Why arguments moved twice when constructing std::thread

Question

Consider this program that essentially creates std::thread that calls the function func() with arg as argument:

#include <thread>
#include <iostream>

struct foo {
   foo() = default;
   foo(const foo&) { std::cout << "copy ctor" << std::endl; }
   foo(foo&&) noexcept { std::cout << "move ctor" << std::endl; }
};

void func(foo){}

int main() {
   foo arg;
   std::thread th(func, arg);
   th.join();
}

My output is

copy ctor
move ctor
move ctor

As far as I understand arg is copied internally in the thread object and then passed to func() as an rvalue (moved). So, I expect one copy construction and one move construction.

Why is there a second move construction?

Ah, a dupe in fact. Shame it can't be closed as such due to having no answers (because n.m. wrote their answer in the comments section for some reason) — Lightness Races in Orbit

bipll bipll · Accepted Answer · 2020-01-05T17:32:23

You pass argument to func by value which should constitute the second move. Apparently std::thread stores it internally one more time before calling func, which AFAIK is absolutely legal in terms of the Standard.

Why arguments moved twice when constructing std::thread

2 Answers