How does std::forward work, especially when passing lvalue/rvalue references?

Question

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What are the main purposes of std::forward and which problems does it solve?

I know what it does and when to use it but I still can't wrap my head around how it works. Please be as detailed as possible and explain when std::forward would be incorrect if it was allowed to use template argument deduction.

Part of my confusion is this: "If it has a name, it's an lvalue" - if that's the case why does std::forward behave differently when I pass thing&& x vs thing& x?

Answer 1

First, let's take a look at what std::forward does according to the standard:

§20.2.3 [forward] p2

Returns: static_cast<T&&>(t)

(Where T is the explicitly specified template parameter and t is the passed argument.)

Now remember the reference collapsing rules:

TR   R

T&   &  -> T&  // lvalue reference to cv TR -> lvalue reference to T
T&   && -> T&  // rvalue reference to cv TR -> TR (lvalue reference to T)
T&&  &  -> T&  // lvalue reference to cv TR -> lvalue reference to T
T&&  && -> T&& // rvalue reference to cv TR -> TR (rvalue reference to T)

(Shamelessly stolen from this answer.)

And then let's take a look at a class that wants to employ perfect forwarding:

template<class T>
struct some_struct{
  T _v;
  template<class U>
  some_struct(U&& v)
    : _v(static_cast<U&&>(v)) {} // perfect forwarding here
                                 // std::forward is just syntactic sugar for this
};

And now an example invocation:

int main(){
  some_struct<int> s1(5);
  // in ctor: '5' is rvalue (int&&), so 'U' is deduced as 'int', giving 'int&&'
  // ctor after deduction: 'some_struct(int&& v)' ('U' == 'int')
  // with rvalue reference 'v' bound to rvalue '5'
  // now we 'static_cast' 'v' to 'U&&', giving 'static_cast<int&&>(v)'
  // this just turns 'v' back into an rvalue
  // (named rvalue references, 'v' in this case, are lvalues)
  // huzzah, we forwarded an rvalue to the constructor of '_v'!

  // attention, real magic happens here
  int i = 5;
  some_struct<int> s2(i);
  // in ctor: 'i' is an lvalue ('int&'), so 'U' is deduced as 'int&', giving 'int& &&'
  // applying the reference collapsing rules yields 'int&' (& + && -> &)
  // ctor after deduction and collapsing: 'some_struct(int& v)' ('U' == 'int&')
  // with lvalue reference 'v' bound to lvalue 'i'
  // now we 'static_cast' 'v' to 'U&&', giving 'static_cast<int& &&>(v)'
  // after collapsing rules: 'static_cast<int&>(v)'
  // this is a no-op, 'v' is already 'int&'
  // huzzah, we forwarded an lvalue to the constructor of '_v'!
}

I hope this step-by-step answer helps you and others understand just how std::forward works.

Answer 2

I think the explanation of std::forward as static_cast<T&&> is confusing. Our intuition for a cast is that it converts a type to some other type -- in this case it would be a conversion to an rvalue reference. It's not! So we are explaining one mysterious thing using another mysterious thing. This particular cast is defined by a table in Xeo's answer. But the question is: Why? So here's my understanding:

Suppose I want to pass you an std::vector<T> v that you're supposed to store in your data structure as data member _v. The naive (and safe) solution would be to always copy the vector into its final destination. So if you are doing this through an intermediary function (method), that function should be declared as taking a reference. (If you declare it as taking a vector by value, you'll be performing an additional totally unnecessary copy.)

void set(const std::vector<T> & v) { _v = v; }

This is all fine if you have an lvalue in your hand, but what about an rvalue? Suppose that the vector is the result of calling a function makeAndFillVector(). If you performed a direct assignment:

_v = makeAndFillVector();

the compiler would move the vector rather than copy it. But if you introduce an intermediary, set(), the information about the rvalue nature of your argument would be lost and a copy would be made.

set(makeAndFillVector()); // set will still make a copy

In order to avoid this copy, you need "perfect forwarding", which would result in optimal code every time. If you're given an lvalue, you want your function to treat it as an lvalue and make a copy. If you're given an rvalue, you want your function to treat it as an rvalue and move it.

Normally you would do it by overloading the function set() separately for lvalues and rvalues:

set(const std::vector<T> & lv) { _v = v; }
set(std::vector<T> && rv) { _v = std::move(rv); }

But now imagine that you're writing a template function that accepts T and calls set() with that T (don't worry about the fact that our set() is only defined for vectors). The trick is that you want this template to call the first version of set() when the template function is instantiated with an lvalue, and the second when it's initialized with an rvalue.

First of all, what should the signature of this function be? The answer is this:

template<class T>
void perfectSet(T && t);

Depending on how you call this template function, the type T will be somewhat magically deduced differently. If you call it with an lvalue:

std::vector<T> v;
perfectSet(v);

the vector v will be passed by reference. But if you call it with an rvalue:

perfectSet(makeAndFillVector());

the (anonymous) vector will be passed by rvalue reference. So the C++11 magic is purposefully set up in such a way as to preserve the rvalue nature of arguments if possible.

Now, inside perfectSet, you want to perfectly pass the argument to the correct overload of set(). This is where std::forward is necessary:

template<class T>
void perfectSet(T && t) {
    set(std::forward<T>(t));
}

Without std::forward the compiler would have to assume that we want to pass t by reference. To convince yourself that this is true, compare this code:

void perfectSet(T && t) {
    set(t);
    set(t); // t still unchanged
}

to this:

void perfectSet(T && t) {
    set(std::forward<T>(t));
    set(t); // t is now empty
}

If you don't explicitly forward t, the compiler has to defensively assume that you might be accessing t again and chose the lvalue reference version of set. But if you forward t, the compiler will preserve the rvalue-ness of it and the rvalue reference version of set() will be called. This version moves the contents of t, which means that the original becomes empty.

This answer turned out much longer than what I initially assumed ;-)

Answer 3

It works because when perfect forwarding is invoked, the type T is not the value type, it may also be a reference type.

For example:

template<typename T> void f(T&&);
int main() {
    std::string s;
    f(s); // T is std::string&
    const std::string s2;
    f(s2); // T is a const std::string&
}

As such, forward can simply look at the explicit type T to see what you really passed it. Of course, the exact implementation of doing this is non-trival, if I recall, but that's where the information is.

When you refer to a named rvalue reference, then that is indeed an lvalue. However, forward detects through the means above that it is actually an rvalue, and correctly returns an rvalue to be forwarded.