no match for ‘operator=’ - c++

Question

In the following example, I am trying if I try to return by reference from fun() then it gives me warning as it should.
However when I try to return by value, I am expecting 'operator=()' to get triggered, but it is throwing the following error which I am not able to figure out why:
error: no match for ‘operator=’ (operand types are ‘Auto_ptr2<Resource>’ and ‘Auto_ptr2<Resource>’)
note: no known conversion for argument 1 from ‘Auto_ptr2<Resource>’ to
‘Auto_ptr2<Resource>&’
Surprisingly, res2=res1 invoked 'operator=()' just fine!!!
template<class T>
class Auto_ptr2
{
T* m_ptr;
public:
Auto_ptr2(T* ptr=nullptr)
:m_ptr(ptr)
{std::cout<<"\nAuto_ptr CTOR called ";}
~Auto_ptr2()
{
std::cout<<"\n~Auto_ptr2 called\n";
delete m_ptr;
}
// A copy constructor that implements move semantics
Auto_ptr2(Auto_ptr2&
{
std::cout<<"\nAuto_PTR2 copy constructor called";
m_ptr = a.m_ptr;
a.m_ptr = nullptr;
}
Auto_ptr2& operator=(Auto_ptr2& a)
{
std::cout<<"\nAuto_ptr2 operator = called";
if (&a == this)
return *this;
delete m_ptr;
m_ptr = a.m_ptr;
a.m_ptr = nullptr;
return *this;
}
T& operator*() const { return *m_ptr; }
T* operator->() const { return m_ptr; }
bool isNull() const { return m_ptr == nullptr; }
};
class Resource
{
public:
Resource() { std::cout << "\nResource acquired"; }
~Resource() { std::cout << "\nResource destroyed"; }
};
Auto_ptr2<Resource> fun(Auto_ptr2<Resource> res3)
{
std::cout << "\nres1 is " << (res3.isNull() ? "null" : "not null");
return res3;
}
int main()
{
Auto_ptr2<Resource> res1(new Resource);
Auto_ptr2<Resource> res2;
res2 = res1; // res2 assumes ownership, res1 is set to null
res2=fun(res2); //ERROR : error: no match for ‘operator=’ (operand types are ‘Auto_ptr2<Resource>’ and ‘Auto_ptr2<Resource>’)
return 0;
}

Answer 1

If I understand correctly, you're writing something similar to std::unique_ptr.
So, as for std::unique_ptr, the operator=() should works with move semantics, so receiving a r-vale reference, Auto_ptr &&, not an l-value reference
Auto_ptr2& operator=(Auto_ptr2 && a)
{
std::cout<<"\nAuto_ptr2 operator = called";
if (&a == this)
return *this;
delete m_ptr;
m_ptr = a.m_ptr;
a.m_ptr = nullptr;
return *this;
}
And the use should pass through std::move()
res2=fun(std::move(res2));
Same problem with constructors: avoid copy constructor (maybe delete it) and write a move constructor.
Auto_ptr2 (Auto_ptr2 const &) = delete;
Auto_ptr2 (Auto_ptr2 && a)
{
std::cout<<"\nAuto_PTR2 move constructor called";
m_ptr = a.m_ptr;
a.m_ptr = nullptr;
}