Copy constructor of thread is deleted by design

If we want to create a vector of threads, we cannot do this.

auto thread_function = []() {
  std::cout << "This is thread function" << std::endl;
};

int n = 100;
std::vector<std::thread> threads(n, std::thread(thread_function));

This leads to compilation errors.

error: calling a private constructor of class 'std::__1::thread'
            ::new((void*)__p) _Up(_VSTD::forward<_Args>(__args)...);

It is because the copy constructor of thread is deleted by design. i.e. thread is movable only but not copyable.

thread() noexcept;(since C++11)
thread( thread&& other ) noexcept; (since C++11)
template< class Function, class... Args >
explicit thread( Function&& f, Args&&... args ); (since C++11)
thread(const thread&) = delete; (since C++11)

The correct way is

auto thread_function = []() {
  std::cout << "This is thread function" << std::endl;
};

std::vector<std::thread> threads;
for (int i = 0; i < 100; i++) {
  threads.push_back(thread(thread_function));
}

push_back function of vector has both copyable and movable version, that’s why it compiles.

void push_back( const T& value );
void push_back( T&& value ); (since C++11)

Still you can use std::move to produce the rvalue reference of thread(thread_function).

How about copy assignment operator of thread?
There is no copy assignment operator but move assignment operator in thread class.
The copy assignment operator is generated only for classes lacking an explicitly declared copy assignment operator, and it won’t generated if a move operation is declared.

thread& operator=( thread&& other ) noexcept; (since C++11)