c++ - C++中的移动构造函数和复制构造函数

Question

我的理解是，当我们从函数返回局部对象时，如果移动构造函数存在，它就会被调用。但是，我遇到了调用复制构造函数的情况，如下面函数 foo2() 中的示例所示。为什么会这样？

#include <cstdio>
#include <memory>
#include <thread>
#include <chrono>

class tNode
{
public:
    tNode(int b = 10)
    {
        a = b;
        printf("a: %d, default constructor %s() is called at %s:%d \n", a, __func__, __FILE__, __LINE__);
    }

    tNode(const tNode& node)
    {
        a = node.a;
        printf("a: %d, copy constructor %s() is called at %s:%d \n", a, __func__, __FILE__, __LINE__);
    }

    tNode& operator=(const tNode& node)
    {
        a = node.a;
        printf("a: %d, copy assignment %s() is called at %s:%d \n", a, __func__, __FILE__, __LINE__);
    }

    tNode(tNode&& node)
    {
        a = node.a;
        printf("a: %d, move constructor %s() is called at %s:%d \n", a, __func__, __FILE__, __LINE__);
    }

    tNode& operator=(tNode&& node)
    {
        a = node.a;
        printf("a: %d, move assignment %s() is called at %s:%d \n", a, __func__, __FILE__, __LINE__);
    }

    ~tNode() { printf("a: %d, destructor %s() is called at %s:%d \n", a, __func__, __FILE__, __LINE__); }

private:
    int a = 0;
};

tNode foo()
{
    tNode node;
    return node;
}

tNode foo2()
{
    std::unique_ptr<tNode> up = std::make_unique<tNode>(20);
    return *up;
}

int main()
{
    {
        tNode n1 = foo();
        tNode n2 = foo2();
    }

    // we pause here to watch how objects are created, copied/moved, and destroyed.
    while (true)
    {
        std::this_thread::sleep_for(std::chrono::seconds(1));
    }

    return 0;
}

上面的代码是用g++ --std=c++17 -fno-elide-constructors编译的输出是:

a: 10, default constructor tNode() is called at testCopyControl.cpp:13
a: 10, move constructor tNode() is called at testCopyControl.cpp:31
a: 10, destructor ~tNode() is called at testCopyControl.cpp:40
a: 10, move constructor tNode() is called at testCopyControl.cpp:31
a: 10, destructor ~tNode() is called at testCopyControl.cpp:40
a: 20, default constructor tNode() is called at testCopyControl.cpp:13
a: 20, copy constructor tNode() is called at testCopyControl.cpp:19
a: 20, destructor ~tNode() is called at testCopyControl.cpp:40
a: 20, move constructor tNode() is called at testCopyControl.cpp:31
a: 20, destructor ~tNode() is called at testCopyControl.cpp:40
a: 20, destructor ~tNode() is called at testCopyControl.cpp:40
a: 10, destructor ~tNode() is called at testCopyControl.cpp:40

从输出中，我们知道当foo2()返回*up时调用了一个拷贝构造函数来初始化一个临时的tNode对象；为什么没有调用移动构造函数？

Answer 1

tNode foo()
{
    tNode node;
    return node;
}

和

tNode n1 = foo();

负责输出

a: 10,  tNode() is called at testCopyControl.cpp:13
a: 10,  move constructor tNode() is called at testCopyControl.cpp:31
a: 10, destructor ~tNode() is called at testCopyControl.cpp:40
a: 10,  move constructor tNode() is called at testCopyControl.cpp:31
a: 10, destructor ~tNode() is called at testCopyControl.cpp:40

您看到的是默认构造函数被调用，然后 node 开始在 return 语句中被视为右值以将其移动到返回值中，然后从返回值移动到 n1

与

tNode foo2()
{
    std::unique_ptr<tNode> up = std::make_unique<tNode>(20);
    return *up;
}

行为不同，因为您没有返回函数局部对象。 *up 为您提供了一个 tNode&，因此 return 语句不能将其视为右值。因为它是一个左值，所以你必须调用复制构造函数将它复制到返回值中。然后，与第一个示例一样，调用移动构造函数将对象从返回值移动到 n2。