How Can I avoid implicit move constructor inside a lambda function - c++

Question

I am using "emplace" method to avoid memory copy.
But, when I am using the "emplace" inside a Lambda function. It always call implicit move constructor.
How I can avoid memory copy inside a Lambda function?
This sample program should not print “I am being moved.”
#include <vector>
#include <iostream>
struct A
{
int a;
A(int t) : a(t)
{
std::cout << "I am being constructed.\n";
}
A(A&& other) : a(std::move(other.a))
{
std::cout << "I am being moved.\n";
}
};
std::vector<A> g_a;
int main()
{
std::cout << "emplace_back:\n";
g_a.emplace_back(1);
std::cout << "emplace_back in lambda:\n";
auto f1 = [](int x) { g_a.emplace_back(x); };
f1(2);
std::cout << "\nContents: ";
for (A const& t : g_a)
std::cout << t.a << " ";
std::cout << std::endl;
}

Answer 1

It's not about the lambda function but rather about the vector that reallocates its memory. You can ammend this with std::vector::reserve.
int main() {
g_a.reserve(10);
^^^^^^^^^^^^^^^^
std::cout << "emplace_back:\n";
g_a.emplace_back(1);
std::cout << "emplace_back in lambda:\n";
auto f1 = [](int x) { g_a.emplace_back(x); };
f1(2);
std::cout << "\nContents: ";
for (A const& t : g_a)
std::cout << t.a << " ";
std::cout << std::endl;
}
Live Demo