`threads[i] = thread(exec_produce, i + 1)`这行代码为什么不会报错？

Question

void exec_produce(int duration) {
    //阻止线程运行到duration秒
    this_thread::sleep_for(chrono::seconds(duration));
    //this_thread::get_id()获取当前线程id
    cout << "exec_produce thread " << this_thread::get_id()
        << " has sleeped " << duration << " seconds" << endl;
}

int main(int argc, const char *argv[])
{
    thread threads[5];
    cout << "create 5 threads ..." << endl;
    for (int i = 0; i < 5; i++) {
        threads[i] = thread(exec_produce, i + 1);
    }
    cout << "finished creating 5 threads, and waiting for joining" << endl;
    //下面代码会报错,原因就是copy操作不可用，相当于是delete操作，所以报错
    /*for(auto it : threads) {
        it.join();
    }*/
    for (auto& it: threads) {
        it.join();
    }
    cout << "Finished!!!" << endl;
    system("pause");
    return 0;
}

我想知道为什么threads[i] = thread(exec_produce, i + 1);这行代码不会报错，但是这里

    for(auto it : threads) {
        it.join();
    }

就报错了？这两个地方不都是调用了赋值运算符吗？

我知道有下面这两个规则：

1）move 赋值操作：thread& operator= (thread&& rhs) noexcept，如果当前对象不可 joinable，需要传递一个右值引用(rhs)给 move 赋值操作；如果当前对象可被 joinable，则 terminate() 报错。

2）拷贝赋值操作被禁用：thread& operator= (const thread&) = delete，thread 对象不可被拷贝。

对于move可能还有些疑惑，谢谢大家~~