All About Lambda Functions in C++: from C++11 to C++17

/!\: Originally published @ .

Title of this article is a bit misleading. Because lambda doesn't always synthesize to function pointer. It's an expression(precisely unique closure). But I have kept it that way for simplicity. So from now on, I will use lambda function & expression interchangeably.

What is lambda function?

• not worth naming(unnamed, anonymous, disposable, etc. Whatever you can call it),
• and also not reused.

[ capture list ] (parameters) -> return-type  
{   
    method definition
} 

• Usually, compiler evaluates a return type of a lambda function itself. So we don't need to specify a trailing return type explicitly i.e. 

  -> return-type

  .
• But in some complex cases, the compiler unable to deduce the return type and we need to specify that.

Why we should use a lambda function?

• C++ includes many useful generic functions like 

  std::for_each

  , which can be handy. Unfortunately, they can also be quite cumbersome to use, particularly if the functor you would like to apply is unique to the particular function. Consider the following code for an example:

struct print
{
    void operator()(int element)
    {
        cout << element << endl;
    }
};
int main(void)
{
    std::vector<int> v = {1, 2, 3, 4, 5};
    std::for_each(v.begin(), v.end(), print());
    return 0;
}

• If you use print once, in that specific place, it seems overkill to be writing a whole class just to do something trivial and one-off.
• However, for this kind of situation inline code would be more suitable & appropriate which can be achieved by lambda function as follows:

std::for_each(v.begin(), v.end(), [](int element) { cout << element << endl; });

How lambda functions works internally?

[&i] ( ) { std::cout << i; }

// is equivalent to

struct anonymous
{
    int &m_i;
    anonymous(int &i) : m_i(i) {}
    inline auto operator()() const
    {
        std::cout << i;
    }
};

• The compiler generates unique closure as above for each lambda function. Finally, the secret revealed.
• Capture list will become a constructor argument in closure, If you capture argument as value then corresponding type data member is created within the closure.
• Moreover, you can declare variable/object in the lambda function argument, which will become an argument to call operator i.e. 

  operator()

  .

Benefits of using a lambda function

• Zero cost abstraction. Yes! you read it right. lambda doesn't cost you performance & as fast as a normal function.
• In addition, code becomes compact, structured & expressive.

Learning lambda expression

int main()
{
    int x = 100, y = 200;
    auto print = [&] { // Capturing object by reference
        std::cout << __PRETTY_FUNCTION__ << " : " << x << " , " << y << std::endl;
    };
    print();
    return 0;
}

main()::<lambda()> : 100 , 200

• In the above example, I have mentioned 

  &

   in capture list. which captures variable 

  x

   & 

  y

   as reference. Similarly, 

  =

   denotes captured by value, which will create data member of the same type within the closure and copy assignment will take place.
• Note that, the parameter list is optional, you can omit the empty parentheses if you do not pass arguments to the lambda expression.

• The following table shows different use cases for the same:

template <typename Functor>
void f(Functor functor)
{
    std::cout << __PRETTY_FUNCTION__ << std::endl;
}
/* Or alternatively you can use this
void f(std::function<int(int)> functor)
{
    std::cout << __PRETTY_FUNCTION__ << std::endl;
} 
*/
int g() { static int i = 0; return i++; }
int main()
{
    auto lambda_func = [i = 0]() mutable { return i++; };
    f(lambda_func); // Pass lambda
    f(g);           // Pass function
}

Function Type : void f(Functor) [with Functor = main()::<lambda(int)>]
Function Type : void f(Functor) [with Functor = int (*)(int)]

• You can also pass lambda function as an argument to other function just like a normal function which I have coded above.
• If you noticed, here I have declared variable i in capture list which will become data member. As a result, every time you call lambda_func, it will be returned and incremented.

class Example
{
public:
    Example() : m_var(10) {}
    void func()
    {
        [=]() { std::cout << m_var << std::endl; }(); // IIFE
    }
private:
    int m_var;
};
int main()
{
    Example e;
    e.func();
}

• this

   pointer can also be captured using 

  [this]

  , 

  [=]

   or 

  [&]

  . In any of these cases, class data members(including 

  private

  ) can be accessed as you do in a normal method.
• If you see the lambda expression line, I have used extra 

  ()

   at the end of the lambda function declaration which used to calls it right thereafter declaration. It is called  (Immediately Invoked Function Expression).

C++ lambda function types

const auto l = [](auto a, auto b, auto c) {};


// is equivalent to
struct anonymous
{
    template <class T0, class T1, class T2>
    auto operator()(T0 a, T1 b, T2 c) const
    {
    }
};

• Generic lambda introduced in C++14 which can captures parameters with 

  auto

   specifier.

void print() {}
template <typename First, typename... Rest>
void print(const First &first, Rest &&... args)
{
    std::cout << first << std::endl;
    print(args...);
}
int main()
{
    auto variadic_generic_lambda = [](auto... param) {
        print(param...);
    };
    variadic_generic_lambda(1, "lol", 1.1);
}

• Lambda with variable parameter pack will be useful in many scenarios like debugging, repeated operation with different data input, etc.

mutable

• Typically, a lambda's function call operator is const-by-value which means lambda requires mutable  keyword if you are capturing anything by-value.

[]() mutable {}

// is equivalent to

struct anonymous
{
    auto operator()()  // call operator
    {
    }
};

• We have already seen an example of this above. I hope you noticed it.

#include <iostream>
#include <type_traits>

int main()
{
    auto funcPtr = +[] {};
    static_assert(std::is_same<decltype(funcPtr), void (*)()>::value);
}

• You can force the compiler to generate lambda as a function pointer rather than closure by adding 

  +

   infront of it as above.

const auto less_than = [](auto x) {
    return [x](auto y) {
        return y < x;
    };
};

int main(void)
{
    auto less_than_five = less_than(5);
    std::cout << less_than_five(3) << std::endl;
    std::cout << less_than_five(10) << std::endl;
    return 0;
}

• Going a bit further, lambda function can also return another lambda function. This will open the doors of endless possibility for customization, code expressiveness & compactibility(BTW, there is no word like this) of code.

constexpr

• Since C++17, a lambda expression can be declared as .

constexpr auto sum = [](const auto &a, const auto &b) { return a + b; };
/*
    is equivalent to
    constexpr struct anonymous
    {
        template <class T1, class T2>
        constexpr auto operator()(T1 a, T2 b) const
        {
            return a + b;
        }
    };
*/
constexpr int answer = sum(10, 10);

• Even if you don't specify 

  constexpr

   , the function call operator will be  

  constexpr

   anyway, if it happens to satisfy all .

Closing words