c++ - 在 C++0x 中未调用移动构造函数

Question

请在下面找到我的代码，我曾经调用移动构造函数(代码灵感来自其他站点)并让我知道它有什么问题，我使用的是 GCC 4.5.3

#include <iostream>
#include <vector>

class Int_Smart_Pointer {

  int *m_p;

public:
  Int_Smart_Pointer() {
    std::cout<<"Derfault Constructor"<< std::endl;
    m_p = NULL;
  }

  explicit Int_Smart_Pointer(int n) {
    std::cout<<"Explicit Constructor: " << n <<std::endl;
    m_p = new int(n);
  }

  Int_Smart_Pointer(const Int_Smart_Pointer& other) {
    std::cout<<"Copy Constructor: "<<std::endl;
    if(other.m_p)
      m_p = new int(*other.m_p);
    else
      m_p = NULL;
  }

  Int_Smart_Pointer(Int_Smart_Pointer&& other) {
    std::cout<<"Move Constructor: "<<std::endl;
    m_p = other.m_p;
    other.m_p = NULL;
  }

  Int_Smart_Pointer& operator=(const Int_Smart_Pointer& other) {
    std::cout<<"Copy Assignment"<< std::endl;
    if(this != &other) {
         delete m_p;
         if(other.m_p)
           m_p = new int(*other.m_p);
         else
           m_p = NULL;
    }

      return *this;
  }

  Int_Smart_Pointer& operator=(Int_Smart_Pointer&& other) {
    std::cout<<"Move Assignment"<< std::endl;
    if(this != &other) {
      delete m_p;
      m_p = other.m_p;
      other.m_p = NULL;
    }

      return *this;
  }

  ~Int_Smart_Pointer() {
    std::cout<<"Default Destructor"<< std::endl;
    delete m_p;
  }

  int get() const{
    if(m_p)
      return *m_p;
    else
      return 0;
  }

};

Int_Smart_Pointer square(const Int_Smart_Pointer& r) {
    const int i = r.get();
    return Int_Smart_Pointer(i * i);
}

Int_Smart_Pointer getMove_Constructor() {
  return Int_Smart_Pointer(100);
}


int main()
{
  Int_Smart_Pointer a(10);
  Int_Smart_Pointer b(a);
  b = square(a);

  std::cout<< std::endl;

  Int_Smart_Pointer c(30);
  Int_Smart_Pointer d(square(c)); //Move constructor Should be called (rvalue)

  std::cout<< std::endl;

  Int_Smart_Pointer e(getMove_Constructor()); //Move constructor Should be called (rvalue)


  std::cout<< std::endl;

  std::vector<Int_Smart_Pointer> ptr_vec;
  ptr_vec.push_back(getMove_Constructor());
  ptr_vec.push_back(getMove_Constructor());

  std::cout<< std::endl;

  return 0;
}

输出是

Explicit Constructor: 10
Copy Constructor:
Explicit Constructor: 100
Move Assignment
Default Destructor

Explicit Constructor: 30
Explicit Constructor: 900

Explicit Constructor: 100

Explicit Constructor: 100
Move Constructor:
Default Destructor
Explicit Constructor: 100
Move Constructor:
Move Constructor:
Default Destructor
Default Destructor

Default Destructor
Default Destructor
Default Destructor
Default Destructor
Default Destructor
Default Destructor
Default Destructor

当我们在构造时使用 std::move 时，它​​正在调用移动构造函数。

Int_Smart_Pointer d(std::move(square(c))); //used std::move and Move constructor called
Int_Smart_Pointer e(std::move(getMove_Constructor())); //used std::move works as above

但是即使我们不使用，gerMove_Constructor 和 square 返回值在构造对象时也会变成右值，因为我们找不到地址空间或对它们的引用，

如果我的理解有问题，请告诉我，如果没有，那么为什么不调用移动构造函数。

提前致谢。萨提亚

Answer 1

当按值返回局部变量时，编译器仍然允许像 C++03 中那样省略复制构造函数(返回值优化)。