标准::向量<t> ::emplace_back 在右值引用上</t>

Question

I have the following function inside a class

  void add_state(std::string &&st) {
    state.emplace_back(st);           // state is a vector
  }

st is an l-value (rvalue reference to a string in this case) based on my understanding. If, I want to move st into the last element in state , should I be using state.emplace_back(std::move(st)) ? What happens if I leave it the way it's written above?

EDIT 1 (example of how add_state is called):

// do a series of operations to acquire std::string str
add_state(std::move(str)); 
// also note that str will never be used again after passing it into add_state

Would it be better if I made add_state(std::string &st) instead? In this case, I think I can just simply call it with add_state(str) ?

Answer 1

[...] should I be using state.emplace_back(std::move(st)) ?

Yes.

What happens if I leave it the way it's written above?

You have correctly identified that st is an lvalue, so it gets copied.

Would it be better if I made add_state(std::string &st) instead? In this case, I think I can just simply call it with add_state(str) ?

You could, but you shouldn't. std::move is just type juggling, it does not do anything by itself (in particular, it doesn't move anything). The actual operation happens inside std::string::basic_string(std::string &&) which is called by emplace_back after it receives the std::string && . Making your own parameter an lvalue reference has no effect other than surprising the caller, who gets their std::string eaten without an std::move on their part.