8 Expressions [expr]

8.1 Primary expressions [expr.prim]

8.1.5 Lambda expressions [expr.prim.lambda]

8.1.5.1 Closure types [expr.prim.lambda.closure]

The type of a lambda-expression (which is also the type of the closure object) is a unique, unnamed non-union class type, called the closure type, whose properties are described below.
The closure type is declared in the smallest block scope, class scope, or namespace scope that contains the corresponding lambda-expression.
[Note
:
This determines the set of namespaces and classes associated with the closure type ([basic.lookup.argdep]).
The parameter types of a lambda-declarator do not affect these associated namespaces and classes.
end note
]
The closure type is not an aggregate type ([dcl.init.aggr]).
An implementation may define the closure type differently from what is described below provided this does not alter the observable behavior of the program other than by changing:
An implementation shall not add members of rvalue reference type to the closure type.
The closure type for a non-generic lambda-expression has a public inline function call operator whose parameters and return type are described by the lambda-expression's parameter-declaration-clause and trailing-return-type respectively.
For a generic lambda, the closure type has a public inline function call operator member template whose template-parameter-list consists of the specified template-parameter-list, if any, to which is appended one invented type template-parameter for each occurrence of auto in the lambda's parameter-declaration-clause, in order of appearance.
The invented type template-parameter is a parameter pack if the corresponding parameter-declaration declares a function parameter pack ([dcl.fct]).
The return type and function parameters of the function call operator template are derived from the lambda-expression's trailing-return-type and parameter-declaration-clause by replacing each occurrence of auto in the decl-specifiers of the parameter-declaration-clause with the name of the corresponding invented template-parameter.
[Example
:
auto glambda = [](auto a, auto&& b) { return a < b; };
bool b = glambda(3, 3.14);                             // OK

auto vglambda = [](auto printer) {
  return [=](auto&& ... ts) {                          // OK: ts is a function parameter pack
    printer(std::forward<decltype(ts)>(ts)...);

    return [=]() {
      printer(ts ...);
    };
  };
};
auto p = vglambda( [](auto v1, auto v2, auto v3)
                   { std::cout << v1 << v2 << v3; } );
auto q = p(1, 'a', 3.14);                              // OK: outputs 1a3.14
q();                                                   // OK: outputs 1a3.14
end example
]
The function call operator or operator template is declared const ([class.mfct.non-static]) if and only if the lambda-expression's parameter-declaration-clause is not followed by mutable.
It is neither virtual nor declared volatile.
Any noexcept-specifier specified on a lambda-expression applies to the corresponding function call operator or operator template.
An attribute-specifier-seq in a lambda-declarator appertains to the type of the corresponding function call operator or operator template.
The function call operator or any given operator template specialization is a constexpr function if either the corresponding lambda-expression's parameter-declaration-clause is followed by constexpr, or it satisfies the requirements for a constexpr function.
[Note
:
Names referenced in the lambda-declarator are looked up in the context in which the lambda-expression appears.
end note
]
[Example
:
auto ID = [](auto a) { return a; };
static_assert(ID(3) == 3); // OK

struct NonLiteral {
  NonLiteral(int n) : n(n) { }
  int n;
};
static_assert(ID(NonLiteral{3}).n == 3); // ill-formed
end example
]
[Example
:
auto monoid = [](auto v) { return [=] { return v; }; };
auto add = [](auto m1) constexpr {
  auto ret = m1();
  return [=](auto m2) mutable {
    auto m1val = m1();
    auto plus = [=](auto m2val) mutable constexpr
                   { return m1val += m2val; };
    ret = plus(m2());
    return monoid(ret);
  };
};
constexpr auto zero = monoid(0);
constexpr auto one = monoid(1);
static_assert(add(one)(zero)() == one()); // OK

// Since two below is not declared constexpr, an evaluation of its constexpr member function call operator
// cannot perform an lvalue-to-rvalue conversion on one of its subobjects (that represents its capture)
// in a constant expression.
auto two = monoid(2);
assert(two() == 2); // OK, not a constant expression.
static_assert(add(one)(one)() == two()); // ill-formed: two() is not a constant expression
static_assert(add(one)(one)() == monoid(2)()); // OK
end example
]
The closure type for a non-generic lambda-expression with no lambda-capture has a conversion function to pointer to function with C++ language linkage having the same parameter and return types as the closure type's function call operator.
The conversion is to “pointer to noexcept function” if the function call operator has a non-throwing exception specification.
The value returned by this conversion function is the address of a function F that, when invoked, has the same effect as invoking the closure type's function call operator.
F is a constexpr function if the function call operator is a constexpr function.
For a generic lambda with no lambda-capture, the closure type has a conversion function template to pointer to function.
The conversion function template has the same invented template parameter list, and the pointer to function has the same parameter types, as the function call operator template.
The return type of the pointer to function shall behave as if it were a decltype-specifier denoting the return type of the corresponding function call operator template specialization.
[Note
:
If the generic lambda has no trailing-return-type or the trailing-return-type contains a placeholder type, return type deduction of the corresponding function call operator template specialization has to be done.
The corresponding specialization is that instantiation of the function call operator template with the same template arguments as those deduced for the conversion function template.
Consider the following:
auto glambda = [](auto a) { return a; };
int (*fp)(int) = glambda;
The behavior of the conversion function of glambda above is like that of the following conversion function:
struct Closure {
  template<class T> auto operator()(T t) const { ... }
  template<class T> static auto lambda_call_operator_invoker(T a) {
    // forwards execution to operator()(a) and therefore has
    // the same return type deduced
    ...
  }
  template<class T> using fptr_t =
     decltype(lambda_call_operator_invoker(declval<T>())) (*)(T);

  template<class T> operator fptr_t<T>() const
    { return &lambda_call_operator_invoker; }
};
end note
]
[Example
:
void f1(int (*)(int))   { }
void f2(char (*)(int))  { }

void g(int (*)(int))    { }  // #1
void g(char (*)(char))  { }  // #2

void h(int (*)(int))    { }  // #3
void h(char (*)(int))   { }  // #4

auto glambda = [](auto a) { return a; };
f1(glambda);  // OK
f2(glambda);  // error: ID is not convertible
g(glambda);   // error: ambiguous
h(glambda);   // OK: calls #3 since it is convertible from ID
int& (*fpi)(int*) = [](auto* a) -> auto& { return *a; }; // OK
end example
]
The value returned by any given specialization of this conversion function template is the address of a function F that, when invoked, has the same effect as invoking the generic lambda's corresponding function call operator template specialization.
F is a constexpr function if the corresponding specialization is a constexpr function.
[Note
:
This will result in the implicit instantiation of the generic lambda's body.
The instantiated generic lambda's return type and parameter types shall match the return type and parameter types of the pointer to function.
end note
]
[Example
:
auto GL = [](auto a) { std::cout << a; return a; };
int (*GL_int)(int) = GL;  // OK: through conversion function template
GL_int(3);                // OK: same as GL(3)
end example
]
The conversion function or conversion function template is public, constexpr, non-virtual, non-explicit, const, and has a non-throwing exception specification.
[Example
:
auto Fwd = [](int (*fp)(int), auto a) { return fp(a); };
auto C = [](auto a) { return a; };

static_assert(Fwd(C,3) == 3); // OK

// No specialization of the function call operator template can be constexpr (due to the local static).
auto NC = [](auto a) { static int s; return a; };
static_assert(Fwd(NC,3) == 3); // ill-formed
end example
]
The lambda-expression's compound-statement yields the function-body ([dcl.fct.def]) of the function call operator, but for purposes of name lookup, determining the type and value of this and transforming id-expressions referring to non-static class members into class member access expressions using (*this) ([class.mfct.non-static]), the compound-statement is considered in the context of the lambda-expression.
[Example
:
struct S1 {
  int x, y;
  int operator()(int);
  void f() {
    [=]()->int {
      return operator()(this->x + y); // equivalent to S1​::​operator()(this->x + (*this).y)
                                      // this has type S1*
    };
  }
};
end example
]
Further, a variable _­_­func_­_­ is implicitly defined at the beginning of the compound-statement of the lambda-expression, with semantics as described in [dcl.fct.def.general].
The closure type associated with a lambda-expression has no default constructor and a deleted copy assignment operator.
It has a defaulted copy constructor and a defaulted move constructor ([class.copy]).
[Note
:
These special member functions are implicitly defined as usual, and might therefore be defined as deleted.
end note
]
The closure type associated with a lambda-expression has an implicitly-declared destructor ([class.dtor]).
A member of a closure type shall not be explicitly instantiated ([temp.explicit]), explicitly specialized ([temp.expl.spec]), or named in a friend declaration ([class.friend]).