Instantiating C++ lambda by its type

Question

I want a way to make functor from function. Now I trying to wrap function call by lambda function and instantiate it later. But compiler says than lambda constructor is dele

Answer 1

We can assume that the lambda will always be empty, thus, we can just do a cast from another empty type, since they both have the same memory layout. So we build a wrapper class that makes a default constructible function object:

template
struct wrapper
{
    static_assert(std::is_empty(), "Lambdas must be empty");
    template
    auto operator()(Ts&&... xs) const -> decltype(reinterpret_cast(*this)(std::forward(xs)...))
    {
        return reinterpret_cast(*this)(std::forward(xs)...);
    }
};

Now a static assert is added to make sure that the lambda is always empty. This should always be the case(since its required to decay to a function pointer), but the standard doesn’t explicitly guarantee it. So we use the assert to at least catch the insane implementation of lambdas. Then we just cast the wrapper class to the lambda since both of them are empty.

Finally the lambda could be constructed like this:

template 
void function_caller()
{
    wrapper f;
    f();
}