5 Curious C++ Lambda Examples: Recursion, constexpr, Containers and More

Table of Contents

In this blog post I’ll show you a couple of interesting examples with lambda expressions. Do you know how to write a recursive lambda? Store them in a container? Or invoke at compile time?

See in the article.

Updated in August 2022: Added C++23 improvements.

1. Recursive Lambda with `std::function`

Writing a recursive function is relatively straightforward: inside a function definition, you can call the same function by its name. How about lambdas?

int main() {
    auto factorial = [](int n) {
        return n > 1 ? n * factorial(n - 1) : 1;
    };
    return factorial(5);
}

This, unfortunately, doesn’t compile…

How can we fix this?

One way is to use std::function:

#include <functional>

int main() {
    const std::function<int(int)> factorial = [&factorial](int n) {
        return n > 1 ? n * factorial(n - 1) : 1;
    };
    return factorial(5);
}

This time we need to capture factorial and then we can refer to it inside the lambda body.

And since C++14 we can also leverage generic lambdas and write the following code:

int main() {
    const auto factorial = [](int n) {
        const auto fact_impl = [](int n, const auto& impl) -> int {
            return n > 1 ? n * impl(n - 1, impl) : 1;
        };
        return fact_impl(n, fact_impl);
    };
    return factorial(5);
}

This time it’s even more complicated (but doesn’t require heavy use of std::function). It uses internal lambda for the main computation and then it’s passed as a generic argument.

But I wonder: have you ever used recursive lambdas? Or it’s better to rely on recursive functions (which seems to be far more comfortable to use and write).

1.1 Recursive Lambdas in C++23

As you can see creating a recursive lambda requires some tricks in C++11 or C++14. Fortunately it will change in C++23! Thanks to a powerful feature called “deducing this” we have a way to access the lambda object from its call operator.

Have a look:

int main()
{
    auto factorial23 = [](this auto&& self, int n) {
        if (n <= 1)
            return 1;
        return n * self(n - 1);
    };
   
    return factorial23(5);
}

Run at Compiler Explorer

The main idea is that each member function (including the call operator of a lambda object), can get explicit this parameter. The compiler passes this argument anyway, so C++23 just added a way to expose it and allow to manipulate. The argument is always passed as the first one and in our cases it’s called self. Notice the keyword this auto to indicate that this is this :)

We can also write an alternative version:

auto factorial23_2 = [](this auto&& self, int n) -> int{
    if (n > 1)          
        return n * self(n - 1);
    return 1;
};

But this time the compiler needs some help with return type deduction - that’s why I added -> int.

Read more in the following article at the MSVC Team blog: C++23’s Deducing this: what it is, why it is, how to use it - C++ Team Blog.

2. `constexpr` Lambdas

But that’s not all with recursion… :)

Since C++17 we can write lambdas that have the call operator defined as constexpr. We can use this property and expand the recursive example into:

int main() {
    constexpr auto factorial = [](int n) {
        constexpr auto fact_impl = [](int n, const auto& impl) -> int {
            return n > 1 ? n * impl(n - 1, impl) : 1;
        };
        return fact_impl(n, fact_impl);
    };
    static_assert(factorial(5) == 120);
}

And in C++20 you can even apply consteval to mark lambdas which can be evaluated only at compile time.

Plus in C++23 you can apply constexpr or consteval along with passing this:

int main()
{
    constexpr auto factorial23 = [](this auto&& self, int n) {
        if (n <= 1)
            return 1;
        return n * self(n - 1);
    };
    static_assert(factorial23(5) == 120);
}

See here @Compiler Explorer.

3. Storing Lambdas in a Container

This might be a bit cheating… but we can theoretically store lambdas in a container.

While closure types have default constructors deleted (unless it’s stateless lambda in C++20), we can do a little hack and store all lambdas as std::function objects. For example:

#include <functional>
#include <iostream>
#include <vector>

int main() {
    std::vector<std::function<std::string(const std::string&)>> vecFilters;
    
    vecFilters.emplace_back([](const std::string& x) { 
        return x + " Amazing"; 
    });
    vecFilters.emplace_back([](const std::string& x) { 
        return x + " Modern"; 
    });
    vecFilters.emplace_back([](const std::string& x) { 
        return x + " C++"; 
    });
    vecFilters.emplace_back([](const std::string& x) { 
        return x + " World!"; 
    });
    
    const std::string str = "Hello";
    auto temp = str;
    
    for (auto &entryFunc : vecFilters)  
        temp = entryFunc(temp);
    
    std::cout << temp;
}

4.Generic lambdas and Help with Deduction

C++14 brought an important addition to lambdas: generic lambda arguments. Here’s one example that shows why is it useful:

#include <algorithm>
#include <iostream>
#include <map>
#include <string>

int main() {
    const std::map<std::string, int> numbers { 
        { "one", 1 }, {"two", 2 }, { "three", 3 }
    };
    
    std::for_each(std::begin(numbers), std::end(numbers), 
         [](const std::pair<std::string, int>& entry) {
             std::cout << entry.first << " = " << entry.second << '\n';
         }
    );
}

Do you know what’s the mistake here? Is the argument type appropriately specified in the inner lambda for for_each?

I specified: const std::pair<std::string, int>& entry.

But it’s wrong as the type of the key/value pair inside a map is:

std::pair<const std::string, int>

That’s why the compiler has to create unwanted temporary copies and then pass them to my lambda.

We can quickly fix this by using a generic lambda from C++14.

std::for_each(std::begin(numbers), std::end(numbers), 
    [](const auto& entry) {
        std::cout << entry.first << " = " << entry.second << '\n';
    }
);

Now the types match, and no additional copies are created.

5. Returning a lambda

If you want to return a lambda from a function (for example for partial function application, currying), then it’s not straightforward because you don’t know the exact type of the closure object.

In C++11 one way was to use std::function:

#include <functional>

std::function<int(int)> CreateAddLambda(int y) {
    return [y](int x) { return x + y; };
}

int main() {
    auto tenPlus = CreateAddLambda(10);
    return tenPlus(32);
}

Run @Compiler Explorer.

Since C++14, we can leverage the auto type deduction for return types and just write:

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

int main() {
    auto tenPlus = CreateAddLambda(10);
    return tenPlus(32);
}

Run @Compiler Explorer.

The above code is far simpler and cheaper as we don’t need to use std::function (and even include <functional>).

Side note: Thanks to the r/cpp comment about the wrong capture of [&y].

Summary

In this quick article, I showed you five interesting lambda examples. They might not be common, but shows flexibility and sometimes even complexity of the closure types.

Do you use lambdas in such contexts? Or maybe you have even more complicated examples? Share your experience in comments below the article.