tinlyx tinlyx - 1 year ago 95
C++ Question

what is the type signature of a c++11/1y lambda function?

I was wondering if there is a standard way to get the type signature (i.e. the return type and the types) of its parameters of any given lambda?

The reason I ask is that I've always wondered what exactly is the type

in the declaration like
auto l =[](int x,int y)->int{return x+y;}
. In other use cases of
, it's a convenience and shorter alternative for a longer type name. But for lambdas, is there even an alternative way to declare the lambda variable?

My understanding is that a standard lambda is nothing more than a function object, and it is its own type. So, even if two lambdas have the same return type and parameter types, they are still two different, unrelated classes/functors. But this there a way to capture the fact that they are the same in terms of type signature?

I think the type signature I am looking for can be something like a
object of the correct types.

A more useful/involved question is, if it's possible to extract the type signature, this is possible to write a general wrapper function to convert any lambda function to a
object of the same type signature.

Answer Source

According to Can the 'type' of a lambda expression be expressed?, there is actually a simple way in current c++ (without needing c++1y) to figure out the return_type and parameter types of a lambda. Adapting this, it is not difficult to assemble a std::function typed signature type (called f_type below) for each lambda.

I. With this abstract type, it is actually possible to have an alternative way to auto for expressing the type signature of a lambda, namely function_traits<..>::f_type below. Note: the f_type is not the real type of a lambda, but rather a summary of a lambda's type signature in functional terms. It is however, probably more useful than the real type of a lambda because every single lambda is its own type.

As shown in the code below, just like one can use vector<int>::iterator_type i = v.begin(), one can also do function_traits<lambda>::f_type f = lambda, which is an alternative to the mysterious auto. Of course, this similarity is only formal. The code below involves converting the lambda to a std::function with the cost of type erasure on construction of std::function object and a small cost for making indirect call through the std::function object. But these implementation issues for using std::function aside (which I don't believe are fundamental and should stand forever), it is possible, after all, to explicitly express the (abstract) type signature of any given lambda.

II. It is also possible to write a make_function wrapper (pretty much like std::make_pair and std::make_tuple) to automatically convert a lambda f ( and other callables like function pointers/functors) to std::function, with the same type-deduction capabilities.

Test code is below:

#include <cstdlib>
#include <tuple>
#include <functional>
#include <iostream>
using namespace std;

// For generic types that are functors, delegate to its 'operator()'
template <typename T>
struct function_traits
    : public function_traits<decltype(&T::operator())>

// for pointers to member function
template <typename ClassType, typename ReturnType, typename... Args>
struct function_traits<ReturnType(ClassType::*)(Args...) const> {
    //enum { arity = sizeof...(Args) };
    typedef function<ReturnType (Args...)> f_type;

// for pointers to member function
template <typename ClassType, typename ReturnType, typename... Args>
struct function_traits<ReturnType(ClassType::*)(Args...) > {
    typedef function<ReturnType (Args...)> f_type;

// for function pointers
template <typename ReturnType, typename... Args>
struct function_traits<ReturnType (*)(Args...)>  {
  typedef function<ReturnType (Args...)> f_type;

template <typename L> 
typename function_traits<L>::f_type make_function(L l){
  return (typename function_traits<L>::f_type)(l);

long times10(int i) { return long(i*10); }

struct X {
  double operator () (float f, double d) { return d*f; } 

// test code
int main()
    auto lambda = [](int i) { return long(i*10); };
    typedef function_traits<decltype(lambda)> traits;
    traits::f_type ff = lambda;

    cout << make_function([](int i) { return long(i*10); })(2) << ", " << make_function(times10)(2) << ", " << ff(2) << endl;
    cout << make_function(X{})(2,3.0) << endl;

    return 0;
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