Usage of std::forward vs std::move

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.