Can a defaulted template type be universal reference?

Question

In the following, is && a universal reference ?

template <class Function = std::greater<int> > void f(Function&& f = Function());

score 7 · Accepted Answer · answered Mar 03 '13 at 02:39

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The term universal reference is a made up term by Scott Meyers to make a distinction between normal rvalue references, i.e. int&& and rvalue references in template code, i.e. T&&. This is important because of the reference collapsing rules that come into play with template code, so this term is used to aid in teaching. The reason it is called a unversal reference is because it can bind to anything.

Just keep in mind that there is no such thing as a universal reference in the C++ language, Function&& is a rvalue reference as far as the language is concerned. However, yes, in Scott Meyer's made up terminology, Function&& is a universal reference in your example.

answered Mar 03 '13 at 02:39

Jesse Good

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+1 It is almost like we were typing at the same time off of the same script. Well done. – Howard Hinnant Mar 03 '13 at 02:42
Thanks, +1 to you also, since I probably learned most of it from reading your answers on SO. – Jesse Good Mar 03 '13 at 02:44
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@StephenLin: Keep checking those answers with code! I did so too with this question. It is irritating when people don't check their answers with a bit of code. – Howard Hinnant Mar 03 '13 at 02:51

score 3 · Answer 2 · answered Mar 03 '13 at 02:39

is && a universal reference ?

Sort of. Like the Easter bunny and Santa Claus, there is actually no such thing as a universal reference (sorry if I spoiled that for you). The term universal reference does not show up in the standard. It is term coined by Scott Meyers to describe templated rvalue references when "special deduction rules" are active.

What you show is valid code. And the special deduction rules do apply in this case. That is, when you pass an lvalue argument A to f, Function will deduce to A&. So by Scott's definition, yes, Function is a "universal reference."

Stephen Lin · Answer 3 · 2013-03-03T03:00:04.100

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Looks like it, std::forward<Function> correctly forwards the argument type when called in your function.

edited Mar 03 '13 at 03:00

answered Mar 03 '13 at 02:43

Stephen Lin

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Can a defaulted template type be universal reference?

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