相关疑难解决方法(0)

在这种情况下

struct Foo {};
Foo meh() {
  return std::move(Foo());
}

我很确定移动是不必要的,因为新创建的Foo将是一个xvalue.

但在这种情况下呢？

struct Foo {};
Foo meh() {
  Foo foo;
  //do something, but knowing that foo can safely be disposed of
  //but does the compiler necessarily know it?
  //we may have references/pointers to foo. how could the compiler know?
  return std::move(foo); //so here the move is needed, right?
}

我认为需要采取行动吗？