如何在lambda中捕获此对象的变量？

Question

I have seen many answers on SO asking about capturing this by reference but I have a different question. What if I want to capture a specific variable owned by this object?

For example:

auto rel_pose = [this->_last_pose["main_pose"],&pose](Eigen::VectorXd pose1, Eigen::VectorXd pose2)
    {
        // Some code
        return pose;
    };

I want to capture the specific variable of this by value and use it inside my lambda expression. Why this is not possible?

Answer 1

It's possible:

struct S
{
    int i = 7;
    char c = 0;
};

int main(int argc, char* argv[])
{
    S s;
    auto l = [integer = s.i]() {
        return integer;
    };

    return l();
}

Answer 2

You can apply by-copy capture with an initializer (since C++14) (or by-reference capture with an initializer, depends on your demand), eg

auto rel_pose = [some_pose = this->_last_pose["main_pose"], &pose](Eigen::VectorXd pose1, Eigen::VectorXd pose2)
{
    // Some code using some_pose
    return pose;
};

Note that we can only capture identifiers in lambda, we can't capture expressions like this->_last_pose["main_pose"] directly. The captures with an initializer just solve such issues straightforward.

Answer 3

Why this is not possible?

It's possible, like other answers show. But you must do it explicitly . Accessing any member of the current object in the lambda is transformed automatically to an access through the this pointer. When you write a plain [this->_last_pose["main_pose"],&pose] , what's really captured is this , and the access to _last_pose goes through it.

It's simply how lambda captures are specified for member variables. Be grateful you are compiling C++14. In C++11 capturing members by value was not as easy as adding an init capture that does an copy.