c++ - 如何在C++ Lambda中捕获 vector 中的元素？

Question

场景:
我用Lambda初始化了一个实例，其中Lambda可以通过按引用捕获实例来修改实例状态。
当我将实例放入 vector 中并调用Lambda时，Lambda会修改原始初始化的实例，而不是 vector 中的实例。
最小的C++程序来说明:

#include <iostream>
#include <functional>
#include <vector>

struct BoolState {
    bool _stateChangeByInstanceReference = false;
    bool _stateChangeByContainerReference = false;
    std::function<void()> _changeStateByInstanceReference;
    std::function<void()> _changeStateByContainerReference;
};

int main()
{
    using namespace std;
    vector<BoolState> bss;
    BoolState bs;
    bs._changeStateByInstanceReference = [&bs]() -> void { bs._stateChangeByInstanceReference = true; };
    bs._changeStateByContainerReference = [&bss]() -> void { bss.at(0)._stateChangeByContainerReference = true; };
    bss.emplace_back(bs);
    bss.back()._changeStateByInstanceReference();
    bss.back()._changeStateByContainerReference();
    cout << "The original instance by instance reference: " << bs._stateChangeByInstanceReference << endl;
    cout << "The original instance by container reference? " << bs._stateChangeByContainerReference << endl;
    cout << "The container instance by instance reference? " << bss.back()._stateChangeByInstanceReference << endl;
    cout << "The container instance by container reference: " << bss.back()._stateChangeByContainerReference << endl;
    /*
        The original instance by instance reference: 1
        The original instance by container reference? 0
        The container instance by instance reference? 0
        The container instance by container reference: 1
    */
}

题:

由于我是在Lambda中通过引用捕获实例的，因此
原始初始化实例和vector中的实例均为2
不同的实例？ (我可以打印出这2个实例在内存中有2个不同的地址，但我不明白:为什么是这种情况？这将如何影响结果？)

是否可以通过Lambda捕获原始引用来修改容器实例状态，反之亦然？

根据@Asteroids With Wings，@ idclev 463035818和@Caleth的解决方案:
非常感谢大家!这对我很有帮助，可以了解这里发生的情况以及如何解决它。
根据@Asteroids With Wings的push_back()版本:

int main()
{
    using namespace std;
    vector<BoolState> bss;
    BoolState bs;
    bs._changeStateByContainerReference = [&bss]() -> void { bss.at(0)._stateChangeByContainerReference = true; };
    // Only copy container version Lambda
    bss.push_back(bs);
    // Modify in place with instance version Lambda
    BoolState& bsInbss = bss.at(0);
    bss.at(0)._changeStateByInstanceReference = [&bsInbss]() -> void { bsInbss._stateChangeByInstanceReference = true; };
    bss.at(0)._changeStateByInstanceReference();
    bss.at(0)._changeStateByContainerReference();
    cout << "The container instance by instance reference:(solved) " << bss.back()._stateChangeByInstanceReference << endl;
    cout << "The container instance by container reference: " << bss.back()._stateChangeByContainerReference << endl;
}

@idclev 463035818的“真实” emplace_back()版本

int main() {
    using namespace std;
    vector<BoolState> bss;
    bss.emplace_back(BoolState());
    BoolState& bsInbss = bss.at(0);
    bss.at(0)._changeStateByInstanceReference = [&bsInbss]() -> void { bsInbss._stateChangeByInstanceReference = true; };
    bss.at(0)._changeStateByContainerReference = [&bss]() -> void { bss.at(0)._stateChangeByContainerReference = true; };
    bss.at(0)._changeStateByInstanceReference();
    bss.at(0)._changeStateByContainerReference();
    cout << "The container instance by instance reference:(solved) " << bss.back()._stateChangeByInstanceReference << endl;
    cout << "The container instance by container reference: " << bss.back()._stateChangeByContainerReference << endl;
}

感谢@Caleth，我没有尝试再次修改原始实例，如果您仍然对互相引用的方法感兴趣，请尝试一下。

Answer 1

这是真正的最小示例😉:

#include <iostream>
#include <vector>

int main()
{
    std::vector<int> v;
    
    int x = 42;
    v.emplace_back(x);
    
    std::cout << &x << '\n';
    std::cout << &v[0] << '\n';
}

// 0x7ffd008d9d14
// 0x1f7fc20

(live demo)
vector 拥有其内容。他们通过复制您给他们的东西来做到这一点。main中的对象与 vector 中的对象不同。修改一个不会修改另一个。
这并不意味着您不能做您想做的事，但是您必须给 vector 中的对象起一个名字才能捕获它:

BoolState& bsInVector = bss.at(0);
auto func = [&bsInVector]() { /* ... */ };