Can the 'type' of a lambda expression be expressed?

Question

Thinking of lambda expressions as \'syntactic sugar\' for callable objects, can the unnamed underlying type be expressed?

An example:

struct gt {         


        

Answer 1

You could use a small class lambda_wrapper<>, to wrap a lambda at low costs. Its much more faster than std::function because there are no virtual function call and a dynamic memory alloc. Wrapper works by deducing the lambda arguments list and return type.

#include 
#include 
#include 

template 
struct lambda_wrapper : public lambda_wrapper {};

template 
struct lambda_wrapper {
private:
    L lambda;

public:
    lambda_wrapper(const L & obj) : lambda(obj) {}

    template
    typename std::result_of::type operator()(Args... a) {
        return this->lambda.operator()(std::forward(a)...);
    }

    template typename
    std::result_of::type operator()(Args... a) const {
        return this->lambda.operator()(std::forward(a)...);
    }
};
template 
auto make_lambda_wrapper(T&&t) {
    return lambda_wrapper(std::forward(t));
}
int main(int argc, char ** argv) 
{
    auto func = make_lambda_wrapper([](int y, int x) -> bool { return x>y; });
    std::set ss(func);
    std::cout << func(2, 4) << std::endl;
}