Can I typically/always use std::forward instead of std::move?
C++C++11Move SemanticsRvalue ReferencePerfect ForwardingC++ Problem Overview
I've been watching Scott Meyers' talk on Universal References from the C++ and Beyond 2012 conference, and everything makes sense so far. However, an audience member asks a question at around 50 minutes in that I was also wondering about. Meyers says that he does not care about the answer because it is non-idiomatic and would silly his mind, but I'm still interested.
The code presented is as follows:
// Typical function bodies with overloading:
void doWork(const Widget& param) // copy
{
// ops and exprs using param
}
void doWork(Widget&& param) // move
{
// ops and exprs using std::move(param)
}
// Typical function implementations with universal reference:
template <typename T>
void doWork(T&& param) // forward => copy and move
{
// ops and exprs using std::forward<T>(param)
}
The point being that when we take an rvalue reference, we know we have an rvalue, so we should std::move
it to preserve the fact that it's an rvalue. When we take a universal reference (T&&
, where T
is a deduced type), we want std::forward
to preserve the fact that it may have been an lvalue or an rvalue.
So the question is: since std::forward
preserves whether the value passed into the function was either an lvalue or an rvalue, and std::move
simply casts its argument to an rvalue, could we just use std::forward
everywhere? Would std::forward
behave like std::move
in all cases where we would use std::move
, or are there some important differences in behaviour that are missed out by Meyers' generalisation?
I'm not suggesting that anybody should do it because, as Meyers correctly says, it's completely non-idiomatic, but is the following also a valid use of std::move
:
void doWork(Widget&& param) // move
{
// ops and exprs using std::forward<Widget>(param)
}
C++ Solutions
Solution 1 - C++
The two are very different and complementary tools.
-
std::move
deduces the argument and unconditionally creates an rvalue expression. This makes sense to apply to an actual object or variable. -
std::forward
takes a mandatory template argument (you must specify this!) and magically creates an lvalue or an rvalue expression depending on what the type was (by virtue of adding&&
and the collapsing rules). This only makes sense to apply to a deduced, templated function argument.
Maybe the following examples illustrate this a bit better:
#include <utility>
#include <memory>
#include <vector>
#include "foo.hpp"
std::vector<std::unique_ptr<Foo>> v;
template <typename T, typename ...Args>
std::unique_ptr<T> make_unique(Args &&... args)
{
return std::unique_ptr<T>(new T(std::forward<Args>(args)...)); // #1
}
int main()
{
{
std::unique_ptr<Foo> p(new Foo('a', true, Bar(1,2,3)));
v.push_back(std::move(p)); // #2
}
{
v.push_back(make_unique<Foo>('b', false, Bar(5,6,7))); // #3
}
{
Bar b(4,5,6);
char c = 'x';
v.push_back(make_unique<Foo>(c, b.ready(), b)); // #4
}
}
In situation #2, we have an existing, concrete object p
, and we want to move from it, unconditionally. Only std::move
makes sense. There's nothing to "forward" here. We have a named variable and we want to move from it.
On the other hand, situation #1 accepts a list of any sort of arguments, and each argument needs to be forwarded as the same value category as it was in the original call. For example, in #3 the arguments are temporary expressions, and thus they will be forwarded as rvalues. But we could also have mixed in named objects in the constructor call, as in situation #4, and then we need forwarding as lvalues.
Solution 2 - C++
Yes, if param
is a Widget&&
, then the following three expressions are equivalent (assuming that Widget
is not a reference type):
std::move(param)
std::forward<Widget>(param)
static_cast<Widget&&>(param)
In general (when Widget
may be a reference), std::move(param)
is equivalent to both of the following expressions:
std::forward<std::remove_reference<Widget>::type>(param)
static_cast<std::remove_reference<Widget>::type&&>(param)
Note how much nicer std::move
is for moving stuff. The point of std::forward
is that it mixes well with template type deduction rules:
template<typename T>
void foo(T&& t) {
std::forward<T>(t);
std::move(t);
}
int main() {
int a{};
int const b{};
//Deduced T Signature Result of `forward<T>` Result of `move`
foo(a); //int& foo(int&) lvalue int xvalue int
foo(b); //int const& foo(int const&) lvalue int const xvalue int const
foo(int{});//int foo(int&&) xvalue int xvalue int
}