Usage of std::forward vs std::move

Question

I always read that std::forward is only for use with template parameters. However, I was asking myself why. See the following example:

void Imag         


        

Answer 1

I recommend reading "Effective Modern C ++" by Scott Meyers, specifically:

• Item 23: Understand std::move and std::forward.
• Item 24: Distinguish universal references for rvalue references.

From a purely technical perspective, the answer is yes: std::forward can do it all. std::move isn’t necessary. Of course, neither function is really necessary, because we could write casts everywhere, but I hope we agree that that would be, well, yucky. std::move’s attractions are convenience, reduced likelihood of error, and greater clarity.

rvalue-reference

This function accepts rvalues and cannot accept lvalues.

void ImageView::setImage(Image&& image){
    _image = std::forward(image);        // error 
    _image = std::move(image);           // conventional
    _image = std::forward<Image>(image); // unconventional
}

Note first that std::move requires only a function argument, while std::forward requires both a function argument and a template type argument.

Universal references (forwarding references)

This function accepts all and does perfect forwarding.

template <typename T> void ImageView::setImage(T&& image){
    _image = std::forward<T>(image);
}

Answer 2

You cannot use std::forward without explicitly specifying its template argument. It is intentionally used in a non-deduced context.

To understand this, you need to really understand how forwarding references (T&& for a deduced T) work internally, and not wave them away as "it's magic." So let's look at that.

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

Let's say we call foo like this:

foo(42);

• 42 is an rvalue of type int.
• T is deduced to int.
• The call to bar therefore uses int as the template argument for std::forward.
• The return type of std::forward<U> is U && (in this case, that's int &&) so t is forwarded as an rvalue.

Now, let's call foo like this:

int i = 42;
foo(i);

• i is an lvalue of type int.
• Because of the special rule for perfect forwarding, when an lvalue of type V is used to deduce T in a parameter of type T &&, V & is used for deduction. Therefore, in our case, T is deduced to be int &.

Therefore, we specify int & as the template argument to std::forward. Its return type will therefore be "int & &&", which collapses to int &. That's an lvalue, so i is forwarded as an lvalue.

Summary

Why this works with templates is when you do std::forward<T>, T is sometimes a reference (when the original is an lvalue) and sometimes not (when the original is an rvalue). std::forward will therefore cast to an lvalue or rvalue reference as appropriate.

You cannot make this work in the non-template version precisely because you'll have only one type available. Not to mention the fact that setImage(Image&& image) would not accept lvalues at all—an lvalue cannot bind to rvalue references.

Answer 3

You have to specify the template type in std::forward.

In this context Image&& image is always an r-value reference and std::forward<Image> will always move so you might as well use std::move.

Your function accepting an r-value reference cannot accept l-values so it is not equivalent to the first two functions.