7 Expressions [expr]

7.5 Primary expressions [expr.prim]

7.5.6 Lambda expressions [expr.prim.lambda]

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

1
#
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.
2
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The closure type is declared in the smallest block scope, class scope, or namespace scope that contains the corresponding lambda-expression.
[Note 1: 
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]
3
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The closure type is not an aggregate type ([dcl.init.aggr]); it is a structural type ([temp.param]) if and only if the lambda has no lambda-capture.
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.
4
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The closure type for a lambda-expression has a public inline function call operator (for a non-generic lambda) or function call operator template (for a generic lambda) ([over.call]) whose parameters and return type are those of the lambda-expression's parameter-declaration-clause and trailing-return-type respectively, and whose template-parameter-list consists of the specified template-parameter-list, if any.
The requires-clause of the function call operator template is the requires-clause immediately following < template-parameter-list >, if any.
The trailing requires-clause of the function call operator or operator template is the requires-clause of the lambda-declarator, if any.
[Note 2: 
The function call operator template for a generic lambda can be an abbreviated function template ([dcl.fct]).
— end note]
[Example 1: 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 auto fact = [](this auto self, int n) -> int { // OK, explicit object parameter return (n <= 1) ? 1 : n * self(n-1); }; std::cout << fact(5); // OK, outputs 120 — end example]
5
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Given a lambda with a lambda-capture, the type of the explicit object parameter, if any, of the lambda's function call operator (possibly instantiated from a function call operator template) shall be either:
  • (5.1)
    the closure type,
  • (5.2)
    a class type publicly and unambiguously derived from the closure type, or
  • (5.3)
    a reference to a possibly cv-qualified such type.
[Example 2: struct C { template <typename T> C(T); }; void func(int i) { int x = [=](this auto&&) { return i; }(); // OK int y = [=](this C) { return i; }(); // error int z = [](this C) { return 42; }(); // OK } — end example]
6
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The function call operator or operator template is a static member function or static member function template ([class.static.mfct]) if the lambda-expression's parameter-declaration-clause is followed by static.
Otherwise, it is a non-static member function or member function template ([class.mfct.non.static]) that is declared const ([class.mfct.non.static]) if and only if the lambda-expression's parameter-declaration-clause is not followed by mutable and the lambda-declarator does not contain an explicit object parameter.
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.
An attribute-specifier-seq in a lambda-expression preceding a lambda-declarator appertains to 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 consteval, or it is constexpr-suitable ([dcl.constexpr]).
It is an immediate function ([dcl.constexpr]) if the corresponding lambda-expression's parameter-declaration-clause is followed by consteval.
[Example 3: 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); // error — end example]
7
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[Example 4: 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()); // error: two() is not a constant expression static_assert(add(one)(one)() == monoid(2)()); // OK — end example]
8
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[Note 3: 
The function call operator or operator template can be constrained ([temp.constr.decl]) by a type-constraint ([temp.param]), a requires-clause ([temp.pre]), or a trailing requires-clause ([dcl.decl]).
[Example 5: template <typename T> concept C1 = /* ... */; template <std::size_t N> concept C2 = /* ... */; template <typename A, typename B> concept C3 = /* ... */; auto f = []<typename T1, C1 T2> requires C2<sizeof(T1) + sizeof(T2)> (T1 a1, T1 b1, T2 a2, auto a3, auto a4) requires C3<decltype(a4), T2> { // T2 is constrained by a type-constraint. // T1 and T2 are constrained by a requires-clause, and // T2 and the type of a4 are constrained by a trailing requires-clause. }; — end example]
— end note]
9
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The closure type for a non-generic lambda-expression with no lambda-capture and no explicit object parameter ([dcl.fct]) whose constraints (if any) are satisfied 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.
If the function call operator is a static member function, then the value returned by this conversion function is a pointer to the function call operator.
Otherwise, the value returned by this conversion function is a pointer to a function F that, when invoked, has the same effect as invoking the closure type's function call operator on a default-constructed instance of the closure type.
F is a constexpr function if the function call operator is a constexpr function and is an immediate function if the function call operator is an immediate function.
10
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For a generic lambda with no lambda-capture and no explicit object parameter ([dcl.fct]), 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.
11
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[Note 4: 
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 6: 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]
12
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If the function call operator template is a static member function template, then the value returned by any given specialization of this conversion function template is a pointer to the corresponding function call operator template specialization.
Otherwise, the value returned by any given specialization of this conversion function template is a pointer to a function F that, when invoked, has the same effect as invoking the generic lambda's corresponding function call operator template specialization on a default-constructed instance of the closure type.
F is a constexpr function if the corresponding specialization is a constexpr function and F is an immediate function if the function call operator template specialization is an immediate function.
[Note 5: 
This will result in the implicit instantiation of the generic lambda's body.
The instantiated generic lambda's return type and parameter types need to match the return type and parameter types of the pointer to function.
— end note]
[Example 7: 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]
13
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The conversion function or conversion function template is public, constexpr, non-virtual, non-explicit, const, and has a non-throwing exception specification.
[Example 8: 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); // error — end example]
14
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The lambda-expression's compound-statement yields the function-body ([dcl.fct.def]) of the function call operator, but it is not within the scope of the closure type.
[Example 9: 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].
15
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The closure type associated with a lambda-expression has no default constructor if the lambda-expression has a lambda-capture and a defaulted default constructor otherwise.
It has a defaulted copy constructor and a defaulted move constructor ([class.copy.ctor]).
It has a deleted copy assignment operator if the lambda-expression has a lambda-capture and defaulted copy and move assignment operators otherwise ([class.copy.assign]).
[Note 6: 
These special member functions are implicitly defined as usual, which can result in them being defined as deleted.
— end note]
16
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The closure type associated with a lambda-expression has an implicitly-declared destructor ([class.dtor]).
17
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A member of a closure type shall not be explicitly instantiated, explicitly specialized, or named in a friend declaration.