struct A
{
A();
A(const A&);
A& operator =(const A&);
A(A&&) = delete;
A& operator =(A&&) = delete;
};
struct B
{
B();
B(const B&);
B& operator =(const B&);
};
int main()
{
A a;
a = A(); // error C2280
B b;
b = B(); // OK
}
My compiler is VC++ 2013 RC.
error C2280: 'A &A::operator =(A &&)' : attempting to reference a
deleted function
I just wonder why the compiler doesn't try A& operator =(const A&);
when A& operator =(A&&)
is deleted?
Is this behavior defined by the C++ standard?
a = A(); // error C2280
The expression on the right is a temporary which means it will look for operator=(A&&)
and sees it is deleted. Hence the error. There is no further search.
=delete
does not mean "don't use me, instead use next best one". It rather means, "don't use me when you need me — instead be alone in the wild."
Here is another example. If I want the instances of my class X
to be created with only long
and no other type (even if it converts into long!), then I would declare class X
as:
struct X
{
X(long arg); //ONLY long - NO int, short, char, double, etc!
template<typename T>
X(T) = delete;
};
X a(1); //error - 1 is int
X b(1L); //ok - 1L is long
That means, the overload resolution is performed before the compiler sees the =delete
part — and thus results in an error because the selected overload is found deleted.
Hope that helps.
When you =delete
a function, you actually are deleting its definition.
8.4.3 Deleted definitions [dcl.fct.def.delete]
1 A function definition of the form:
attribute-specifier-seqopt decl-specifier-seqopt declarator = delete ;
is called a deleted definition. A function with a deleted definition is also called a deleted function.
But by doing so, you are also declaring that function. Quoting from the standard [1]:
4 A deleted function is implicitly inline. [ Note: The one-definition rule (3.2) applies to deleted definitions.
—end note ] A deleted definition of a function shall be the first declaration of the function [...]
And so by doing a = A()
, the compiler actually resolves to A::operator=(A&&)
because it has been declared (not A::operator(const A&)
, because A&&
is "more binding" to r-values). However with its definition being deleted, the line is ill-formed.
2 A program that refers to a deleted function implicitly or explicitly, other than to declare it, is ill-formed.
[1] The tone of the emphasized sentence here is actually imperative. The standard directs that declaring a function =delete
d must first appear before other declarations of it. But still, it supports the fact that deleting a function also declares the function.