Question

为什么调用emplace_back（）后立即调用〜Base（）
为什么在析构函数调用后可以访问sayHello（）
为什么〜Base（）被再次调用

#include <iostream>
#include <vector>

class Base
{
    private:

        static int m_count;

    public:

        Base()
        {
            std::cout << " Base created. Count = " << ++m_count << std::endl;
        }

        ~Base()
        {
            std::cout << " Base destroyed. Count = " << --m_count << std::endl;
        }

        void sayHello() const
        {
            std::cout << " Base says hello" << std::endl;
        }
};

int Base::m_count = 0;

int main()
{
    {
        std::vector< Base > vBase;

        vBase.emplace_back ( Base() );  // <- Why does ~Base() get called here

        vBase[0].sayHello(); // <- Why is this function accessible after call to dtor
    }
    return 0;
}

程序输出...

Base created. Count = 1  
Base destroyed. Count = 0  
Base says hello
Base destroyed. Count = -1

Answer 1

您错过了重点。 Emplace函数从给定的参数就地构造对象，而不是例如push_back从先前存在的对象复制构造它。您应该已经编写了vBase.emplace_back()，它可以在向量内部就位构造对象，而没有构造函数参数（即默认构造）。

就目前情况而言，您实际上是通过Base默认构造一个Base()对象，并将其传递给emplace，这会调用带有Base对象的构造函数（即move构造函数），复制它，然后破坏原始对象Base()。

其副本仍在引导程序中，这就是为什么您仍然可以访问它的原因。被摧毁的是临时的。第二个析构函数调用是当向量超出范围时销毁副本。

因此，您基本上只是在做与push_back相同的事情。

Answer 2

在调用vBase.emplace_back ( Base() );中，您首先创建一个Base对象。该向量在适当的位置创建了另一个Base，第一个Base拥有的资源随后被移到新的资源中。然后删除第一个碱基。在向量中，您现在有了一个移动的构造Base，这就是对sayHello()的调用起作用的原因。

您可能想要做的是让emplace_back实际构造对象，而无需手动创建临时对象。通过仅提供构造Base所需的参数来做到这一点。像这样：

vBase.emplace_back();

Answer 3

您不会在计数器中包括由move和copy构造函数创建的对象，也不会记录调用。如果您更改日志记录以解决违反“三五规则”的情况，则会看到以下内容：

#include <typeinfo>
#include <iostream>

/// noisy
///
/// A class logs all of the calls to Big Five and the default constructor
/// The name of the template parameter, received as if by call
/// to `typeid(T).name()`, is displayed in the logs.
template<typename T>
struct noisy
{
    noisy& operator=(noisy&& other) noexcept { std::cout << "MOVE ASSIGNMENT<" << typeid(T).name() << ">(this = " << this << ", other = " << &other << ")\n"; return *this; }
    noisy& operator=(const noisy& other) { std::cout << "COPY ASSIGNMENT<" << typeid(T).name() << ">(this = " << this << ", other = " << &other << ")\n"; return *this; }
    noisy(const noisy& other) { std::cout << "COPY CONSTRUCTOR<" << typeid(T).name() << ">(this = " << this << ", other = " << &other << ")\n"; }
    noisy(noisy&& other) noexcept { std::cout << "MOVE CONSTRUCTOR<" << typeid(T).name() << ">(this = " << this << ", other = " << &other << ")\n"; }
    ~noisy() { std::cout << "DESTRUCTOR<" << typeid(T).name() << ">(this = " << this << ")\n"; }
    noisy() { std::cout << "CONSTRUCTOR<" << typeid(T).name() << ">(this = " << this << ")\n"; }
};

#include <iostream>
#include <vector>

class Base : public noisy<Base>
{
    public:

        void sayHello() const
        {
            std::cout << "Base says hello" << "(this = " << this << ")" << std::endl;
        }
};

int main()
{
    {
        std::vector< Base > vBase;

        vBase.emplace_back ( Base() );  // <- Why does ~Base() get called here

        vBase[0].sayHello(); // <- Why is this function accessible after call to dtor
    }
    return 0;
}

输出：

CONSTRUCTOR<4Base>(this = 0x7fff300b580f)
MOVE CONSTRUCTOR<4Base>(this = 0x18a6c30, other = 0x7fff300b580f)
DESTRUCTOR<4Base>(this = 0x7fff300b580f)
Base says hello(this = 0x18a6c30)
DESTRUCTOR<4Base>(this = 0x18a6c30)

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为什么emplace_back（）的行为如此？

3 个答案: