为什么不是这里调用的std :: string＆＃39移动构造函数？

Question

我有这段代码：

#include <iostream>
#include <string>

using namespace std;

class S
{
public:

  S(std::string && s)
  : str(std::move(s))
  {
  }

  /* Why does a copy of s happen somewhere in here? */
  S(std::string && s, bool dummyArg)
  : str(s)
  {
  }

  void printVal() const
  {
    cout << "S::s is \"" << str << "\"\n";
  }

private:
  string str;
};


int main(int argc, const char * argv[])
{
  string str1{"Some string that I want to move into S"};
  string str2{"Some other string that I want to move into S"};

  // S s0{str1}; // Does not compile because no constructor matches
  S s1{std::move(str1)};
  S s2{std::move(str2), false};

  s1.printVal();
  cout << "str1 is: \"" << str1 << "\"\n";

  cout << "\n";

  s2.printVal();
  cout << "str2 is : \"" << str2 << "\"\n";
}

产生以下输出:(使用Clang（XCode 9）编译）

S::s is "Some string that I want to move into S"
str1 is: ""

S::s is "Some other string that I want to move into S"
str2 is : "Some other string that I want to move into S"

因此，我们可以清楚地看到，在使用第二个构造函数时，str2并未移至S::str。我不明白为什么添加std::move会产生影响，因为std::move只是对其参数执行static_cast<T&&>，这已经是构造函数的类型。