push_back到std :: vector,重复调用复制构造函数

Question

有了代码,我有以下输出:

A::A() is called
test #1
A::A(const A & other) is called
test #2
A::A(const A & other) is called
A::A(const A & other) is called
test #3
A::A(const A & other) is called
A::A(const A & other) is called
A::A(const A & other) is called

在调试代码时,对于3个测试用例,我发现第一次复制构造函数的调用是相同的(我认为这是有意义的):制作对象的副本并推送到向量.

但是,通过"_Umove_if_noexcept"对复制构造函数进行额外调用.

对于测试#2,当vec已经有一个条目时,它将进一步调用复制构造函数的一次.

对于测试#3,当vec已经有两个条目时,它将进一步调用复制构造函数的两倍.

这在Visual Studio 2017和gcc 5.4.0上是可重现的.

为什么会这样？是否存在性能问题？

谢谢

#include <iostream>
#include <vector>
class A
{
public:
    //constructor
    A()
    {
        a = 10;
        std::cout << "A::A() is called" << std::endl;
    }

    //copy constructor
    A(const A& other) : a(other.a)
    {
        std::cout << "A::A(const A & other) is called" << std::endl;
    }

    //assignment operator
    A& operator=(const A& other)
    {
        std::cout << "A::operator=(const A & other) is called" << std::endl;
        a = other.a;
        return *this;
    }
public:
    int a;
};

int main()
{
    std::vector<A> vec;
    //A::A() is called
    A a;

    std::cout << "test #1" << std::endl;
    //A::A(const A & other) is called by push_back
    vec.push_back(a);

    std::cout << "test #2" << std::endl;
    //A::A(const A & other) is called 
    //A::A(const A & other) is called from _Umove_if_noexcept
    vec.push_back(a);

    std::cout << "test #3" << std::endl;
    //A::A(const A & other) is called
    //A::A(const A & other) is called from _Umove_if_noexcept
    //A::A(const A & other) is called from _Umove_if_noexcept
    vec.push_back(a);   

     std::cin.get();

    return 0;
}

Answer 1

首先,你必须记住,除非你为向量保留内存,否则它需要分配和重新分配内存,因为它的容量需要增加.

重新分配基本上分配新内存,然后将向量中的元素复制到新内存.

此外,当您将值推入向量时,该值需要复制到向量中.您可以通过放置值(这意味着它在向量中构造到位)或通过在将其推回时移动值来避免这种情况.

Answer 2

第一个副本是明确的 - push_back()需要复制对象.

在第二个副本中,发生了两件事:

1. 首先,因为向量不包含足够的空间,所以需要扩展,并且在该操作期间,它需要复制所有存储的对象
2. 第二,复制对象