Lambdas

Syntax#

• [default-capture, capture-list] (argument-list) mutable throw-specification attributes -> return-type { lambda-body } // Order of lambda specifiers and attributes.
• [capture-list] (argument-list) { lambda-body } // Common lambda definition.
• [=] (argument-list) { lambda-body } // Captures all needed local variables by value.
• [&] (argument-list) { lambda-body } // Captures all needed local variables by reference.
• [capture-list] { lambda-body } // Argument list and specifiers can be omitted.

Parameters#

//Explicitly define this lambdas as constexpr
[]() constexpr {
    //Do stuff
}

What is a lambda expression?

A lambda expression provides a concise way to create simple function objects. A lambda expression is a prvalue whose result object is called closure object, which behaves like a function object.

The name ‘lambda expression’ originates from lambda calculus, which is a mathematical formalism invented in the 1930s by Alonzo Church to investigate questions about logic and computability. Lambda calculus formed the basis of LISP, a functional programming language. Compared to lambda calculus and LISP, C++ lambda expressions share the properties of being unnamed, and to capture variables from the surrounding context, but they lack the ability to operate on and return functions.

A lambda expression is often used as an argument to functions that take a callable object. That can be simpler than creating a named function, which would be only used when passed as the argument. In such cases, lambda expressions are generally preferred because they allow defining the function objects inline.

A lambda consists typically of three parts: a capture list [], an optional parameter list () and a body {}, all of which can be empty:

[](){}                // An empty lambda, which does and returns nothing

Capture list

[] is the capture list. By default, variables of the enclosing scope cannot be accessed by a lambda. Capturing a variable makes it accessible inside the lambda, either as a copy or as a reference. Captured variables become a part of the lambda; in contrast to function arguments, they do not have to be passed when calling the lambda.

int a = 0;                       // Define an integer variable
auto f = []()   { return a*9; }; // Error: 'a' cannot be accessed
auto f = [a]()  { return a*9; }; // OK, 'a' is "captured" by value
auto f = [&a]() { return a++; }; // OK, 'a' is "captured" by reference
                                 //      Note: It is the responsibility of the programmer
                                 //      to ensure that a is not destroyed before the
                                 //      lambda is called.
auto b = f();                    // Call the lambda function. a is taken from the capture list and not passed here.

Parameter list

() is the parameter list, which is almost the same as in regular functions. If the lambda takes no arguments, these parentheses can be omitted (except if you need to declare the lambda mutable). These two lambdas are equivalent:

auto call_foo  = [x](){ x.foo(); };
auto call_foo2 = [x]{ x.foo(); };

The parameter list can use the placeholder type auto instead of actual types. By doing so, this argument behaves like a template parameter of a function template. Following lambdas are equivalent when you want to sort a vector in generic code:

auto sort_cpp11 = [](std::vector<T>::const_reference lhs, std::vector<T>::const_reference rhs) { return lhs < rhs; }; 
auto sort_cpp14 = [](const auto &lhs, const auto &rhs) { return lhs < rhs; }; 

Function body

{} is the body, which is the same as in regular functions.

Calling a lambda

A lambda expression’s result object is a closure, which can be called using the operator() (as with other function objects):

int multiplier = 5;
auto timesFive = [multiplier](int a) { return a * multiplier; }; 
std::out << timesFive(2); // Prints 10

multiplier = 15;
std::out << timesFive(2); // Still prints 2*5 == 10

Return Type

By default, the return type of a lambda expression is deduced.

[](){ return true; };

In this case the return type is bool.

You can also manually specify the return type using the following syntax:

[]() -> bool { return true; };

Mutable Lambda

Objects captured by value in the lambda are by default immutable. This is because the operator() of the generated closure object is const by default.

auto func = [c = 0](){++c; std::cout << c;};  // fails to compile because ++c
                                              // tries to mutate the state of
                                              // the lambda.

Modifying can be allowed by using the keyword mutable, which make the closer object’s operator() non-const:

auto func = [c = 0]() mutable {++c; std::cout << c;};

If used together with the return type, mutable comes before it.

auto func = [c = 0]() mutable -> int {++c; std::cout << c; return c;};

An example to illustrate the usefulness of lambdas

Before C++11:

// Generic functor used for comparison
struct islessthan
{
    islessthan(int threshold) : _threshold(threshold) {}

    bool operator()(int value) const
    {
        return value < _threshold;
    }
private:
    int _threshold;
};

// Declare a vector
const int arr[] = { 1, 2, 3, 4, 5 };
std::vector<int> vec(arr, arr+5);

// Find a number that's less than a given input (assume this would have been function input)
int threshold = 10;
std::vector<int>::iterator it = std::find_if(vec.begin(), vec.end(), islessthan(threshold));

Since C++11:

// Declare a vector
std::vector<int> vec{ 1, 2, 3, 4, 5 };

// Find a number that's less than a given input (assume this would have been function input)
int threshold = 10;
auto it = std::find_if(vec.begin(), vec.end(), [threshold](int value) { return value < threshold; });

Specifying the return type

For lambdas with a single return statement, or multiple return statements whose expressions are of the same type, the compiler can deduce the return type:

// Returns bool, because "value > 10" is a comparison which yields a Boolean result
auto l = [](int value) {
    return value > 10;
}

For lambdas with multiple return statements of different types, the compiler can’t deduce the return type:

// error: return types must match if lambda has unspecified return type
auto l = [](int value) {
    if (value < 10) {
        return 1;
    } else {
        return 1.5;
    }
};

In this case you have to specify the return type explicitly:

// The return type is specified explicitly as 'double'
auto l = [](int value) -> double {
    if (value < 10) {
        return 1;
    } else {
        return 1.5;
    }
};

The rules for this match the rules for auto type deduction. Lambdas without explicitly specified return types never return references, so if a reference type is desired it must be explicitly specified as well:

auto copy = [](X& x) { return x; };       // 'copy' returns an X, so copies its input
auto ref  = [](X& x) -> X& { return x; }; // 'ref' returns an X&, no copy

Capture by value

If you specify the variable’s name in the capture list, the lambda will capture it by value. This means that the generated closure type for the lambda stores a copy of the variable. This also requires that the variable’s type be copy-constructible:

int a = 0;

[a]() {
    return a;   // Ok, 'a' is captured by value
};

auto p = std::unique_ptr<T>(...);

[p]() {         // Compile error; `unique_ptr` is not copy-constructible
    return p->createWidget(); 
};

From C++14 on, it is possible to initialize variables on the spot. This allows move only types to be captured in the lambda.

auto p = std::make_unique<T>(...);

[p = std::move(p)]() {
    return p->createWidget(); 
};

Even though a lambda captures variables by value when they are given by their name, such variables cannot be modified within the lambda body by default. This is because the closure type puts the lambda body in a declaration of operator() const.

The const applies to accesses to member variables of the closure type, and captured variables that are members of the closure (all appearances to the contrary):

int a = 0;

[a]() {
    a = 2;      // Illegal, 'a' is accessed via `const`

    decltype(a) a1 = 1; 
    a1 = 2; // valid: variable 'a1' is not const
};

To remove the const, you have to specify the keyword mutable on the lambda:

int a = 0;

[a]() mutable {
    a = 2;      // OK, 'a' can be modified
    return a;
};

Because a was captured by value, any modifications done by calling the lambda will not affect a. The value of a was copied into the lambda when it was constructed, so the lambda’s copy of a is separate from the external a variable.

int a = 5 ; 
auto plus5Val = [a] (void) { return a + 5 ; } ; 
auto plus5Ref = [&a] (void) {return a + 5 ; } ; 

a = 7 ; 
std::cout << a << ", value " << plus5Val() << ", reference " << plus5Ref() ;
// The result will be "7, value 10, reference 12"

Generalized capture

Lambdas can capture expressions, rather than just variables. This permits lambdas to store move-only types:

auto p = std::make_unique<T>(...);

auto lamb = [p = std::move(p)]() //Overrides capture-by-value of `p`.
{
  p->SomeFunc();
};

This moves the outer p variable into the lambda capture variable, also called p. lamb now owns the memory allocated by make_unique. Because the closure contains a type that is non-copyable, this means that lamb is itself non-copyable. But it can be moved:

auto lamb_copy = lamb; //Illegal
auto lamb_move = std::move(lamb); //legal.

Now lamb_move owns the memory.

Note that std::function<> requires that the values stored be copyable. You can write your own move-only-requiring std::function, or you could just stuff the lambda into a shared_ptr wrapper:

auto shared_lambda = [](auto&& f){
  return [spf = std::make_shared<std::decay_t<decltype(f)>>(decltype(f)(f))]
  (auto&&...args)->decltype(auto) {
    return (*spf)(decltype(args)(args)...);
  };
};
auto lamb_shared = shared_lambda(std::move(lamb_move));

takes our move-only lambda and stuffs its state into a shared pointer then returns a lambda that can be copied, and then stored in a std::function or similar.

Generalized capture uses auto type deduction for the variable’s type. It will declare these captures as values by default, but they can be references as well:

int a = 0;

auto lamb = [&v = a](int add) //Note that `a` and `v` have different names
{
  v += add; //Modifies `a`
};

lamb(20); //`a` becomes 20.

Generalize capture does not need to capture an external variable at all. It can capture an arbitrary expression:

auto lamb = [p = std::make_unique<T>(...)]()
{
    p->SomeFunc();
}

This is useful for giving lambdas arbitrary values that they can hold and potentially modify, without having to declare them externally to the lambda. Of course, that is only useful if you do not intend to access those variables after the lambda has completed its work.

Capture by reference

If you precede a local variable’s name with an &, then the variable will be captured by reference. Conceptually, this means that the lambda’s closure type will have a reference variable, initialized as a reference to the corresponding variable from outside of the lambda’s scope. Any use of the variable in the lambda body will refer to the original variable:

// Declare variable 'a'
int a = 0;

// Declare a lambda which captures 'a' by reference
auto set = [&a]() {
    a = 1;
};

set();
assert(a == 1);

The keyword mutable is not needed, because a itself is not const.

Of course, capturing by reference means that the lambda must not escape the scope of the variables it captures. So you could call functions that take a function, but you must not call a function that will store the lambda beyond the scope of your references. And you must not return the lambda.

Default capture

By default, local variables that are not explicitly specified in the capture list, cannot be accessed from within the lambda body. However, it is possible to implicitly capture variables named by the lambda body:

int a = 1;
int b = 2;

// Default capture by value
[=]() { return a + b; }; // OK; a and b are captured by value

// Default capture by reference
[&]() { return a + b; }; // OK; a and b are captured by reference

Explicit capturing can still be done alongside implicit default capturing. The explicit capture definition will override the default capture:

int a = 0;
int b = 1;

[=, &b]() {
    a = 2; // Illegal; 'a' is capture by value, and lambda is not 'mutable'
    b = 2; // OK; 'b' is captured by reference
};

Generic lambdas

Lambda functions can take arguments of arbitrary types. This allows a lambda to be more generic:

auto twice = [](auto x){ return x+x; };

int i = twice(2); // i == 4
std::string s = twice("hello"); // s == "hellohello"

This is implemented in C++ by making the closure type’s operator() overload a template function. The following type has equivalent behavior to the above lambda closure:

struct _unique_lambda_type
{
  template<typename T>
  auto operator() (T x) const {return x + x;}
};

Not all parameters in a generic lambda need be generic:

[](auto x, int y) {return x + y;}

Here, x is deduced based on the first function argument, while y will always be int.

Generic lambdas can take arguments by reference as well, using the usual rules for auto and &. If a generic parameter is taken as auto&&, this is a forwarding reference to the passed in argument and not an rvalue reference:

auto lamb1 = [](int &&x) {return x + 5;};
auto lamb2 = [](auto &&x) {return x + 5;};
int x = 10;
lamb1(x); // Illegal; must use `std::move(x)` for `int&&` parameters.
lamb2(x); // Legal; the type of `x` is deduced as `int&`.

Lambda functions can be variadic and perfectly forward their arguments:

auto lam = [](auto&&... args){return f(std::forward<decltype(args)>(args)...);};

or:

auto lam = [](auto&&... args){return f(decltype(args)(args)...);};

which only works “properly” with variables of type auto&&.

A strong reason to use generic lambdas is for visiting syntax.

boost::variant<int, double> value;
apply_visitor(value, [&](auto&& e){
  std::cout << e;
});

Here we are visiting in a polymorphic manner; but in other contexts, the names of the type we are passing isn’t interesting:

mutex_wrapped<std::ostream&> os = std::cout;
os.write([&](auto&& os){
  os << "hello world\n";
});

Repeating the type of std::ostream& is noise here; it would be like having to mention the type of a variable every time you use it. Here we are creating a visitor, but no a polymorphic one; auto is used for the same reason you might use auto in a for(:) loop.

Conversion to function pointer

If a lambda’s capture list is empty, then the lambda has an implicit conversion to a function pointer that takes the same arguments and returns the same return type:

auto sorter = [](int lhs, int rhs) -> bool {return lhs < rhs;};

using func_ptr = bool(*)(int, int);
func_ptr sorter_func = sorter; // implicit conversion

Such a conversion may also be enforced using unary plus operator:

func_ptr sorter_func2 = +sorter; // enforce implicit conversion

Calling this function pointer behaves exactly like invoking operator() on the lambda. This function pointer is in no way reliant on the source lambda closure’s existence. It therefore may outlive the lambda closure.

This feature is mainly useful for using lambdas with APIs that deal in function pointers, rather than C++ function objects.

Conversion to a function pointer is also possible for generic lambdas with an empty capture list. If necessary, template argument deduction will be used to select the correct specialization.

auto sorter = [](auto lhs, auto rhs) { return lhs < rhs; };
using func_ptr = bool(*)(int, int);
func_ptr sorter_func = sorter;  // deduces int, int
// note however that the following is ambiguous
// func_ptr sorter_func2 = +sorter;

Class lambdas and capture of this

A lambda expression evaluated in a class’ member function is implicitly a friend of that class:

class Foo
{
private:
    int i;
    
public:
    Foo(int val) : i(val) {}
    
    // definition of a member function
    void Test()
    {
        auto lamb = [](Foo &foo, int val)
        {
            // modification of a private member variable
            foo.i = val;
        };
        
        // lamb is allowed to access a private member, because it is a friend of Foo
        lamb(*this, 30);
    }
};

Such a lambda is not only a friend of that class, it has the same access as the class it is declared within has.

Lambdas can capture the this pointer which represents the object instance the outer function was called on. This is done by adding this to the capture list:

class Foo
{
private:
    int i;
    
public:
    Foo(int val) : i(val) {}
    
    void Test()
    {
        // capture the this pointer by value
        auto lamb = [this](int val)
        {
            i = val;
        };
        
        lamb(30);
    }
};

When this is captured, the lambda can use member names of its containing class as though it were in its containing class. So an implicit this-> is applied to such members.

Be aware that this is captured by value, but not the value of the type. It is captured by the value of this, which is a pointer. As such, the lambda does not own this. If the lambda out lives the lifetime of the object that created it, the lambda can become invalid.

This also means that the lambda can modify this without being declared mutable. It is the pointer which is const, not the object being pointed to. That is, unless the outer member function was itself a const function.

Also, be aware that the default capture clauses, both [=] and [&], will also capture this implicitly. And they both capture it by the value of the pointer. Indeed, it is an error to specify this in the capture list when a default is given.

Lambdas can capture a copy of the this object, created at the time the lambda is created. This is done by adding *this to the capture list:

class Foo
{
private:
    int i;
    
public:
    Foo(int val) : i(val) {}
    
    void Test()
    {
        // capture a copy of the object given by the this pointer
        auto lamb = [*this](int val) mutable
        {
            i = val;
        };
        
        lamb(30); // does not change this->i
    }
};

Porting lambda functions to C++03 using functors

Lambda functions in C++ are syntactic sugar that provide a very concise syntax for writing functors. As such, equivalent functionality can be obtained in C++03 (albeit much more verbose) by converting the lambda function into a functor:

// Some dummy types:
struct T1 {int dummy;};
struct T2 {int dummy;};
struct R {int dummy;};

// Code using a lambda function (requires C++11)
R use_lambda(T1 val, T2 ref) {
  // Use auto because the type of the lambda is unknown.
  auto lambda = [val, &ref](int arg1, int arg2) -> R {
    /* lambda-body */
    return R();
  };
  return lambda(12, 27);
}

// The functor class (valid C++03)
// Similar to what the compiler generates for the lambda function.
class Functor {
  // Capture list.
  T1 val;
  T2& ref;

public:
  // Constructor
  inline Functor(T1 val, T2& ref) : val(val), ref(ref) {}

  // Functor body
  R operator()(int arg1, int arg2) const {
    /* lambda-body */
    return R();
  }
};

// Equivalent to use_lambda, but uses a functor (valid C++03).
R use_functor(T1 val, T2 ref) {
  Functor functor(val, ref);
  return functor(12, 27);
}

// Make this a self-contained example.
int main() {
  T1 t1;
  T2 t2;
  use_functor(t1,t2);
  use_lambda(t1,t2);
  return 0;
}

If the lambda function is mutable then make the functor’s call-operator non-const, i.e.:

R operator()(int arg1, int arg2) /*non-const*/ {
  /* lambda-body */
  return R();
}

Recursive lambdas

Let’s say we wish to write Euclid’s gcd() as a lambda. As a function, it is:

int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a%b);
}

But a lambda cannot be recursive, it has no way to invoke itself. A lambda has no name and using this within the body of a lambda refers to a captured this (assuming the lambda is created in the body of a member function, otherwise it is an error). So how do we solve this problem?

Use std::function

We can have a lambda capture a reference to a not-yet constructed std::function:

std::function<int(int, int)> gcd = [&](int a, int b){
    return b == 0 ? a : gcd(b, a%b);
};

This works, but should be used sparingly. It’s slow (we’re using type erasure now instead of a direct function call), it’s fragile (copying gcd or returning gcd will break since the lambda refers to the original object), and it won’t work with generic lambdas.

Using two smart pointers:

auto gcd_self = std::make_shared<std::unique_ptr< std::function<int(int, int)> >>();
*gcd_self = std::make_unique<std::function<int(int, int)>>(
  [gcd_self](int a, int b){
    return b == 0 ? a : (**gcd_self)(b, a%b);
  };
};

This adds a lot of indirection (which is overhead), but it can be copied/returned, and all copies share state. It does let you return the lambda, and is otherwise less fragile than the above solution.

Use a Y-combinator

With the help of a short utility struct, we can solve all of these problems:

template <class F>
struct y_combinator {
    F f; // the lambda will be stored here

    // a forwarding operator():
    template <class... Args>
    decltype(auto) operator()(Args&&... args) const {
        // we pass ourselves to f, then the arguments.
        // the lambda should take the first argument as `auto&& recurse` or similar.
        return f(*this, std::forward<Args>(args)...);
    }
};
// helper function that deduces the type of the lambda:
template <class F>
y_combinator<std::decay_t<F>> make_y_combinator(F&& f) {
    return {std::forward<F>(f)};
}
// (Be aware that in C++17 we can do better than a `make_` function)

we can implement our gcd as:

auto gcd = make_y_combinator(
  [](auto&& gcd, int a, int b){
    return b == 0 ? a : gcd(b, a%b);
  }
);

The y_combinator is a concept from the lambda calculus that lets you have recursion without being able to name yourself until you are defined. This is exactly the problem lambdas have.

You create a lambda that takes “recurse” as its first argument. When you want to recurse, you pass the arguments to recurse.

The y_combinator then returns a function object that calls that function with its arguments, but with a suitable “recurse” object (namely the y_combinator itself) as its first argument. It forwards the rest of the arguments you call the y_combinator with to the lambda as well.

In short:

auto foo = make_y_combinator( [&](auto&& recurse, some arguments) {
  // write body that processes some arguments
  // when you want to recurse, call recurse(some other arguments)
});

and you have recursion in a lambda with no serious restrictions or significant overhead.

Using lambdas for inline parameter pack unpacking

Parameter pack unpacking traditionally requires writing a helper function for each time you want to do it.

In this toy example:

template<std::size_t...Is>
void print_indexes( std::index_sequence<Is...> ) {
  using discard=int[];
  (void)discard{0,((void)(
    std::cout << Is << '\n' // here Is is a compile-time constant.
  ),0)...};
}
template<std::size_t I>
void print_indexes_upto() {
  return print_indexes( std::make_index_sequence<I>{} );
}

The print_indexes_upto wants to create and unpack a parameter pack of indexes. In order to do so, it must call a helper function. Every time you want to unpack a parameter pack you created, you end up having to create a custom helper function to do it.

This can be avoided with lambdas.

You can unpack parameter packs into a set of invocations of a lambda, like this:

template<std::size_t I>
using index_t = std::integral_constant<std::size_t, I>;
template<std::size_t I>
constexpr index_t<I> index{};

template<class=void, std::size_t...Is>
auto index_over( std::index_sequence<Is...> ) {
  return [](auto&& f){
    using discard=int[];
    (void)discard{0,(void(
      f( index<Is> )
    ),0)...};
  };
}

template<std::size_t N>
auto index_over(index_t<N> = {}) {
  return index_over( std::make_index_sequence<N>{} );
}

With fold expressions, index_over() can be simplified to:

template<class=void, std::size_t...Is>
auto index_over( std::index_sequence<Is...> ) {
  return [](auto&& f){
    ((void)(f(index<Is>)), ...);
  };
}

Once you have done that, you can use this to replace having to manually unpack parameter packs with a second overload in other code, letting you unpack parameter packs “inline”:

template<class Tup, class F>
void for_each_tuple_element(Tup&& tup, F&& f) {
  using T = std::remove_reference_t<Tup>;
  using std::tuple_size;
  auto from_zero_to_N = index_over< tuple_size<T>{} >();

  from_zero_to_N(
    [&](auto i){
      using std::get;
      f( get<i>( std::forward<Tup>(tup) ) );
    }
  );
}

The auto i passed to the lambda by the index_over is a std::integral_constant<std::size_t, ???>. This has a constexpr conversion to std::size_t that does not depend on the state of this, so we can use it as a compile-time constant, such as when we pass it to std::get<i> above.

To go back to the toy example at the top, rewrite it as:

template<std::size_t I>
void print_indexes_upto() {
  index_over(index<I>)([](auto i){
    std::cout << i << '\n'; // here i is a compile-time constant
  });
}

which is much shorter, and keeps logic in the code that uses it.

Live example to play with.