7 Expressions [expr]

7.5 Primary expressions [expr.prim]

7.5.5 Lambda expressions [expr.prim.lambda]

7.5.5.1 General [expr.prim.lambda.general]

lambda-specifier:
consteval
constexpr
mutable
static
1
#
A lambda-expression provides a concise way to create a simple function object.
[Example 1: #include <algorithm> #include <cmath> void abssort(float* x, unsigned N) { std::sort(x, x + N, [](float a, float b) { return std::abs(a) < std::abs(b); }); } — end example]
2
#
A lambda-expression is a prvalue whose result object is called the closure object.
[Note 1: 
A closure object behaves like a function object.
— end note]
3
#
An ambiguity can arise because a requires-clause can end in an attribute-specifier-seq, which collides with the attribute-specifier-seq in lambda-expression.
In such cases, any attributes are treated as attribute-specifier-seq in lambda-expression.
[Note 2: 
Such ambiguous cases cannot have valid semantics because the constraint expression would not have type bool.
[Example 2: auto x = []<class T> requires T::operator int [[some_attribute]] (int) { } — end example]
— end note]
4
#
A lambda-specifier-seq shall contain at most one of each lambda-specifier and shall not contain both constexpr and consteval.
If the lambda-declarator contains an explicit object parameter ([dcl.fct]), then no lambda-specifier in the lambda-specifier-seq shall be mutable or static.
The lambda-specifier-seq shall not contain both mutable and static.
If the lambda-specifier-seq contains static, there shall be no lambda-capture.
[Note 3: 
The trailing requires-clause is described in [dcl.decl].
— end note]
5
#
If the lambda-declarator does not include a trailing-return-type, it is considered to be -> auto.
[Note 4: 
In that case, the return type is deduced from return statements as described in [dcl.spec.auto].
— end note]
[Example 3: auto x1 = [](int i) { return i; }; // OK, return type is int auto x2 = []{ return { 1, 2 }; }; // error: deducing return type from braced-init-list int j; auto x3 = [&]()->auto&& { return j; }; // OK, return type is int& — end example]
6
#
A lambda is a generic lambda if the lambda-expression has any generic parameter type placeholders ([dcl.spec.auto]), or if the lambda has a template-parameter-list.
[Example 4: int i = [](int i, auto a) { return i; }(3, 4); // OK, a generic lambda int j = []<class T>(T t, int i) { return i; }(3, 4); // OK, a generic lambda — end example]

7.5.5.2 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
#
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
#
The closure type is not an aggregate type ([dcl.init.aggr]) and not a structural type ([temp.param]).
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
#
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
#
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 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
#
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
#
[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
#
[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
#
The closure type for a non-generic lambda-expression with no lambda-capture 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 the address of the function call operator.
Otherwise, 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 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
#
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.
11
#
[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
#
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 the address of the corresponding function call operator template specialization.
Otherwise, 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 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
#
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
#
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
#
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
#
The closure type associated with a lambda-expression has an implicitly-declared destructor ([class.dtor]).
17
#
A member of a closure type shall not be explicitly instantiated, explicitly specialized, or named in a friend declaration.

7.5.5.3 Captures [expr.prim.lambda.capture]

simple-capture:
identifier ...
& identifier ...
this
* this
1
#
The body of a lambda-expression may refer to local entities of enclosing block scopes by capturing those entities, as described below.
2
#
If a lambda-capture includes a capture-default that is &, no identifier in a simple-capture of that lambda-capture shall be preceded by &.
If a lambda-capture includes a capture-default that is =, each simple-capture of that lambda-capture shall be of the form “& identifier ...”, “this”, or “* this”.
[Note 1: 
The form [&,this] is redundant but accepted for compatibility with ISO C++ 2014.
— end note]
Ignoring appearances in initializers of init-captures, an identifier or this shall not appear more than once in a lambda-capture.
[Example 1: struct S2 { void f(int i); }; void S2::f(int i) { [&, i]{ }; // OK [&, this, i]{ }; // OK, equivalent to [&, i] [&, &i]{ }; // error: i preceded by & when & is the default [=, *this]{ }; // OK [=, this]{ }; // OK, equivalent to [=] [i, i]{ }; // error: i repeated [this, *this]{ }; // error: this appears twice } — end example]
3
#
A lambda-expression shall not have a capture-default or simple-capture in its lambda-introducer unless its innermost enclosing scope is a block scope ([basic.scope.block]) or it appears within a default member initializer and its innermost enclosing scope is the corresponding class scope ([basic.scope.class]).
4
#
The identifier in a simple-capture shall denote a local entity ([basic.lookup.unqual], [basic.pre]).
The simple-captures this and * this denote the local entity *this.
An entity that is designated by a simple-capture is said to be explicitly captured.
5
#
If an identifier in a capture appears as the declarator-id of a parameter of the lambda-declarator's parameter-declaration-clause or as the name of a template parameter of the lambda-expression's template-parameter-list, the program is ill-formed.
[Example 2: void f() { int x = 0; auto g = [x](int x) { return 0; }; // error: parameter and capture have the same name auto h = [y = 0]<typename y>(y) { return 0; }; // error: template parameter and capture // have the same name } — end example]
6
#
An init-capture inhabits the lambda scope ([basic.scope.lambda]) of the lambda-expression.
An init-capture without ellipsis behaves as if it declares and explicitly captures a variable of the form “auto init-capture ;”, except that:
  • (6.1)
    if the capture is by copy (see below), the non-static data member declared for the capture and the variable are treated as two different ways of referring to the same object, which has the lifetime of the non-static data member, and no additional copy and destruction is performed, and
  • (6.2)
    if the capture is by reference, the variable's lifetime ends when the closure object's lifetime ends.
[Note 2: 
This enables an init-capture like “x = std​::​move(x)”; the second “x” must bind to a declaration in the surrounding context.
— end note]
[Example 3: int x = 4; auto y = [&r = x, x = x+1]()->int { r += 2; return x+2; }(); // Updates ​::​x to 6, and initializes y to 7. auto z = [a = 42](int a) { return 1; }; // error: parameter and conceptual local variable have the same name auto counter = [i=0]() mutable -> decltype(i) { // OK, returns int return i++; }; — end example]
7
#
For the purposes of lambda capture, an expression potentially references local entities as follows:
If an expression potentially references a local entity within a scope in which it is odr-usable ([basic.def.odr]), and the expression would be potentially evaluated if the effect of any enclosing typeid expressions ([expr.typeid]) were ignored, the entity is said to be implicitly captured by each intervening lambda-expression with an associated capture-default that does not explicitly capture it.
The implicit capture of *this is deprecated when the capture-default is =; see [depr.capture.this].
[Example 4: void f(int, const int (&)[2] = {}); // #1 void f(const int&, const int (&)[1]); // #2 void test() { const int x = 17; auto g = [](auto a) { f(x); // OK, calls #1, does not capture x }; auto g1 = [=](auto a) { f(x); // OK, calls #1, captures x }; auto g2 = [=](auto a) { int selector[sizeof(a) == 1 ? 1 : 2]{}; f(x, selector); // OK, captures x, can call #1 or #2 }; auto g3 = [=](auto a) { typeid(a + x); // captures x regardless of whether a + x is an unevaluated operand }; }
Within g1, an implementation can optimize away the capture of x as it is not odr-used.
— end example]
[Note 4: 
The set of captured entities is determined syntactically, and entities are implicitly captured even if the expression denoting a local entity is within a discarded statement ([stmt.if]).
[Example 5: template<bool B> void f(int n) { [=](auto a) { if constexpr (B && sizeof(a) > 4) { (void)n; // captures n regardless of the value of B and sizeof(int) } }(0); } — end example]
— end note]
8
#
An entity is captured if it is captured explicitly or implicitly.
An entity captured by a lambda-expression is odr-used ([basic.def.odr]) by the lambda-expression.
[Note 5: 
As a consequence, if a lambda-expression explicitly captures an entity that is not odr-usable, the program is ill-formed ([basic.def.odr]).
— end note]
[Example 6: 
🔗
void f1(int i) { int const N = 20; auto m1 = [=]{ int const M = 30; auto m2 = [i]{ int x[N][M]; // OK, N and M are not odr-used x[0][0] = i; // OK, i is explicitly captured by m2 and implicitly captured by m1 }; }; struct s1 { int f; void work(int n) { int m = n*n; int j = 40; auto m3 = [this,m] { auto m4 = [&,j] { // error: j not odr-usable due to intervening lambda m3 int x = n; // error: n is odr-used but not odr-usable due to intervening lambda m3 x += m; // OK, m implicitly captured by m4 and explicitly captured by m3 x += i; // error: i is odr-used but not odr-usable // due to intervening function and class scopes x += f; // OK, this captured implicitly by m4 and explicitly by m3 }; }; } }; } struct s2 { double ohseven = .007; auto f() { return [this] { return [*this] { return ohseven; // OK }; }(); } auto g() { return [] { return [*this] { }; // error: *this not captured by outer lambda-expression }(); } }; — end example]
9
#
[Note 6: 
Because local entities are not odr-usable within a default argument ([basic.def.odr]), a lambda-expression appearing in a default argument cannot implicitly or explicitly capture any local entity.
Such a lambda-expression can still have an init-capture if any full-expression in its initializer satisfies the constraints of an expression appearing in a default argument ([dcl.fct.default]).
— end note]
[Example 7: void f2() { int i = 1; void g1(int = ([i]{ return i; })()); // error void g2(int = ([i]{ return 0; })()); // error void g3(int = ([=]{ return i; })()); // error void g4(int = ([=]{ return 0; })()); // OK void g5(int = ([]{ return sizeof i; })()); // OK void g6(int = ([x=1]{ return x; })()); // OK void g7(int = ([x=i]{ return x; })()); // error } — end example]
10
#
An entity is captured by copy if
For each entity captured by copy, an unnamed non-static data member is declared in the closure type.
The declaration order of these members is unspecified.
The type of such a data member is the referenced type if the entity is a reference to an object, an lvalue reference to the referenced function type if the entity is a reference to a function, or the type of the corresponding captured entity otherwise.
A member of an anonymous union shall not be captured by copy.
11
#
Every id-expression within the compound-statement of a lambda-expression that is an odr-use ([basic.def.odr]) of an entity captured by copy is transformed into an access to the corresponding unnamed data member of the closure type.
[Note 7: 
An id-expression that is not an odr-use refers to the original entity, never to a member of the closure type.
However, such an id-expression can still cause the implicit capture of the entity.
— end note]
If *this is captured by copy, each expression that odr-uses *this is transformed to instead refer to the corresponding unnamed data member of the closure type.
[Example 8: void f(const int*); void g() { const int N = 10; [=] { int arr[N]; // OK, not an odr-use, refers to automatic variable f(&N); // OK, causes N to be captured; &N points to // the corresponding member of the closure type }; } — end example]
12
#
An entity is captured by reference if it is implicitly or explicitly captured but not captured by copy.
It is unspecified whether additional unnamed non-static data members are declared in the closure type for entities captured by reference.
If declared, such non-static data members shall be of literal type.
[Example 9: // The inner closure type must be a literal type regardless of how reference captures are represented. static_assert([](int n) { return [&n] { return ++n; }(); }(3) == 4); — end example]
A bit-field or a member of an anonymous union shall not be captured by reference.
13
#
An id-expression within the compound-statement of a lambda-expression that is an odr-use of a reference captured by reference refers to the entity to which the captured reference is bound and not to the captured reference.
[Note 8: 
The validity of such captures is determined by the lifetime of the object to which the reference refers, not by the lifetime of the reference itself.
— end note]
[Example 10: auto h(int &r) { return [&] { ++r; // Valid after h returns if the lifetime of the // object to which r is bound has not ended }; } — end example]
14
#
If a lambda-expression m2 captures an entity and that entity is captured by an immediately enclosing lambda-expression m1, then m2's capture is transformed as follows:
  • (14.1)
    If m1 captures the entity by copy, m2 captures the corresponding non-static data member of m1's closure type; if m1 is not mutable, the non-static data member is considered to be const-qualified.
  • (14.2)
    If m1 captures the entity by reference, m2 captures the same entity captured by m1.
[Example 11: 
The nested lambda-expressions and invocations below will output 123234.
int a = 1, b = 1, c = 1; auto m1 = [a, &b, &c]() mutable { auto m2 = [a, b, &c]() mutable { std::cout << a << b << c; a = 4; b = 4; c = 4; }; a = 3; b = 3; c = 3; m2(); }; a = 2; b = 2; c = 2; m1(); std::cout << a << b << c; — end example]
15
#
When the lambda-expression is evaluated, the entities that are captured by copy are used to direct-initialize each corresponding non-static data member of the resulting closure object, and the non-static data members corresponding to the init-captures are initialized as indicated by the corresponding initializer (which may be copy- or direct-initialization).
(For array members, the array elements are direct-initialized in increasing subscript order.)
These initializations are performed in the (unspecified) order in which the non-static data members are declared.
[Note 9: 
This ensures that the destructions will occur in the reverse order of the constructions.
— end note]
16
#
[Note 10: 
If a non-reference entity is implicitly or explicitly captured by reference, invoking the function call operator of the corresponding lambda-expression after the lifetime of the entity has ended is likely to result in undefined behavior.
— end note]
17
#
A simple-capture containing an ellipsis is a pack expansion ([temp.variadic]).
An init-capture containing an ellipsis is a pack expansion that declares an init-capture pack ([temp.variadic]).
[Example 12: template<class... Args> void f(Args... args) { auto lm = [&, args...] { return g(args...); }; lm(); auto lm2 = [...xs=std::move(args)] { return g(xs...); }; lm2(); } — end example]