Why copy constructor is called in capture list of lambda

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

using namespace std;

struct B{
    B(){
        std::cout<<"B"<<std::endl;
    }
    
    B(B&& b2){
        std::cout<<"B&& from object moved"<<b2.moved_<<std::endl;
        b2.moved_ = true;
    }
    
    B(const B&){
        std::cout<<"const B&"<<std::endl;
    }
    
    // ~B(){
    //     std::cout<<"~B"<<std::endl;
    // }
    
    int bb=10;
    bool moved_=false;
};

struct FF{
    FF(std::function<void()> ff):ff_(std::move(ff)){}
    std::function<void()> ff_;
};

int main()
{
    std::vector<int> a{1,2,3};
    std::vector<FF> b{};
    B bb;
    std::transform(a.cbegin(),a.cend(),std::back_inserter(b),[x = std::move(bb)](auto const i){
        
        return FF([xx = std::move(x),i](){
            std::cout<<"FF"<<i<<std::endl;
        });
    });
    
    for(auto const& j:b){
      j.ff_();
    }
    return 0;
}

As you can see from the code , my container b is a vector of FF, it accepts a function object as parameter . I want to confirm how lambda capture the parameter~

My question is why const& constructor will be called when xx=std::move(x) in capture list?

I think there should not be extra constructor.

>Solution :

It’s because the lambda is not mutable, contains const B x that can’t be moved.

Easy fix

[x = std::move(bb)](auto const i) mutable {