Move constructor is not getting called in C++0x

Question

Please find my code below, I have used to call move constructor (code Inspired from other sites)and let me know whats wrong with it, I am using 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;
}

And the output is

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

When we use std::move while constructing, it is calling move constructor.

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

But even if we do not use, gerMove_Constructor and square return values becomes rvalue while constructing object as we can not find address space or reference to them,

Please let me know if something is wrong in my understanding, if not then why move constructor is not called.

Thanks In Advance. Satya

Answer 1

When returning a local variable by value, the compiler is still allowed to elide copy constructors as in C++03 (return value optimization).