Return type deduction for normal functions

Introduction

This proposal also resolves core DRs 975 (lambda return type deduction from multiple return statements) and 1048 (inconsistency between auto and lambda return type deduction).

Redeclaration

struct A {
  auto f(); // forward declaration
};
auto A::f() { return 42; }

auto f(int);
template <class T, int = sizeof(f(T()))> void g();
void h1() { g<int>(); } // no match, g() disqualified by SFINAE
auto f(int i) { return i; }
void h2() { g<int>(); } // OK

The proposed wording allows non-defining declarations so long as all declarations have the same declared type, without considering the deduced type.

auto f(); // return type is unknown
auto f() { return 42; } // return type is int
auto f(); // redeclaration
int  f(); // error, declares a different function

template <class T> auto g(T t); // forward declaration
template <class T> auto g(T t) { return t; } // return type is deduced at instantiation time
template <class T> auto g(T t); // redeclaration

Of course, using such a function in an expression when only a forward declaration has been seen is ill-formed:

auto f();    // return type is unknown
int i = f(); // error, return type of f is unknown

template <class T> auto f(T t) { return t; } // #1
template auto f(int); // OK
template char f(char); // error, no matching template
template<> auto f(double); // OK, forward declaration with unknown return type

template <class T> T f(T t) { return t; } // OK, not functionally equivalent to #1
template char f(char); // OK, now there is a matching template
template auto f(float); // OK, matches #1

Multiple returns

The limitation on return type deduction to function bodies consisting of a single return statement is inconvenient for lambdas (DR 975), but is likely to be even more of an annoyance on normal functions.

auto iterate(int len) // error, body isn't "return expr;"
{
  for (int i = 0; i < len; ++i)
    if (search (i))
      return i;
  return -1;
}

Recursion

auto h() { return h(); } // error, return type of h is unknown

auto sum(int i) {
  if (i == 1)
    return i;          // return type deduced to int
  else
    return sum(i-1)+i; // ok to call it now
}

Instantiation

template <class T> auto f(T t) { return t; } // return type deduced at instantiation time
typedef decltype(f(1)) fint_t; // must instantiate f<int> to deduce return type

More general deduction

template <class T> struct A { static T t; };
template <class T> auto& f() { return A::t; } // returns by reference

Such a function must have a return statement, however, since there is no way to get void from auto&.

auto& f() { } // error, no return statement

Difference from decltype

Unfortunately, there is no way to get the effect of decltype with an auto return type; plain auto never deduces to a reference, and auto&& always deduces to a reference. This is a significant problem, as it means that forwarding functions can't use auto. We could consider using decltype semantics instead of the existing auto semantics, but that would mean giving different deduction semantics to auto depending on whether the declaration is of a variable or a function, and making auto functions different from lambdas.

One possibility for addressing this issue would be to change the deduction rules for auto&& specifically; another would be to write decltype(auto) to get the decltype semantics without having to repeat the expression.

Proposed wording

The auto type-specifier designates a placeholder type that will be replaced later, by either deduction from an initializer or signifies that the type of a variable being declared shall be deduced from its initializer or that a function declarator shall include a trailing-return-type.

The auto type-specifier may can appear with a function declarator either in the decl-specifier-seq or type-specifier-seq or as part of a conversion-function-id with a trailing-return-type (8.3.5) in any context where such a declarator is valid. If the function declarator includes a trailing-return-type (8.3.5), that specifies the declared return type of the function, so the return type is not considered to use auto. Otherwise, the return type of the function is deduced from return statements in the body of the function, if any.

Otherwise, the type of the variable The type of a variable declared using auto is deduced from its initializer. The name of the variable being declared shall not appear in the initializer expression. This use of auto is allowed when declaring variables in a block (6.3), in namespace scope (3.3.6), and in a for-init-statement (6.5.3). auto shall appear as one of the decl-specifiers in the decl-specifier-seq and the decl-specifier-seq shall be followed by one or more init-declarators, each of which shall have a non-empty initializer.

[ Example:

auto x = 5;                 // OK: x has type int
const auto *v = &x, u = 6;  // OK: v has type const int*, u has type const int
static auto y = 0.0;        // OK: y has type double
auto int r;                 // error: auto is not a storage-class-specifier
auto f() -> int;             // OK: f returns int
auto g() { return 0.0; }     // OK: g returns double
auto h();                    // OK, h's return type will be deduced when it is defined

— end example ]

The auto type-specifier can also be used in declaring a variable in the condition of a selection statement (6.4) or an iteration statement (6.5), in the type-specifier-seq in the new-type-id or type-id of a new-expression (5.3.4), in a for-range-declaration, and in declaring a static data member with a brace-or-equal-initializer that appears within the member-specification of a class definition (9.4.2).

A program that uses auto in a context not explicitly allowed in this section is ill-formed.

Once the type of a declarator-id has been determined according to 8.3, the type of the declared variable using the declarator-id When a variable declared with a type that uses auto is initialized, or a return statement occurs in a function declared with a return type that uses auto, the deduced return type or variable type is determined from the type of its initializer using the rules for template argument deduction. Let T be the declared type that has been determined for a variable identifier dof the variable or return type of the function. Obtain P from T by replacing the occurrences of auto with either a new invented type template parameter U or, if the initializer is a braced-init-list (8.5.4), with std::initializer_list<U>. The type deduced for the variable d or return type is then the deduced A determined using the rules of template argument deduction from a function call (14.8.2.1), where P is a function template parameter type and the initializer for d is the corresponding argument. In the case of a return with no operand, the corresponding argument is void(). If the deduction fails, the declaration is ill-formed. [ Example:

auto x1 = { 1, 2 };    // decltype(x1) is std::initializer_list<int>
auto x2 = { 1, 2.0 };  // error: cannot deduce element type

— end example ]

If the list of declarators contains more than one declarator, the type of each declared variable is determined as described above. If a function with a declared return type that uses auto has multiple return statements, the return type is deduced for each return statement. If In either case, if the type deduced for the template parameter U is not the same in each deduction, the program is ill-formed. [ Example:

const auto &i = expr;

The type of i is the deduced type of the parameter u in the call f(expr) of the following invented function template:

template <class U> void f(const U& u);

— end example ]

If a function with a declared return type that uses auto has no return statements, the return type is deduced as though from a return statement with no operand at the closing brace of the function body. [ Example:

auto  f() { } // OK, return type is void
auto* g() { } // error, cannot deduce auto* from void()

When a declaration that uses auto in its declared type appears as an expression, the type of the expression is the deduced type of the declaration. If the declaration does not yet have a deduced type, the expression is ill-formed. But once a return statement has been seen in a function, the return type deduced from that statement can be used in the rest of the function, including in other return statements. [ Example:

auto n = n; // error, n's type is unknown
auto f();
void g() { &f; } // error, f's return type is unknown
auto sum(int i) {
  if (i == 1)
    return i;  // sum's return type is int
  else
    return sum(i-1)+i; // OK, sum's return type has been deduced
}

—end example]

Return type deduction for a function template that uses auto in its declared type occurs at instantiation time even if the function body contains a return statement with a non-type-dependent operand. [ Note: So any use of a specialization of the function template will cause an implicit instantiation. —end note ] [ Example:

template <class T> auto f(T t) { return t; } // return type deduced at instantiation time
typedef decltype(f(1)) fint_t; // must instantiate f<int> to deduce return type

—end example]

Redeclarations or specializations of a function or function template with a declared return type that uses auto must also use auto, not a deduced type. [ Example:

auto f();
auto f() { return 42; }
auto f(); // OK
int f();  // error, cannot be overloaded with auto f()

template <typename T> auto g(T t) { return t; } // #1
template auto g(int);      // OK, return type is int
template char g(char);     // error, no matching template
template<> auto g(double); // OK, forward declaration with unknown return type

template <class T> T g(T t) { return t; } // OK, not functionally equivalent to #1
template char g(char);     // OK, now there is a matching template
template auto g(float);    // still matches #1

void h() { return g(42); } // error, ambiguous

If a lambda-expression does not include a lambda-declarator, it is as if the lambda-declarator were (). If a lambda-expression does not include a trailing-return-type, it is as if the trailing-return-type denotes the following type:

• if the compound-statement is of the form

  { attribute-specifier-seqopt return expression ; }

  the type of the returned expression after lvalue-to-rvalue conversion (4.1), array-to-pointer conver- sion (4.2), and function-to-pointer conversion (4.3);
• otherwise, void.

The lambda return type is auto, which is replaced by the trailing-return-type if provided, or otherwise deduced from return statements as described in 7.1.6.4. [ Example:

auto x1 = [](int i){ return i; }; // OK: return type is int
auto x2 = []{ return { 1, 2 }; }; // OK: return type is std::initializer_list<int>error: the return type is void (a
                                  // braced-init-list is not an expression)

— end example ]

References