给定lambda捕获，什么规则决定了生成的闭包成员的类型？

Question

如何从捕获中准确预测将在lambda中创建哪种类型的成员？

在C ++中，我认为按值捕获T类型的对象会创建类型为const T的数据成员，并引用T&。但在编译时：

#include <iostream>

struct A{
    A(){std::cout<<"A\n";}
    A(const A&){std::cout<<"A&\n";}
    void cf()const{}
    void f(){}
};


int main(){
    A a;
    A& ra=a;
    const A& cra=a;
    auto f00 = [ra, cra, &a]()-> void{
        //Fixed:
        //ra is A, cra is const A, a is A&
        //lambda is void operator()()const
        a.cf(); a.f();//pass
        //ra.cf(); ra.f();//ra.f compilation err. 
        //cra.cf(); cra.f();//cra.f compilation err
    };
    //f00(); //A&,A&

    auto f01 = [ra, cra, &a]()mutable-> void{
        //Fixed:
        //ra is A, cra is const A, a is A&
        //lambda is void operator()()mutalbe
        a.cf(); a.f();//pass
        ra.cf(); ra.f();//pass
        cra.cf(); cra.f();//cra.cf pass, but cra.f error, why?
    };
    //f01(); //A&,A&

    auto f02 = [&ra, &cra, &a]()mutable-> void{
        //Fixed:
        //ra is A&, cra is const A&, a is A&
        //lambda is void operator()()mutable
        a.cf(); a.f();//pass
        ra.cf(); ra.f();//pass
        //cra.cf(); cra.f();//cra.cf pass, but cra.f error, why?
    };
    f02(); //
    return 0;
}

我遇到了以下编译错误：

test_lambda.cpp:26:25: error: passing 'const A' as 'this' argument discards qualifiers [-fpermissive]
         cra.cf(); cra.f();//pass, cra.f error
                         ^
test_lambda.cpp:8:10: note:   in call to 'void A::f()'
     void f(){}
          ^

这是否意味着cra确实是通过引用捕获的，而不是像我预期的那样被引用的对象的副本？