关于 C++ 中的右值引用

Question

std::vector<std::string> func(int num) {
  std::vector<std::string> vec;
  int sum = 0;
  for(int i = 0; i < num; i++) {
    sum +=i;
    std::string s = std::to_string(sum);
    vec.emplace_back(s);
  }
  return vec;
}

what's the difference between the code below about using rvalue and lvalue reference.

  std::vector<std::string> res = func(10);      // (case 1)
  std::vector<std::string> &&res = func(10);    // (case 2)
  std::vector<std::string> &res = func(10);   // I got an error!, case3
  const std::vector<std::string> &res = func(10);  // case4

Question:

1. can the rvalue reference(case 2) save a memory copy? than case1
2. why lvalue reference(case3) got an error but it work with the const (case4)?

Answer 1

1. When a prvalue (a "temporary") is bound to an rvalue reference or a const lvalue reference, the lifetime of the temporary is extended to the lifetime of the reference.
2. a non- const lvalue reference does not extend the lifetime of a prvalue.

can the rvalue reference(case 2) save a memory copy? than case1

It preserves the lifetime of the single returned object. See 1.

why lvalue reference(case3) got an error

See 2.

but it work with the const (case4)?

See 1.

Answer 2

1. can the rvalue reference(case 2) save a memory copy? than case1

If you mean, "does case 2 avoid some copy that case 1 has", then the answer is: No. Both are effectively identical if res is an automatic variable. I recommend writing the case 1 because it is simpler and less confusing.

Firstly, there is a move from vec to the returned value. In practice, this move can be elided by an optimiser. Such optimisation is known as NRVO (Named Return Value Optimisation).

Since C++17: The is initialised directly by the move mentioned above (which may have been elided) in both cases.

Pre-C++17: There is another move from the return value to the initialised object in both cases. This move can also be elided in practice.

2. why lvalue reference(case3) got an error but it work with the const (case4)?

Because lvalue references to non-const may not be bound to rvalues.