Question

给定类NamedType，我想调用move-constructor：

#include <cstdlib>
#include <ctime>
#include <iostream>
#include <utility>

template<typename T, typename PARAMETER>
class NamedType {
 public:
  NamedType() { /*std::cout << "default ";*/ }
  NamedType(T const &value) : value_(value) { /*std::cout << "parametrized ";*/ }

  NamedType(NamedType const &other) : value_(other.value_) { std::cout << "copy "; }
  NamedType(NamedType &&other) noexcept : value_(std::exchange(other.value_,0)) { std::cout << "move "; }

  NamedType& operator=(NamedType const &rhs) { value_ = rhs.value_; std::cout << "copyAssign "; return *this; }
  NamedType& operator=(NamedType &&rhs) { value_ = std::exchange(rhs.value_,0); std::cout << "moveAssign "; return *this; }

  ~NamedType() = default;

  T const & Get() const { return value_; }
 private:
  T value_;
};

using Degree = NamedType<int, struct DegreeParameter>;

Degree GetRandom() {
  return Degree{std::rand() % 100};
}

调用以下脚本时

int main() {
  std::srand(std::time(0));
  Degree degreeCopyAssign, degreeMoveAssign;  // no output, since uninitialized
  Degree degreeParametrized(1);  // std::cout << degreeParametrized.Get() << std::endl;

  Degree degreeCopy(degreeParametrized); std::cout << degreeCopy.Get() << std::endl;
  Degree degreeUnknown(GetRandom()); std::cout << degreeUnknown.Get() << std::endl;  // which constructor is called? why no implicit move?
  Degree degreeMove(std::move(GetRandom())); std::cout << degreeMove.Get() << std::endl;

  degreeCopyAssign = degreeCopy; std::cout << degreeCopyAssign.Get() << std::endl;
  degreeMoveAssign = GetRandom(); std::cout << degreeMoveAssign.Get() << std::endl;  // why implicit move here?

  return 0;
}

未调用move-constructor，但我希望degreeUnknown被调用。在那里叫哪个构造函数？

为什么移动命令隐式地被degreeMoveAssign调用，而不被移动构造函数cf调用。 degreeMove？

该脚本的示例输出为

copy 1
64
move 96
copyAssign 1
moveAssign 7

谢谢：）

未知的构造函数，而不是move-constructor调用

0 个答案: