b2;
    b2.setoA(b.getoA());
    std::cout << "**************************" << std::endl;
    createB();

    return 0;
}

当我检查这段代码的输出时：

    Calling constructor

    Calling constructor

    Calling copy constructor

    Calling copy assignment operator

    ++++++++++++++++++++++++++++++++++

    Calling constructor

    Calling copy assignment operator

    ++++++++++++++++++++++++++++++++++

    Calling constructor

    Calling move constructor

    Calling copy assignment operator

    ++++++++++++++++++++++++++++++++++

    Calling constructor

    Calling copy constructor

    Calling copy assignment operator

    ++++++++++++++++++++++++++++++++++

    Calling constructor

    Calling constructor

    Calling copy constructor

    Calling copy assignment operator

我这里有些疑惑：

我以为如果你传递r-value，移动构造函数就会被调用，是吗？这 b.setoA(createA()); 不是 r-value 吗？

我怎样才能让 constructor/operator 被跟注？

Answer 1

First of all in first section, why is constructor being called twice?

因为您构造了一个 B 和一个 A，前者有自己的 A 实例，第一个（意外的）构造函数调用来自该实例。

I thought that if you pass r-value move constructor will be called, is that right? Isn't this b.setoA(createA()); an r-value?

从 createA 中调用构造函数（是的，return 值是 r 值），但是，复制省略发生并且在setoA.

的参数变量中直接实例化对象然而，

在 setoA 中，选择了复制赋值，因为现在 a 是一个左值。如果您想搬家，您需要：

void setoA(A a) { oA_ = std::move(a); }

Answer 2

编译器可能会选择性地省略某些副本和移动。为防止这种情况，请在 GCC 和 Clang 中使用 -fno-elide-constructors。此外，一些移动省略在 C++17 中成为强制性的，因此要强制编译器使用 C++11 移动语义，请同时使用 -std=c++11。

C++11 中的移动语义

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