Can a default function return auto?

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Consider the following class:

class example
{
public:
    auto & operator =(const example &) = default;
    auto & operator =(example &&) = default;
};

Are those declarations considered legal?

language-lawyer c++
3 Answers
auto & operator =(const example &) = default;
auto & operator =(example &&) = default;

Are those declarations considered legal?

No.

[dcl.spec.auto] ... If the declared return type of the function contains a placeholder type, the return type of the function is deduced from non-discarded return statements, if any, in the body of the function ([stmt.if]).

A defaulted function definition does not have a body, which conflicts the quoted rule. There is nothing that the return type could be deduced from and there is no rule that states what that type would be in this case.

operator<=> has an exceptional rule specifying what the return type will be when auto is used as shown in Caleth's answer, but operator= does not have such rule. I see no reason such rule couldn't be introduced to allow auto in defaulted assignment operators.

user_20793030

In C++20, Yes1

Let R be the declared return type of a defaulted three-way comparison operator function, and let xi be the elements of the expanded list of subobjects for an object x of type C`.

  • If R is auto, then let cvi Ri be the type of the expression xi <=> xi. The operator function is defined as deleted if that expression is not usable or if Ri is not a comparison category type ([cmp.categories.pre]) for any i. The return type is deduced as the common comparison type (see below) of R0, R1, …, Rn−1.

[class.spaceship/2]

Before C++20, No

A function definition of the form: attribute-specifier-seq opt decl-specifier-seq opt declarator virt-specifier-seq opt = default ; is called an explicitly-defaulted definition. A function that is explicitly defaulted shall

  • be a special member function,
  • have the same declared function type (except for possibly differing ref-qualifiers and except that in the case of a copy constructor or copy assignment operator, the parameter type may be “reference to non-const T”, where T is the name of the member function's class) as if it had been implicitly declared, and

[dcl.fct.def.default] (emphasis added)

  1. But only <=>. A defaulted == must return bool, and assignment has similar restrictions as previous standards.

if F is an assignment operator, and the return type of T1 differs from the return type of T2 or T1's parameter type is not a reference, the program is ill-formed;

[dcl.fct.def.default]

user_26108100

not a language lawyer answer

From my experience compilers don't accept auto return type for defaulted special member functions, so I assume they are indeed not allowed by the standard.

The only exception I know of is the C++20 default three-way comparison operator:

#include <compare>

struct X
{
    auto operator<=>(const X&) const = default;
};
user_28053050
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