What is the lifetime of a C++ lambda expression?

Question

(I have read What is the lifetime of lambda-derived implicit functors in C++? already and it does not answer this question.)

I understand that C++ lambda syntax is j

Answer 1

In the code that you posted:

std::function meta_add(int x) {
    auto add = [x](int y) { return x + y; };
    return add;
}

The std::function object that is returned by the function actually holds a moved instance of the lambda function object that was assigned to local variable add.

When you define a C++11 lambda that captures by-value or by-reference, the C++ compiler automatically generates a unique functional type, an instance of which is constructed when the lambda is called or assigned to a variable. To illustrate, your C++ compiler might generate the following class type for the lambda defined by [x](int y) { return x + y; }:

class __lambda_373s27a
{
    int x;

public:
    __lambda_373s27a(int x_)
        : x(x_)
    {
    }

    int operator()(int y) const {
        return x + y;
    }
};

Then, the meta_add function is essentially equivalent to:

std::function meta_add(int x) {
    __lambda_373s27a add = __lambda_373s27a(x);
    return add;
}

EDIT: By the way, I am not sure if you know this, but this is an example of function currying in C++11.