How can I give a C++ lambda expression more than one operator()?


Suppose you have stored a C++ lambda expression into a variable, and you want to call it in different ways. This seems impossible, because when you define the lambda expression, you can provide only one operator():

auto lambda = [captures](int v) { return v + 2; };

This lambda has only one way of calling it: You pass an integer and it returns an integer.

But it turns out that you can create a lambda that can be called in multiple ways: Use an auto parameter!

auto lambda = [](auto p)
  if constexpr (std::is_same_v<decltype(p), int>) {
   return p + 1;
  } else {
   return "oops";

auto result1 = lambda(123); // result1 is 124
auto result2 = lambda('x'); // result2 is "oops"

By declaring the parameter as auto, the lambda accepts any single parameter. We then use if constexpr and std::is_same_v to see what type was actually passed, and implement the desired function body for each type.

Notice that the different branches of the if don’t need to agree on the return type. In our example, passing an integer adds one and produces another integer. But passing anything else returns the string "oops"!

You can create a bunch of tag types to make it look almost as if your lambda had member functions.

struct add_tax_t {};
constexpr add_tax_t add_tax;

struct apply_discount_t {};
constexpr apply_discount_t apply_discount;

auto lambda = [total](auto op, auto value) mutable
  using Op = decltype(op);
  if constexpr (std::is_same_v<Op, add_tax_t>) {
   total += total * value; // value is the tax rate
   return total;
  } else if constexpr (std::is_same_v<Op, apply_discount_t>) {
   total -= std::max(value, total); // value is the discount
   return total;
  } else {
   static_assert(!sizeof(Op*), "Don't know what you are asking me to do.");

lambda(apply_discount, 5.00); // apply $5 discount
lambda(add_tax, 0.10); // add 10% tax

So far, all of our “methods” have the same number of parameters, but you can use a parameter pack to permit different numbers of parameters:

auto lambda = [total](auto op, auto... args) mutable
  using Op = decltype(op);
  using ArgsT = std::tuple<decltype(args)...>;
  if constexpr (std::is_same_v<Op, add_tax_t>) {
   auto [tax_rate] = ArgsT(args...);
   total += total * tax_rate;
   return total;
  } else if constexpr (std::is_same_v<Op, apply_discount_t>) {
   auto [amount, expiration] = ArgsT(args...);
   if (expiration < now()) {
     total -= std::max(amount, total);
   return total;
  } else {
   static_assert(!sizeof(Op*), "Don't know what you are asking me to do.");

In this case, the add_tax “method” takes a single parameter, whereas the apply_discount “method” takes two.

You could even dispatch based solely on the types and arity.

auto lambda = [total](auto... args) mutable
  using ArgsT = std::tuple<decltype(args)...>;
  if constexpr (std::is_same_v<ArgsT, std::tuple<int, int>>) {
   // two integers = add to total
   auto [a, b] = ArgsT(args...);
   total += a + b;
  } else if constexpr (std::is_same_v<ArgsT, std::tuple<>>) {
   // no parameters = print
  } else {
   static_assert(!sizeof(Op*), "Don't know what you are asking me to do.");

This might come in handy if you have a lambda that is used to accumulate something: You can pass the lambda to the function that expects to do the accumulating, and then call the lambda using a secret knock to extract the answer.

auto lambda = [limit, total = 0](auto value) mutable
 using T = decltype(value);
 if constexpr (std::is_same_v<T, const char*>) {
  // secret knock: Return total if invoked with const char* 
  return total;
 } else {
  // Otherwise, just add them up until we hit the limit.
  total += value;
  return total <= limit;

auto unused = std::find_if_not(begin, end, std::ref(lambda));
if (unused != end) print("Limit exceeded.");
auto total = lambda("total"); // extract the total

This is basically a complete and utter abuse of the language, and I hope you’re ashamed of yourself.

Bonus chatter: All of this is just a way of defining a struct without having to say the word struct.

 double limit;
 double total = 0.00;

 auto add_tax(auto tax_rate) { total += total * tax_rate; }
 auto apply_discount(auto amount) { total -= std::max(amount, total); }
 auto get_total() const { return total; }
} lambda{1000.00 /* limit */};

Bonus bonus chatter: Java anonymous classes provide a more straightforward syntax:

var lambda = new Object() {
  int total = 0;
  public void add(int value) { total += value; }
  public int get_total() { return total; }

var result = lambda.get_total();


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  • Ben Voigt

    Does that std::find_not_if trick actually work? I thought the functor was passed by value.

    Far better to capture the accumulator local variable by reference instead of storing it inside the lambda, then all copies share the same underlying variable, and the surrounding scope can access it too.

  • Peter Cooper Jr.

    Nitpick/typo: It’s been a long time since I’ve done anything with C++, but I would expect that with that first auto result1 = lambda(123); that result1 would actually be 124, not 3.

    Though this is such an abuse of the language that I feel it’d serve you right if the compiler gave you a 3 there anyway.

  • Daniel Grunwald

    C++17 also has this neat trick for combining multiple lambdas into one:

    template<class… Ts> struct overloaded : Ts… { using Ts::operator()…; };
    template<class… Ts> overloaded(Ts…) -> overloaded<Ts…>;

    overloaded ends up deriving from multiple lambda types and inheriting all their call operators. This uses C++17 class template argument deduction and C++17 aggregate base initialization.

    The downside is that you’ll waste some memory if multiple lambdas capture the same variables.

  • David Ryan

    Really neat example, even if an ‘abuse’ of the language.

    Small nitpick. In a couple of examples here, I would expect the ‘total’ variable to be captured by reference (so that “total +=” has an effect, especially in those versions of the lambda where total is not returned). In those versions of the lambda where total is returned, I would expect the example usage to assign the result to total, e.g. total = lambda(apply_discount, 5.00);

  • Yuri Khan

    Functional programming textbooks give very similar examples in their “and this is how you implement a stateful object with methods in a functional language”: pack your state in a lambda’s closure, and have it accept tag values that tell it which method to execute.

  • Tarun Elankath

    Could you please explain that static_assert on the negation of sizeof(0p*) ? Understood everything else. I wish you would write about useful C++ idioms and techniques that can be used in real code, not just pointless programming poems.

    • Raymond ChenMicrosoft employee

      Yes, this exercise was productive in the sense that it employed multiple tricks which can be useful in real code. We’ll dig into those tricks over time.