C++ 在C+;中未调用移动构造函数+;0x
请在下面找到我的代码,我曾经调用move构造函数(源于其他站点的代码),并让我知道它有什么问题,我使用的是GCC4.5.3C++ 在C+;中未调用移动构造函数+;0x,c++,c++11,C++,C++11,请在下面找到我的代码,我曾经调用move构造函数(源于其他站点的代码),并让我知道它有什么问题,我使用的是GCC4.5.3 #include <iostream> #include <vector> class Int_Smart_Pointer { int *m_p; public: Int_Smart_Pointer() { std::cout<<"Derfault Constructor"<< std::endl;
#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;
}
当我们在构造时使用std::move时,它调用的是move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
但即使我们不使用,gerMove_构造函数和平方返回值在构造对象时变为右值,因为我们找不到地址空间或对它们的引用
请让我知道,如果我的理解有问题,如果没有,那么为什么不调用move构造函数
提前谢谢。
Satya显然,必须在
square
和getMove\u构造函数中调用取int
的构造函数。
显然,在Int\u Smart\u指针d(square(c))
和Int\u Smart\u指针e(getMove\u Constructor())
中,没有调用其他构造函数,因为它们可以被编译器优化掉。当按值返回局部变量时,编译器仍然可以像在c++03(返回值优化)中那样省略复制构造函数谢谢henrik,智能指针(Int&&n){std::cout
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