C++14 Lambdas and Perfect Forwarding

So the joke's on me, I guess.

In my discussion of std::move vs. std::forward, I explained that when you call std::forward, the expectation is that you'll pass a type consistent with the rules for template type deduction, meaning (1) an lvalue reference type for lvalues and (2) a non-reference type for rvalues.  I added,

If you decide to be a smart aleck and write [code passing an rvalue reference type], the reference-collapsing rules will see that you get the same behavior as [you would passing a non-reference type], but with any luck, your team lead will shift you to development in straight C, where you'll have to content yourself with writing bizarre macros.

Well.  As I said, the joke seems to be on me, because the standardization commitee  apparently consists largely of smart alecks.

Let me explain.

The recently-adopted C++14 CD includes beefy additions to lambda capabilities, including the support for polymorphic lambdas that Herb Sutter can't help but mention I've been whining about for years. This means that in C++14, we now have the expressive power that the Boost Lambda library has been offering since 2002. Ahem. But C++14 goes further, supporting also variadic lambdas, generalized captures (including capture-by-move), and, of particular relevance to this post, support for perfect forwarding.

Suppose we want to write a C++14 lambda that takes a parameter and perfect-forwards it to some function f:

auto forwardingLambda = [](auto&& param) { /* perfect-forward param to f */ };

Writing the perfect-forwarding call is easy, but it's probably not obvious how.  The normal way to perfect-forward something it to use std::forward, so we'd expect to write essentially this:

auto forwardingLambda = [](auto&& param) { f(std::forward<T>(param)); };

But, uh oh, there's no T to pass to std::forward.  (In the class generated from the lambda expression, there is, but inside the lambda itself, there's no type for param.)  So what do we pass to std::forward? We can hardly pass auto. (Consider what would happen if we had a lambda taking multiple parameters, each of type auto and each of which we wanted to forward. In that case, each std::forward<auto> would be ambiguous: which auto should std::forward use?)

The solution takes advantage of two observations. First, the type-deduction rules for auto in lambdas are the same as for templates. This means that if an lvalue argument is passed to the lambda, param's type will be an lvalue reference--exactly what we need for std::forward. If an rvalue argument is passed, its type will be an rvalue reference. For such parameters, we can recover the type to pass to std::forward by stripping it of its reference-ness. We could thus write forwardingLambda like this:

auto forwardingLambda = [](auto&& param) {
  f(std::forward<std::conditional<std::is_rvalue_reference<decltype(param)>::value,
                                  std::remove_reference<decltype(param)>::type,
                                  decltype(param)>::type>(param));
};

At least I think we could. I don't have a C++14 compiler to try it with, and, anyway, it's too gross to waste time on. It would be sad, indeed, if this is what the standardization committee expected us to do to effect perfect forwarding inside its spiffy new C++14 lambdas. Fortunately, it doesn't.

Which brings us to observation number two. As I noted near the beginning of this post,

If you decide to be a smart aleck and write [code passing an rvalue reference type to std::forward], the reference-collapsing rules will see that you get the same behavior as [you would passing a non-reference type].

That means that if param's type is an rvalue reference, there is no need to strip off its reference-ocity. Instead, you can smart aleck your way to success by simply passing that type directly to std::forward.  Like so:

auto forwardingLambda = [](auto&& param) { f(std::forward<decltype(param)>(param)); };

Frankly, this is more verbose than I'd prefer. One could imagine a world where you could say something like this:

auto forwardingLambda =
  [](<T1>&& param1, <T2>&& param2) { f(std::forward<T1>(param1), std::forward<T2>(param2)); };

But that's not the world we live in, and given that C++14 gives us polymorphic lambdas, variadic lambdas, and move-enabled lambdas, I'm not going to complain about the world of C++14 lambdas.  Except possibly to Herb :-)

Scott