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Function declaration

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A function declaration introduces the function name and its type. A function definition associates the function name/type with the function body.

Contents

[edit] Function declaration

Function declarations may appear in any scope. A function declaration at class scope introduces a class member function (unless the friend specifier is used), see member functions and friend functions for details.

The type of the function being declared is composed from the return type (provided by the decl-specifier-seq of the declaration syntax) and the function declarator

noptr-declarator ( parameter-list ) cv(optional) ref(optional) except(optional) attr(optional) (1)
noptr-declarator ( parameter-list ) cv(optional) ref(optional) except(optional) attr(optional) -> trailing (2) (since C++11)

(see Declarations for the other forms of the declarator syntax)

1) Regular function declarator syntax
2) Trailing return type declaration. The decl-specifier-seq in this case must contain the keyword auto
noptr-declarator - any valid declarator, but if it begins with *, &, or &&, it has to be surrounded by parentheses.
parameter-list - possibly empty, comma-separated list of the function parameters (see below for details)
attr - (since C++11) optional list of attributes. These attributes are applied to the type of the function, not the function itself. The attributes for the function appear after the identifier within the declarator and are combined with the attributes that appear in the beginning of the declaration, if any.
cv - const/volatile qualification, only allowed in non-static member function declarations
ref - (since C++11) ref-qualification, only allowed in non-static member function declarations
except -

dynamic exception specification

(until C++11)

either dynamic exception specification
or noexcept specification

(since C++11)
(until C++17)

noexcept specification

(since C++17)

Note that the exception specification is not part of the function type

(until C++17)
trailing - Trailing return type, useful if the return type depends on argument names, such as template<class T, class U> auto add(T t, U u) -> decltype(t + u); or is complicated, such as in auto fpif(int)->int(*)(int)


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.

(since C++20)

Function declarators can be mixed with other declarators, where decl-specifier-seq allows:

// declares an int, an int*, a function, and a pointer to a function
int a = 1, *p = NULL, f(), (*pf)(double);
// decl-specifier-seq is int
// declarator f() declares (but doesn't define)
//                a function taking no arguments and returning int
 
struct S
{
    virtual int f(char) const, g(int) &&; // declares two non-static member functions
    virtual int f(char), x; // compile-time error: virtual (in decl-specifier-seq)
                            // is only allowed in declarations of non-static
                            // member functions
};

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

(since C++20)

The return type of a function cannot be a function type or an array type (but can be a pointer or reference to those).

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
(since C++11)

As with any declaration, the type of the function func declared as ret func(params) is ret(params) (except for parameter type rewriting described below): see type naming.

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
(since C++14)

[edit] Parameter list

Parameter list determines the arguments that can be specified when the function is called. It is a comma-separated list of parameter declarations, each of which has the following syntax:

attr(optional) decl-specifier-seq declarator (1)
attr(optional) decl-specifier-seq declarator = initializer (2)
attr(optional) decl-specifier-seq abstract-declarator(optional) (3)
attr(optional) decl-specifier-seq abstract-declarator(optional) = initializer (4)
void (5)
1) Declares a named (formal) parameter. For the meanings of decl-specifier-seq and declarator, see declarations.
int f(int a, int* p, int (*(*x)(double))[3]);
2) Declares a named (formal) parameter with a default value.
int f(int a = 7, int* p = nullptr, int (*(*x)(double))[3] = nullptr);
3) Declares an unnamed parameter.
int f(int, int*, int (*(*)(double))[3]);
4) Declares an unnamed parameter with a default value.
int f(int = 7, int* = nullptr, int (*(*)(double))[3] = nullptr);
5) Indicates that the function takes no parameters, it is the exact synonym for an empty parameter list: int f(void); and int f(); declare the same function. Note that the type void (possibly cv-qualified) cannot be used in a parameter list otherwise: int f(void, int); and int f(const void); are errors (although derived types, such as void* can be used). In a template, only non-dependent void type can be used (a function taking a single parameter of type T does not become a no-parameter function if instantiated with T = void).

An ellipsis ... may appear at the end of the parameter list; this declares a variadic function:

int printf(const char* fmt ...);

For compatibility with C89, an optional comma may appear before the ellipsis if the parameter list contains at least one parameter:

int printf(const char* fmt, ...); // OK, same as above

Although decl-specifier-seq implies there can exist specifiers other than type specifiers, the only other specifier allowed is register as well as auto (until C++11), and it has no effect.

(until C++17)

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
(since C++20)

Parameter names declared in function declarations are usually for only self-documenting purposes. They are used (but remain optional) in function definitions.

The type of each function parameter in the parameter list is determined according to the following rules:

1) First, decl-specifier-seq and the declarator are combined as in any declaration to determine the type.
2) If the type is "array of T" or "array of unknown bound of T", it is replaced by the type "pointer to T"
3) If the type is a function type F, it is replaced by the type "pointer to F"
4) Top-level cv-qualifiers are dropped from the parameter type (This adjustment only affects the function type, but doesn't modify the property of the parameter: int f(const int p, decltype(p)*); and int f(int, const int*); declare the same function)

Because of these rules, the following function declarations declare exactly the same function:

int f(char s[3]);
int f(char[]);
int f(char* s);
int f(char* const);
int f(char* volatile s);

The following declarations also declare exactly the same function:

int f(int());
int f(int (*g)());

An ambiguity arises in a parameter list when a type name is nested in parentheses (including lambda expressions) (since C++11). In this case, the choice is between the declaration of a parameter of type pointer to function and the declaration of a parameter with redundant parentheses around the identifier of the declarator. The resolution is to consider the type name as a simple type specifier (which is the pointer to function type):

class C {};
 
void f(int(C)) {} // void f(int(*fp)(C param)) {}
                  // NOT void f(int C) {}
 
void g(int *(C[10])); // void g(int *(*fp)(C param[10]));
                      // NOT void g(int *C[10]);

Parameter type cannot be a type that includes a reference or a pointer to array of unknown bound, including a multi-level pointers/arrays of such types, or a pointer to functions whose parameters are such types.

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
(since C++11)

[edit] Function definition

A non-member function definition may appear at namespace scope only (there are no nested functions). A member function definition may also appear in the body of a class definition. They have the following syntax:

attr(optional) decl-specifier-seq(optional) declarator virt-specifier-seq(optional) function-body

where function-body is one of the following

ctor-initializer(optional) compound-statement (1)
function-try-block (2)
= delete ; (3) (since C++11)
= default ; (4) (since C++11)
1) regular function body
2) function-try-block (which is a regular function body wrapped in a try/catch block)
3) explicitly deleted function definition
4) explicitly defaulted function definition, only allowed for special member functions and comparison operator functions (since C++20)
attr - (since C++11) optional list of attributes. These attributes are combined with the attributes after the identifier in the declarator (see top of this page), if any.
decl-specifier-seq - the return type with specifiers, as in the declaration grammar
declarator - function declarator, same as in the function declaration grammar above (can be parenthesized). as with function declaration, it may be followed by a requires-clause (since C++20)
virt-specifier-seq - (since C++11) override, final, or their combination in any order (only allowed for non-static member functions)
ctor-initializer - member initializer list, only allowed in constructors
compound-statement - the brace-enclosed sequence of statements that constitutes the body of a function
int max(int a, int b, int c)
{
    int m = (a > b) ? a : b;
    return (m > c) ? m : c;
}
 
// decl-specifier-seq is "int"
// declarator is "max(int a, int b, int c)"
// body is { ... }

The function body is a compound statement (sequence of zero or more statements surrounded by a pair of curly braces), which is executed when the function call is made.

The parameter types, as well as the return type of a function definition cannot be (possibly cv-qualified) incomplete class types unless the function is defined as deleted (since C++11). The completeness check is only made in the function body, which allows member functions to return the class in which they are defined (or its enclosing class), even if it is incomplete at the point of definition (it is complete in the function body).

The parameters declared in the declarator of a function definition are in scope within the body. If a parameter is not used in the function body, it does not need to be named (it's sufficient to use an abstract declarator):

void print(int a, int) // second parameter is not used
{
    std::printf("a = %d\n", a);
}

Even though top-level cv-qualifiers on the parameters are discarded in function declarations, they modify the type of the parameter as visible in the body of a function:

void f(const int n) // declares function of type void(int)
{
    // but in the body, the type of n is const int
}

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
(since C++11)

[edit] Notes

In case of ambiguity between a variable declaration using the direct-initialization syntax and a function declaration, the compiler always chooses function declaration; see direct-initialization.

Feature-test macro Value Std Comment
__cpp_decltype_auto 201304L (C++14) decltype(auto)
__cpp_return_type_deduction 201304L (C++14) Return type deduction for normal functions

[edit] Example

#include <iostream>
#include <string>
 
// simple function with a default argument, returning nothing
void f0(const std::string& arg = "world!")
{
    std::cout << "Hello, " << arg << '\n';
}
 
// the declaration is in namespace (file) scope
// (the definition is provided later)
int f1();
 
// function returning a pointer to f0, pre-C++11 style
void (*fp03())(const std::string&)
{
    return f0;
}
 
// function returning a pointer to f0, with C++11 trailing return type
auto fp11() -> void(*)(const std::string&)
{
    return f0;
}
 
int main()
{
    f0();
    fp03()("test!");
    fp11()("again!");
    int f2(std::string) noexcept; // declaration in function scope
    std::cout << "f2(\"bad\"): " << f2("bad") << '\n';
    std::cout << "f2(\"42\"): " << f2("42") << '\n';
}
 
// simple non-member function returning int
int f1()
{
    return 007;
}
 
// function with an exception specification and a function try block
int f2(std::string str) noexcept
try
{
    return std::stoi(str);
}
catch (const std::exception& e)
{
    std::cerr << "stoi() failed!\n";
    return 0;
}

Possible output:

stoi() failed!
Hello, world!
Hello, test!
Hello, again!
f2("bad"): 0
f2("42"): 42

[edit] Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

DR Applied to Behavior as published Correct behavior
CWG 135 C++98 member functions defined in class
could not have a parameter of or return
its own class because it is incomplete
allowed
CWG 332 C++98 a parameter could have cv-qualified void type prohibited
CWG 393 C++98 types that include pointers/references to
array of unknown bound could not be parameters
such types are allowed
CWG 452 C++98 member initializer list was not a part of function body made it a part of function
body by modifying the syntax
of function definition
CWG 577 C++98 dependent type void could be used to
declare a function taking no parameters
only non-dependent
void is allowed
CWG 1327 C++11 defaulted or deleted functions could not
be specified with override or final
allowed
CWG 1355 C++11 only special member functions could be user-provided extended to all functions
CWG 1394 C++11 deleted functions could not have any parameter of
an incomplete type or return an incomplete type
incomplete type allowed
CWG 1824 C++98 the completeness check on parameter type and
return type of a function definition could be made
outside the context of the function definition
only make the completeness
check in the context of
the function definition
CWG 1877 C++14 return type deduction treated return; as return void(); simply deduce the return
type as void in this case
CWG 2015 C++11 the implicit odr-use of a deleted
virtual function was ill-formed
such odr-uses are exempt
from the use prohibition
CWG 2044 C++14 return type deduction on functions returning void
would fail if the declared return type is decltype(auto)
updated the deduction
rule to handle this case
CWG 2081 C++14 function redeclarations could use return type
deduction even if the initial declaration does not
not allowed
CWG 2145 C++98 the declarator in function definition could not be parenthesized allowed
CWG 2259 C++11 the ambiguity resolution rule regarding parenthesized
type names did not cover lambda expressions
covered
CWG 2430 C++98 in the definition of a member function in a class definition,
the type of that class could not be the return type or
parameter type due to the resolution of CWG issue 1824
only make the completeness
check in the function body

[edit] See also

C documentation for Declaring functions