Lambda expressions (since C++11)

< cpp‎ | language

Constructs a closure: an unnamed function object capable of capturing variables in scope.


[ captures ] <tparams>(optional)(c++20) ( params ) specifiers exception attr -> ret requires(optional)(c++20) { body } (1)
[ captures ] ( params ) -> ret { body } (2)
[ captures ] ( params ) { body } (3)
[ captures ] { body } (4)

1) Full declaration.

2) Declaration of a const lambda: the objects captured by copy are const in the lambda body.

3) Omitted trailing-return-type: the return type of the closure's operator() is deduced from return statements as if for a function whose return type is declared auto.

4) Omitted parameter list: function takes no arguments, as if the parameter list was (). This form can only be used if none of constexpr, mutable, exception specification, attributes, or trailing return type is used.


captures - a comma-separated list of zero or more captures, optionally beginning with a capture-default.

See below for the detailed description of captures.

A lambda expression can use a variable without capturing it if the variable

  • is a non-local variable or has static or thread local storage duration (in which case the variable cannot be captured), or
  • is a reference that has been initialized with a constant expression.

A lambda expression can read the value of a variable without capturing it if the variable

  • has const non-volatile integral or enumeration type and has been initialized with a constant expression, or
  • is constexpr and trivially copy constructible.

Structured bindings cannot be captured.

(since C++17)
<tparams>(C++20) - a template parameter list (in angle brackets), used to provide names to the template parameters of a generic lambda (see ClosureType::operator() below). Like in a template declaration, the template parameter list may be followed by an optional requires-clause, which specifies the constraints on the template arguments. If provided, the template parameter list cannot be empty (<> is not allowed).
params - The list of parameters, as in named functions, except that default arguments are not allowed (until C++14). If auto is used as a type of a parameter, the lambda is a generic lambda. (since C++14)
specifiers - Optional sequence of specifiers.The following specifiers are allowed:
  • mutable: allows body to modify the parameters captured by copy, and to call their non-const member functions
  • constexpr: explicitly specifies that the function call operator is a constexpr function. When this specifier is not present, the function call operator will be constexpr anyway, if it happens to satisfy all constexpr function requirements
(since C++17)
exception - provides the exception specification or the noexcept clause for operator() of the closure type
attr - provides the attribute specification for operator() of the closure type
ret - Return type. If not present it's implied by the function return statements (or void if it doesn't return any value)
requires - adds a constraint to operator() of the closure type
body - Function body

The lambda expression is a prvalue expression of unique unnamed non-union non-aggregate class type, known as closure type, which is declared (for the purposes of ADL) in the smallest block scope, class scope, or namespace scope that contains the lambda expression. The closure type has the following members:


ret operator()(params) const { body }
(the keyword mutable was not used)
ret operator()(params) { body }
(the keyword mutable was used)
ret operator()(params) const { body }
(since C++14)
(generic lambda)
ret operator()(params) { body }
(since C++14)
(generic lambda, the keyword mutable was used)

Executes the body of the lambda-expression, when invoked. When accessing a variable, accesses its captured copy (for the entities captured by copy), or the original object (for the entities captured by reference). Unless the keyword mutable was used in the lambda-expression, the function-call operator is const-qualified and the objects that were captured by copy are non-modifiable from inside this operator(). The function-call operator is never volatile-qualified and never virtual.

The function-call operator is always constexpr if it satisfies the requirements of a constexpr function. It is also constexpr if the keyword constexpr was used in the lambda declaration.

(since C++17)

For every parameter in params whose type is specified as auto, an invented template parameter is added to template-params, in order of appearance. The invented template parameter may be a parameter pack if the corresponding function member of params is a function parameter pack.

// generic lambda, operator() is a template with two parameters
auto glambda = [](auto a, auto&& b) { return a < b; };
bool b = glambda(3, 3.14); // ok
// generic lambda, operator() is a template with one parameter
auto vglambda = [](auto printer) {
    return [=](auto&&... ts) // generic lambda, ts is a parameter pack
        return [=] { printer(ts...); }; // nullary lambda (takes no parameters)
auto p = vglambda([](auto v1, auto v2, auto v3) { std::cout << v1 << v2 << v3; });
auto q = p(1, 'a', 3.14); // outputs 1a3.14
q();                      // outputs 1a3.14

ClosureType's operator() cannot be explicitly instantiated or explicitly specialized.

(since C++14)

If the lambda definition uses an explicit template parameter list, that template parameter list is used with operator(). For every parameter in params whose type is specified as auto, an additional invented template parameter is appended to the end of that template parameter list:

// generic lambda, operator() is a template with two parameters
auto glambda = []<class T>(T a, auto&& b) { return a < b; };
// generic lambda, operator() is a template with one parameter pack
auto f = []<typename ...Ts>(Ts&& ...ts) {
   return foo(std::forward<Ts>(ts)...);
(since C++20)

the exception specification exception on the lambda-expression applies to the function-call operator or operator template.

For the purpose of name lookup, determining the type and value of the this pointer and for accessing non-static class members, the body of the closure type's function call operator is considered in the context of the lambda-expression.

struct X {
    int x, y;
    int operator()(int);
    void f()
        // the context of the following lambda is the member function X::f
            return operator()(this->x + y); // X::operator()(this->x + (*this).y)
                                            // this has type X*

ClosureType's operator() cannot be named in a friend declaration.

Dangling references

If a non-reference entity is captured by reference, implicitly or explicitly, and the function call operator of the closure object is invoked after the entity's lifetime has ended, undefined behavior occurs. The C++ closures do not extend the lifetimes of the captured references.

Same applies to the lifetime of the object pointed to by the captured this pointer.

ClosureType::operator ret(*)(params)()

(capture-less non-generic lambda)
using F = ret(*)(params);
operator F() const;
(until C++17)
using F = ret(*)(params);
constexpr operator F() const;
(since C++17)
(capture-less generic lambda)
template<template-params> using fptr_t = /*see below*/;
template<template-params> operator fptr_t<template-params>() const;
(since C++14)
(until C++17)
template<template-params> using fptr_t = /*see below*/;
template<template-params> operator fptr_t<template-params>() const;
(since C++17)

This user-defined conversion function is only defined if the capture list of the lambda-expression is empty. It is a public, constexpr (since C++17) non-virtual, non-explicit, const noexcept (since C++14) member function of the closure object.

A generic captureless lambda has user-defined conversion function template with the same invented template parameter list as the function-call operator template. If the return type is empty or auto, it is obtained by return type deduction on the function template specialization, which, in turn, is obtained by template argument deduction for conversion function templates.

void f1(int (*)(int)) {}
void f2(char (*)(int)) {}
void h(int (*)(int)) {} // #1
void h(char (*)(int)) {} // #2
auto glambda = [](auto a) { return a; };
f1(glambda); // ok
f2(glambda); // error: not convertible
h(glambda); // ok: calls #1 since #2 is not convertible
int& (*fpi)(int*) = [](auto* a)->auto& { return *a; }; // ok
(since C++14)

The value returned by this conversion function is a pointer to a function with C++ language linkage that, when invoked, has the same effect as invoking the closure object's function call operator directly.

This function is constexpr if the function call operator (or specialization, for generic lambdas) is constexpr.

auto Fwd= [](int(*fp)(int), auto a){return fp(a);};
auto C=[](auto a){return a;};
auto NC=[](auto a){static int s; return a;};
static_assert(Fwd(NC,3)==3); // error: no specialization can be constexpr because of s

If the closure object's operator() has a non-throwing exception specification, then the pointer returned by this function has the type pointer to noexcept function.

(since C++17)


ClosureType() = delete;
(until C++14)
ClosureType() = default;
(since C++20)(only if no captures are specified)
ClosureType(const ClosureType& ) = default;
(since C++14)
ClosureType(ClosureType&& ) = default;
(since C++14)

Closure types are not DefaultConstructible. Closure types have a deleted (until C++14)no (since C++14) default constructor.

(until C++20)

If no captures are specified, the closure type has a defaulted default constructor. Otherwise, it has no default constructor (this includes the case when there is a capture-default, even if it does not actually capture anything).

(since C++20)

The copy constructor and the move constructor are implicitly-declared (until C++14)declared as defaulted (since C++14) and may be implicitly-defined according to the usual rules for copy constructors and move constructors.

ClosureType::operator=(const ClosureType&)

ClosureType& operator=(const ClosureType&) = delete;
(until C++20)
ClosureType& operator=(const ClosureType&) = default;
ClosureType& operator=(ClosureType&&) = default;
(since C++20)
(only if no captures are specified)
ClosureType& operator=(const ClosureType&) = delete;
(since C++20)

The copy assignment operator is defined as deleted (and the move assignment operator is not declared). Closure types are not CopyAssignable.

(until C++20)

If no captures are specified, the closure type has a defaulted copy assignment operator and a defaulted move assignment operator. Otherwise, it has a deleted copy assignment operator (this includes the case when there is a capture-default, even if it does not actually capture anything).

(since C++20)


~ClosureType() = default;

The destructor is implicitly-declared.


T1 a;

T2 b;


If the lambda-expression captures anything by copy (either implicitly with capture clause [=] or explicitly with a capture that does not include the character &, e.g. [a, b, c]), the closure type includes unnamed non-static data members, declared in unspecified order, that hold copies of all entities that were so captured.

Those data members that correspond to captures without initializers are direct-initialized when the lambda-expression is evaluated. Those that correspond to captures with initializers are initialized as the initializer requires (could be copy- or direct-initialization). If an array is captured, array elements are direct-initialized in increasing index order. The order in which the data members are initialized is the order in which they are declared (which is unspecified).

The type of each data member is the type of the corresponding captured entity, except if the entity has reference type (in that case, references to functions are captured as lvalue references to the referenced functions, and references to objects are captured as copies of the referenced objects).

For the entities that are captured by reference (with the default capture [&] or when using the character &, e.g. [&a, &b, &c]), it is unspecified if additional data members are declared in the closure type , but any such additional members must satisfy LiteralType (since C++17).

Lambda-expressions are not allowed in unevaluated expressions, template arguments, alias declarations, typedef declarations, and anywhere in a function (or function template) declaration except the function body and the function's default arguments.

(until C++20)

Lambda capture

The captures is a comma-separated list of zero or more captures, optionally beginning with the capture-default. The only capture defaults are

  • & (implicitly capture the used automatic variables by reference) and
  • = (implicitly capture the used automatic variables by copy).

The current object (*this) can be implicitly captured if either capture default is present. If implicitly captured, it is always captured by reference, even if the capture default is =. The implicit capture of *this when the capture default is = is deprecated. (since C++20)

The syntax of an individual capture in captures is

identifier (1)
identifier ... (2)
identifier initializer (3) (since C++14)
& identifier (4)
& identifier ... (5)
& identifier initializer (6) (since C++14)
this (7)
* this (8) (since C++17)
1) simple by-copy capture
2) simple by-copy capture that is a pack expansion
3) by-copy capture with an initializer
4) simple by-reference capture
5) simple by-reference capture that is a pack expansion
6) by-reference capture with an initializer
7) simple by-reference capture of the current object
8) simple by-copy capture of the current object

If the capture-default is &, subsequent simple captures must not begin with &.

struct S2 { void f(int i); };
void S2::f(int i)
    [&]{};          // OK: by-reference capture default
    [&, i]{};       // OK: by-reference capture, except i is captured by copy
    [&, &i] {};     // Error: by-reference capture when by-reference is the default
    [&, this] {};   // OK, equivalent to [&]
    [&, this, i]{}; // OK, equivalent to [&, i]

If the capture-default is =, subsequent simple captures must begin with & or be *this (since C++17) or this (since C++20).

struct S2 { void f(int i); };
void S2::f(int i)
    [=]{};          // OK: by-copy capture default
    [=, &i]{};      // OK: by-copy capture, except i is captured by reference
    [=, *this]{};   // until C++17: Error: invalid syntax
                    // since c++17: OK: captures the enclosing S2 by copy
    [=, this] {};   // until C++20: Error: this when = is the default
                    // since C++20: OK, same as [=]

Any capture may appear only once:

struct S2 { void f(int i); };
void S2::f(int i)
    [i, i] {};        // Error: i repeated
    [this, *this] {}; // Error: "this" repeated (C++17)

Only lambda-expressions defined at block scope or in a default member initializer may have a capture-default or captures without initializers. For such lambda-expression, the reaching scope is defined as the set of enclosing scopes up to and including the innermost enclosing function (and its parameters). This includes nested block scopes and the scopes of enclosing lambdas if this lambda is nested.

The identifier in any capture without an initializer (other than the this-capture) is looked up using usual unqualified name lookup in the reaching scope of the lambda. The result of the lookup must be a variable with automatic storage duration declared in the reaching scope. The variable (or this) is explicitly captured.

A capture with an initializer acts as if it declares and explicitly captures a variable declared with type auto, whose declarative region is the body of the lambda expression (that is, it is not in scope within its initializer), except that:

  • if the capture is by-copy, the non-static data member of the closure object is another way to refer to that auto variable.
  • if the capture is by-reference, the reference variable's lifetime ends when the lifetime of the closure object ends

This is used to capture move-only types with a capture such as x = std::move(x).

This also makes it possible to capture by const reference, with &cr = std::as_const(x) or similar.

int x = 4;
auto y = [&r = x, x = x + 1]()->int
        r += 2;
        return x * x;
    }(); // updates ::x to 6 and initializes y to 25.
(since C++14)

If a capture list has a capture-default and does not explicitly capture the enclosing object (as this or *this) or an automatic variable, it captures it implicitly if

  • the body of the lambda odr-uses the variable or the this pointer
  • or the variable or the this pointer is named in a potentially-evaluated expression within an expression that depends on a generic lambda parameter (until C++17) (including when the implicit this-> is added before a use of non-static class member). For this purpose, the operand of typeid is always considered potentially-evaluated. Entities might be implicitly captured even if they are only named within a discarded statement. (since C++17)
void f(int, const int (&)[2] = {}) {} // #1
void f(const int&, const int (&)[1]) {} // #2
void test()
    const int x = 17;
    auto g0 = [](auto a) { f(x); }; // ok: calls #1, does not capture x
    auto g1 = [=](auto a) { f(x); }; // does not capture x in C++14, captures x in C++17
                                     // the capture can be optimized away
    auto g2 = [=](auto a) {
            int selector[sizeof(a) == 1 ? 1 : 2] = {};
            f(x, selector); // ok: is a dependent expression, so captures x
    auto g3 = [=](auto a) {
      typeid(a + x);  // captures x regardless of whether a + x is an unevaluated operand
(since C++14)

If the body of a lambda odr-uses an entity captured by copy, the member of the closure type is accessed. If it is not odr-using the entity, the access is to the original object:

void f(const int*);
void g()
    const int N = 10;
        int arr[N]; // not an odr-use: refers to g's const int N
        f(&N); // odr-use: causes N to be captured (by copy)
               // &N is the address of the closure object's member N, not g's N
If a lambda odr-uses a reference that is captured by reference, it is using the object referred-to by the original reference, not the captured reference itself:
#include <iostream>
auto make_function(int& x) {
  return [&]{ std::cout << x << '\n'; };
int main() {
  int i = 3;
  auto f = make_function(i); // the use of x in f binds directly to i
  i = 5;
  f(); // OK; prints 5

Within the body of a lambda, any use of decltype on any variable with automatic storage duration is as if it were captured and odr-used, even though decltype itself isn't an odr-use and no actual capture takes place:

void f3() {
    float x, &r = x;
    { // x and r are not captured (appearance in a decltype operand is not an odr-use)
        decltype(x) y1; // y1 has type float
        decltype((x)) y2 = y1; // y2 has type float const& because this lambda
                               // is not mutable and x is an lvalue
        decltype(r) r1 = y1;   // r1 has type float& (transformation not considered)
        decltype((r)) r2 = y2; // r2 has type float const&

Any entity captured by a lambda (implicitly or explicitly) is odr-used by the lambda-expression (therefore, implicit capture by a nested lambda triggers implicit capture in the enclosing lambda).

All implicitly-captured variables must be declared within the reaching scope of the lambda.

If a lambda captures the enclosing object (as this or *this), either the nearest enclosing function must be a non-static member function or the lambda must be in a default member initializer:

struct s2 {
  double ohseven = .007;
  auto f() { // nearest enclosing function for the following two lambdas
    return [this] { // capture the enclosing s2 by reference
      return [*this] { // capture the enclosing s2 by copy (C++17)
          return ohseven;// OK
  auto g() {
     return []{ // capture nothing
         return [*this]{};// error: *this not captured by outer lambda-expression

If a lambda expression (or an instantiation of a generic lambda's function call operator) ODR-uses this or any variable with automatic storage duration, it must be captured by the lambda expression.

void f1(int i)
    int const N = 20;
    auto m1 = [=] {
            int const M = 30;
            auto m2 = [i] {
                    int x[N][M]; // N and M are not odr-used 
                                 // (ok that they are not captured)
                    x[0][0] = i; // i is explicitly captured by m2
                                 // and implicitly captured by m1
    struct s1 // local class within f1()
        int f;
        void work(int n) // non-static member function
            int m = n * n;
            int j = 40;
            auto m3 = [this, m] {
                auto m4 = [&, j] { // error: j is not captured by m3
                        int x = n; // error: n is implicitly captured by m4
                                   // but not captured by m3
                        x += m;    // ok: m is implicitly captured by m4
                                   // and explicitly captured by m3
                        x += i;    // error: i is outside of the reaching scope
                                   // (which ends at work())
                        x += f;    // ok: this is captured implicitly by m4
                                   // and explicitly captured by m3

Class members cannot be captured explicitly by a capture without initializer (as mentioned above, only variables are permitted in the capture list):

class S {
  int x = 0;
  void f() {
    int i = 0;
//  auto l1 = [i, x]{ use(i, x); };    // error: x is not a variable
    auto l2 = [i, x=x]{ use(i, x); };  // OK, copy capture
    i = 1; x = 1; l2(); // calls use(0,0)
    auto l3 = [i, &x=x]{ use(i, x); }; // OK, reference capture
    i = 2; x = 2; l3(); // calls use(1,2)

When a lambda captures a member using implicit by-copy capture, it does not make a copy of that member variable: the use of a member variable m is treated as an expression (*this).m, and *this is always implicitly captured by reference:

class S {
  int x = 0;
  void f() {
    int i = 0;
    auto l1 = [=]{ use(i, x); }; // captures a copy of i and a copy of the this pointer
    i = 1; x = 1; l1(); // calls use(0,1), as if i by copy and x by reference
    auto l2 = [i, this]{ use(i, x); }; // same as above, made explicit
    i = 2; x = 2; l2(); // calls use(1,2), as if i by copy and x by reference
    auto l3 = [&]{ use(i, x); }; // captures i by reference and a copy of the this pointer
    i = 3; x = 2; l3(); // calls use(3,2), as if i and x are both by reference
    auto l4 = [i, *this]{ use(i, x); }; // makes a copy of *this, including a copy of x
    i = 4; x = 4; l4(); // calls use(3,2), as if i and x are both by copy

If a lambda-expression appears in a default argument, it cannot explicitly or implicitly capture anything.

Members of anonymous unions cannot be captured.

If a nested lambda m2 captures something that is also captured by the immediately enclosing lambda m1, then m2's capture is transformed as follows:

  • if the enclosing lambda m1 captures by copy, m2 is capturing the non-static member of m1's closure type, not the original variable or this.
  • if the enclosing lambda m1 by reference, m2 is capturing the original variable or this.
#include <iostream>
int main()
    int a = 1, b = 1, c = 1;
    auto m1 = [a, &b, &c]() mutable {
        auto m2 = [a, b, &c]() mutable {
            std::cout << a << b << c << '\n';
            a = 4; b = 4; c = 4;
        a = 3; b = 3; c = 3;
    a = 2; b = 2; c = 2;
    m1();                             // calls m2() and prints 123
    std::cout << a << b << c << '\n'; // prints 234


This example shows how to pass a lambda to a generic algorithm and how objects resulting from a lambda declaration can be stored in std::function objects.

#include <vector>
#include <iostream>
#include <algorithm>
#include <functional>
int main()
    std::vector<int> c = {1, 2, 3, 4, 5, 6, 7};
    int x = 5;
    c.erase(std::remove_if(c.begin(), c.end(), [x](int n) { return n < x; }), c.end());
    std::cout << "c: ";
    std::for_each(c.begin(), c.end(), [](int i){ std::cout << i << ' '; });
    std::cout << '\n';
    // the type of a closure cannot be named, but can be inferred with auto
    // since C++14, lambda could own default arguments
    auto func1 = [](int i = 6) { return i + 4; };
    std::cout << "func1: " << func1() << '\n';
    // like all callable objects, closures can be captured in std::function
    // (this may incur unnecessary overhead)
    std::function<int(int)> func2 = [](int i) { return i + 4; };
    std::cout << "func2: " << func2(6) << '\n';


c: 5 6 7
func1: 10
func2: 10

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 975 C++11 the return type of closure's operator() was only deduced if lambda body contains a single return deduced as if for C++14 auto-returning function
CWG 1891 C++14 closure had a deleted default ctor and implicit copy/move ctors no default and defaulted copy/move
CWG 1722 C++14 the conversion function for captureless lambdas had unspecified exception specification conversion function is noexcept

See also

auto specifier specifies a type defined by an expression (C++11)
wraps callable object of any type with specified function call signature
(class template)