As mentioned in Declarations, the declarator can be followed by a requires clause, which declares the associated constraints for the function, which must be satisfied in order for the function to be selected by overload resolution. (example: void f1(int a) requires true;) Note that the associated constraint is part of function signature, but not part of function type.

Using a volatile-qualified object type as parameter type or return type is deprecated.

As with any declaration, attributes that appear before the declaration and the attributes that appear immediately after the identifier within the declarator both apply to the entity being declared or defined (in this case, to the function):

[[noreturn]] void f [[noreturn]] (); // okay: both attributes apply to the function f

However, the attributes that appear after the declarator (in the syntax above), apply to the type of the function, not to the function itself:

void f() [[noreturn]]; // error: this attribute has no effect on the function itself

Return type deduction

If the decl-specifier-seq of the function declaration contains the keyword auto, trailing return type may be omitted, and will be deduced by the compiler from the type of the expression used in the return statement. If the return type does not use decltype(auto), the deduction follows the rules of template argument deduction:

int x = 1;
auto f() { return x; }        // return type is int
const auto& f() { return x; } // return type is const int&

If the return type is decltype(auto), the return type is as what would be obtained if the expression used in the return statement were wrapped in decltype:

int x = 1;
decltype(auto) f() { return x; }  // return type is int, same as decltype(x)
decltype(auto) f() { return(x); } // return type is int&, same as decltype((x))

(note: "const decltype(auto)&" is an error, decltype(auto) must be used on its own).

If there are multiple return statements, they must all deduce to the same type:

auto f(bool val)
{
    if (val) return 123; // deduces return type int
    else return 3.14f;   // error: deduces return type float
}

If there is no return statement or if the argument of the return statement is a void expression, the declared return type must be either decltype(auto), in which case the deduced return type is void, or (possibly cv-qualified) auto, in which case the deduced return type is then (identically cv-qualified) void:

auto f() {}              // returns void
auto g() { return f(); } // returns void
auto* x() {}             // error: cannot deduce auto* from void

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:

auto sum(int i)
{
    if (i == 1)
        return i;              // sum’s return type is int
    else
        return sum(i - 1) + i; // okay: sum’s return type is already known
}

If the return statement uses a brace-init-list, deduction is not allowed:

auto func () { return {1, 2, 3}; } // error

Virtual functions and coroutines (since C++20)cannot use return type deduction:

struct F
{
    virtual auto f() { return 2; } // error
};

Function templates other than user-defined conversion functions can use return type deduction. The deduction takes place at instantiation even if the expression in the return statement is not dependent. This instantiation is not in an immediate context for the purposes of SFINAE.

template<class T>
auto f(T t) { return t; }
typedef decltype(f(1)) fint_t;    // instantiates f<int> to deduce return type
 
template<class T>
auto f(T* t) { return *t; }
void g() { int (*p)(int*) = &f; } // instantiates both fs to determine return types,
                                  // chooses second template overload

Redeclarations or specializations of functions or function templates that use return type deduction must use the same return type placeholders:

auto f(int num) { return num; }
// int f(int num);            // error: no placeholder return type
// decltype(auto) f(int num); // error: different placeholder
 
template<typename T>
auto g(T t) { return t; }
template auto g(int);     // okay: return type is int
// template char g(char); // error: not a specialization of the primary template g

Similarly, redeclarations or specializations of functions or function templates that do not use return type deduction must not use a placeholder:

int f(int num);
// auto f(int num) { return num; } // error: not a redeclaration of f
 
template<typename T>
T g(T t) { return t; }
template int g(int);      // okay: specialize T as int
// template auto g(char); // error: not a specialization of the primary template g

Explicit instantiation declarations do not themselves instantiate function templates that use return type deduction:

template<typename T>
auto f(T t) { return t; }
extern template auto f(int); // does not instantiate f<int>
 
int (*p)(int) = f; // instantiates f<int> to determine its return type,
                   // but an explicit instantiation definition 
                   // is still required somewhere in the program

If any of the function parameters uses a placeholder (either auto or a concept type), the function declaration is instead an abbreviated function template declaration:

void f1(auto);    // same as template<class T> void f(T)
void f2(C1 auto); // same as template<C1 T> void f7(T), if C1 is a concept

The ellipsis that indicates variadic arguments need not be preceded by a comma, even if it follows the ellipsis that indicates a parameter pack expansion, so the following function templates are exactly the same:

template<typename... Args>
void f(Args..., ...);
 
template<typename... Args>
void f(Args... ...);
 
template<typename... Args>
void f(Args......);

An example of when such declaration might be used is the possible implementation of std::is_function.

#include <cstdio>
 
template<typename... Variadic, typename... Args>
constexpr void invoke(auto (*fun)(Variadic......), Args... args)
{
    fun(args...);
}
 
int main()
{
    invoke(std::printf, "%dm•%dm•%dm = %d%s%c", 2,3,7, 2*3*7, "m³", '\n');
}

Output:

2m•3m•7m = 42m³

Deleted functions

If, instead of a function body, the special syntax = delete ; is used, the function is defined as deleted. Any use of a deleted function is ill-formed (the program will not compile). This includes calls, both explicit (with a function call operator) and implicit (a call to deleted overloaded operator, special member function, allocation function etc), constructing a pointer or pointer-to-member to a deleted function, and even the use of a deleted function in an unevaluated expression. However, implicit ODR-use of a non-pure virtual member function that happens to be deleted is allowed.

If the function is overloaded, overload resolution takes place first, and the program is only ill-formed if the deleted function was selected:

struct sometype
{
    void* operator new(std::size_t) = delete;
    void* operator new[](std::size_t) = delete;
};
sometype* p = new sometype; // error: attempts to call deleted sometype::operator new

The deleted definition of a function must be the first declaration in a translation unit: a previously-declared function cannot be redeclared as deleted:

struct sometype { sometype(); };
sometype::sometype() = delete; // error: must be deleted on the first declaration

User-provided functions

A function is user-provided if it is user-declared and not explicitly defaulted or deleted on its first declaration. A user-provided explicitly-defaulted function (i.e., explicitly defaulted after its first declaration) is defined at the point where it is explicitly defaulted; if such a function is implicitly defined as deleted, the program is ill-formed. Declaring a function as defaulted after its first declaration can provide efficient execution and concise definition while enabling a stable binary interface to an evolving code base.

// All special member functions of `trivial` are
// defaulted on their first declarations respectively,
// they are not user-provided
struct trivial
{
    trivial() = default;
    trivial(const trivial&) = default;
    trivial(trivial&&) = default;
    trivial& operator=(const trivial&) = default;
    trivial& operator=(trivial&&) = default;
    ~trivial() = default;
};
 
struct nontrivial
{
    nontrivial(); // first declaration
};
 
// not defaulted on the first declaration,
// it is user-provided and is defined here
nontrivial::nontrivial() = default;

__func__

Within the function body, the function-local predefined variable __func__ is defined as if by.

static const char __func__[] = "function-name";

This variable has block scope and static storage duration:

struct S
{
    S(): s(__func__) {} // okay: initializer-list is part of function body
    const char* s;
};
void f(const char* s = __func__); // error: parameter-list is part of declarator

#include <iostream>
 
void Foo() { std::cout << __func__ << ' '; }
 
struct Bar
{
    Bar() { std::cout << __func__ << ' '; }
    ~Bar() { std::cout << __func__ << ' '; }
    struct Pub { Pub() { std::cout << __func__ << ' '; } };
};
 
int main()
{
    Foo();
    Bar bar;
    Bar::Pub pub;
}

Possible output:

Foo Bar Pub ~Bar