How can I std::move a vector into a lambda function?

Question

#include <iostream>
#include <functional>


int main() {
    std::vector<int> integers = {0, 1, 2};

    std::function<void()> lambda = [integers] {
        printf("vector size (in lambda function): %zu\n", integers.size());
    };

    printf("vector size (in main()): %zu\n", integers.size());

    lambda();

    return 0;
}

In this simple program, vector integers is sent into a lambda function by copy, so it is unchanged after this lambda function created.

Now how do I destroy/clear this vector the same moment it sent into this lambda function? Like when a vector is sent into a function by std::move()?

The closet solution I have so far is to achieve above is:

std::vector<int> integers = {0, 1, 2};

std::function<void()> lambda = [integers] {
    printf("vector size (in lambda function): %zu\n", integers.size());
};
integers.clear();

printf("vector size (in main()): %zu\n", integers.size());

lambda();

But even so, the vector is still cleared AFTER lambda function created.

So how do I destroy this vector the moment when lambda function created?

Answer 1

While C++11 only allows references or copied values in lambda captures, C++14 introduces Lambda capture expressions. You can move your vector in one of these expressions.

#include <cstdio>
#include <functional>
#include <utility>    

int main() {
    std::vector<int> integers = {0, 1, 2};

    std::function<void()> lambda = [integers=std::move(integers)] {
                       // Initializing move ^^^^^^^^^^^^^^^^^^^^
       std::printf("vector size (in lambda function): %zu\n", integers.size());
    };

    std::printf("vector size (in main()): %zu\n", integers.size());

    lambda();

    return 0;
}

See it work.

vector size (in main()): 0
vector size (in lambda function): 3