I have stumbled over code today, that I don't understand. Please consider the following example:
#include <iostream>
#include <string>
class A
{
public:
template <class Type>
Type& operator=(Type&& theOther)
{
text = std::forward<Type>(theOther).text;
return *this;
}
private:
std::string text;
};
class B
{
public:
B& operator=(B&& theOther)
{
text = std::forward<B>(theOther).text;
return *this;
}
private:
std::string text;
};
int main()
{
A a1;
A a2;
a2 = a1;
B b1;
B b2;
b2 = b1;
return 0;
}
When compiling, MinGW-w64/g++ 10.2 states:
..\src\Main.cpp: In function 'int main()':
..\src\Main.cpp:41:7: error: use of deleted function 'B& B::operator=(const B&)'
41 | b2 = b1;
| ^~
..\src\Main.cpp:19:7: note: 'B& B::operator=(const B&)' is implicitly declared as deleted because 'B' declares a move constructor or move assignment operator
19 | class B
| ^
mingw32-make: *** [Makefile:419: Main.o] Error 1
I fully understand the error message. But I don't understand why I don't get the same message with class A
. Isn't the templated move assignment operator also a move assignment operator? Why then is the copy assignment operator not deleted? Is this well-written code?
Isn't the templated move assignment operator also a move assignment operator?
No, it's not considered as move assignment operator.
(emphasis mine)
A move assignment operator of class T is a non-template non-static member function with the name
operator=
that takes exactly one parameter of typeT&&
,const T&&
,volatile T&&
, orconst volatile T&&
.
As the effect, A
still has the implicitly-declared copy/move assignment operator.
BTW: Your template assignment operator takes forwarding reference, it could accept both lvalue and rvalue. In a2 = a1;
, it wins against the generated copy assignment operator in overload resolution and gets called.