移动构造函数没有在C++中调用0x

请在下面找到我的代码，我曾经调用移动构造函数（代码从其他网站启发），并让我知道它有什么问题，我正在使用GCC 4.5.3移动构造函数没有在C++中调用0x

#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

当我们使用std ::移动，同时施工，它是调用移动构造函数。

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和方形的返回值变为右值而构建的对象，因为我们找不到地址空间或引用他们，

请让我知道什么是错的我的理解，如果不是那么为什么移动构造函数不被调用。

在此先感谢。 Satya

来源

2011-10-04 Satya Ravipati

显然，必须在square和getMove_Constructor之内调用采用int的构造函数。显然，在Int_Smart_Pointer d(square(c))和Int_Smart_Pointer e(getMove_Constructor())中，没有其他构造函数被调用，因为它们可以被编译器优化掉。

来源

2011-10-04 09:56:29 Henrik

谢谢你henrik， Int_Smart_Pointer（int && n）{ std :: cout <<“移动构造函数：”<< std :: endl; m_p = new int（n）; } 我得到一个错误和显然，需要将现有构造函数（int n）更改为（int＆n）并解决问题。 –

当通过值返回一个局部变量时，编译器仍然允许像C++ 03（返回值优化）那样复制构造函数。

来源

2011-10-04 09:59:58 visitor

移动构造函数没有在C++中调用0x

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