Does 'const' disqualify universal reference?

Question

I have a human class with ctor using universal reference

class Human {
 public:
  template<typename T>
  explicit Human(T&& rhs) {
    // do some initialization work
  }

  Human(const Human& rhs);  // the default ctor I don't care about
}

Now if I have a const Human object

const Human one_I_do_not_care; // then play with that
Human the_human_I_care(one_I_do_not_care)  // now create another one

Does the last line use the template ctor or the default ctor? My understanding is the "const" will disqualify the template ctor, am I correct?

Human the_human_I_care(one_I_do_not_care)  // line in question

By const disqualify the template ctor, I mean adding const then it would not match the template ctor, not it still matches two but compiler picks one.

What's preventing you from adding a few debugging statements, and figuring it out on your own? — Sam Varshavchik, Jan 20 '18 at 17:55
I did, what I got is different than what's mentioned in the book — WhatABeautifulWorld, Jan 20 '18 at 18:04
If I tell the compiler to use the move constructor, it works. `Human the_human_I_care{std::move(one_I_do_not_care)};` — Eljay, Jan 20 '18 at 18:23
@Eljay how come? one_I_do_not_care is const, how could you move it? I think you still just do a copy — WhatABeautifulWorld, Jan 20 '18 at 18:27
@WhatABeautifulWorld • std::move performs a cast to a universal reference. — Eljay, Jan 20 '18 at 18:29
@Eljay I know, I mean after you cast you still do a copy. But this has nothing to do with my question. I just want to know if the *const* makes the template ctor not qualified. — WhatABeautifulWorld, Jan 20 '18 at 18:31
@WhatABeautifulWorld • Once you have a `Human const&&` from the std::move, that universal reference will work in the explicit constructor as a `T&&` where T is `Human const`. — Eljay, Jan 20 '18 at 18:44

JFMR · Accepted Answer · 2018-01-20T18:29:07.930

8

Does the last line use the template ctor or the default ctor? My understanding is the const will disqualify the template ctor, am I correct?

No. You are passing a const-qualified Human as an argument. Since both constructors would match equally well (i.e.: if the template would be instantiated to take a const Human&), then the non-template constructor is preferred over the template (and no template instantiation for a const Human& parameter ever occur).

edited Jan 20 '18 at 18:29

answered Jan 20 '18 at 17:56

JFMR

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I did try to test it out. I think what you said is half correct: 1) Adding a const does pick the default ctor; 2) marking the default ctor *delete* makes it not be able to find a ctor, which means it's wrong to say "the template would be instantiated to take a const Human&". – WhatABeautifulWorld Jan 20 '18 at 18:05
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@WhatABeautifulWorld by *would be instantiated* I mean it is **not** instantiated: instantiation does not happen. – JFMR Jan 20 '18 at 18:08
@水飲み鳥 "both constructors match equally well " this is not correct. The template ctor will not match after adding const. Please revise your answer and I will upvote :) – WhatABeautifulWorld Jan 20 '18 at 18:09
@WhatABeautifulWorld It should be "both ctors **would** match equally well* – JFMR Jan 20 '18 at 18:12
@水飲み鳥 I tested on my local machine, it seems if I mark the default ctor *delete*, the compiler could not find a ctor. If both match equally well, it would find the template ctor, right? That's all the question is about and the part I want to know. – WhatABeautifulWorld Jan 20 '18 at 18:15
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@WhatABeautifulWorld if you delete the default ctor and you've not defined a non-template ctor, then you have no ctor at all. – JFMR Jan 20 '18 at 18:16
But I have the template ctor? I am confused? – WhatABeautifulWorld Jan 20 '18 at 18:17
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@WhatABeautifulWorld, no, deleted constructors still participate in overload resolution. – Mário Feroldi Jan 20 '18 at 18:19
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@WhatABeautifulWorld Deleting things aren't missing. They follow usual overload resolution rules. If they are selected, an error occurs. This is *different* than them not being there. – Yakk - Adam Nevraumont Jan 20 '18 at 18:20
@WhatABeautifulWorld the way to get the template "ctor" called is by kind of "playing" with the process overloading resolution: you must have at least one non-template ctor, so that it matches worse than the template ctor. That way the template ctor is elected. – JFMR Jan 20 '18 at 18:23
Thanks all! To summaries, T&& could still bind to const, const doesn't disqualify that. Seems hard to digest because of this: https://stackoverflow.com/questions/34556808/const-applied-to-universal-reference-parameter – WhatABeautifulWorld Jan 20 '18 at 18:23
Oh, @水飲み鳥 btw, your answer should be "No". My question is: does the const DISQUALIFY the template ctor or not. From the discussion above, it doesn't disqualify that. – WhatABeautifulWorld Jan 20 '18 at 18:28

Does 'const' disqualify universal reference?

1 Answers1