移出某些元素后无法从矢量中删除

Question

你好，我正在尝试理解如何优化从一个向量到另一个向量的移动元素，检查这是不是在调用复制或移动构造函数。

但是，当我试图擦除移出到另一个向量v2上的std::vector v1中的元素时，编译器会报告以下消息：

vector-move-elements.cpp:19:5: note: copy assignment operator is implicitly deleted because 'Foo' has a
      user-declared move constructor
    Foo(Foo&& other) {
    ^

以下是代码

#include <vector>
#include <string>
#include <iostream>
// #include <algorithm> std::copy_if

// inspired from https://stackoverflow.com/questions/15004517/moving-elements-from-stdvector-to-another-one
struct Foo
{
    Foo(std::string&& s, uint32_t idx) {
        this->name = std::move(s);
        this->index = std::move(idx);
        std::cout << "Calling default constructor " << name << idx << std::endl;
    }
    Foo(const Foo& other) {
        this->name = other.name;
        this->index = other.index;
        std::cout << "Calling copy constructor " << other.name << other.index << std::endl;
    }
    Foo(Foo&& other) {
        this->name = std::move(other.name);
        this->index = std::move(other.index);
        std::cout << "Calling move constructor " << other.name << other.index << std::endl;
    }
    std::string name;
    uint32_t index;
};

int main() {
    std::vector<Foo> v1, v2;
    v1.reserve(25); // https://stackoverflow.com/questions/52109542/push-back-to-stdvector-the-copy-constructor-is-repeatedly-called
    v2.reserve(25); // without this the push_back instruction causes a call to copy constructor

    // fill v1 with dummy elements
    for (uint32_t i = 0; i < 25; ++i ) {
        std::string s = std::string("name_") + std::to_string(i);
        std::cout << s << std::endl;
        v1.emplace_back(Foo(std::move(s), i));
    }

    std::cout << "End of construction" << std::endl;

    // populate v1 with at least 17 elements...
    const auto it = std::next(v1.begin(), 17); // same as v1.begin()+17;
    std::move(v1.begin(), it, std::back_inserter(v2));
    // std::copy_if(std::make_move_iterator(begin(v1)),
    //          std::make_move_iterator(end(v1)),
    //          std::back_inserter(v2),
    //          [](const Foo& v){ return v.index <= 17; });

    // this line DOESN'T COMPILE 
    v1.erase(v1.begin(), it); // copy assignment operator is implicitly deleted because 'Foo' has a
      user-declared move constructor

    // if I don't loop on auto& the for loop will call copy constructor
    for (auto const& e : v1) {
        std::cout << "v1: " << e.name << " " << e.index << std::endl;
    }
    for (auto const& e : v2) {
        std::cout << "v2: " << e.name << " " << e.index << std::endl;
    }
}

Answer 1

您需要检查的是three/five/zero的规则。

你需要一个复制赋值操作符来编译你的代码：

Foo& operator=(const Foo& other) {
    this->name = other.name;
    this->index = other.index;
    std::cout << "Calling copy assignment " << other.name << other.index << std::endl;
    return *this;
}

以及符合规则5的移动赋值运算符。

在godbolt上直播