请在下面找到我的代码,我曾经调用移动构造函数(代码受其他站点启发)并让我知道它有什么问题,我正在使用 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
当按值返回局部变量时,仍然允许编译器像 C++03 中那样省略复制构造函数(返回值优化)。
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