Why does this call the default constructor?

Question

struct X
{
    X()    { std::cout << "X()\n";    }
    X(int) { std::cout << "X(int)\n"; }
};

const int answer = 42;

int main()
{
    X(answer);
}

I would have expected this to print either

X(int), because X(answer); could be interpreted as a cast from int to X, or
nothing at all, because X(answer); could be interpreted as the declaration of a variable.

However, it prints X(), and I have no idea why X(answer); would call the default constructor.

BONUS POINTS: What would I have to change to get a temporary instead of a variable declaration?

It does the same thing if you change `int answer` -> any other type. — Flexo, Jul 27 '12 at 15:35
I'm not sure what X(answer) is supposed to do, but if you create an instance initialized with an int, say X xx(answer), you'll get the X(int) print. — Michael, Jul 27 '12 at 15:37
@JTA: Because that one can never be interpreted as a declaration, so it has to be a constructor invocation. — Xeo, Jul 27 '12 at 15:39
@JTA And finally, `X(int(answer));` doesn't print anything, because it's a function declaration :) — fredoverflow, Jul 27 '12 at 15:51
*nothing at all, because X(answer); could be interpreted as the declaration of a variable.* That declaration would also be a definition, and it triggers the execution of the default constructor... which in turn means that you answered your own question. — David Rodríguez - dribeas, Jul 27 '12 at 15:58
@David You're a bit late to the party, can I fetch you a drink? ;) — fredoverflow, Jul 27 '12 at 16:01
@David `double(expresso);` there you go, declared just for you ;) — fredoverflow, Jul 27 '12 at 16:06
@FredOverflow: I must need a *definition* to use it, because I am feeling no effect... — David Rodríguez - dribeas, Jul 27 '12 at 16:56

Xeo · Accepted Answer · 2012-09-23T11:51:44.233

73

nothing at all, because X(answer); could be interpreted as the declaration of a variable.

Your answer is hidden in here. If you declare a variable, you invoke its default ctor (if non-POD and all that stuff).

On your edit: To get a temporary, you have a few options:

(X(answer));
(X)answer;
static_cast<X>(answer)
X{answer}; (C++11)
[]{ return X(answer); }(); (C++11, may incur copy)
void(), X(answer);
X((void(),answer));
true ? X(answer) : X();
if(X(answer), false){}
for(;X(answer), false;);
X(+answer);

edited Sep 23 '12 at 11:51

answered Jul 27 '12 at 15:36

Xeo

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The `static_cast(answer)` feels the "most C++" answer -- it's even recommended by an old [GCC documentation](http://gcc.gnu.org/onlinedocs/gcc-4.0.2/gcc/Volatiles.html) as a way to force an rvalue. – Kerrek SB Jul 27 '12 at 16:23
Won't the brace initializer also maybe incur a copy? – rubenvb Jul 27 '12 at 16:45
@rubenvb: Why would it? It's just a fancy new way of saying `X(answer)` and guarantees a ctor call. – Xeo Jul 27 '12 at 16:50
@Xeo: because brace initializer syntax takes its arguments by value? (<- note the question mark) – rubenvb Jul 27 '12 at 16:51
@rubenvb: Oh, I didn't mean that the `answer` is copied, I meant when the `X` is returned from inside the lambda. – Xeo Jul 27 '12 at 16:52
4

@KerrekSB But surely only before C++11, no? Now, the canonical answer will be `X{answer}`. – Konrad Rudolph Jul 28 '12 at 11:48
Can anybody please explain how `X(answer)` is equivalent to `X answer` ? – Sam Aug 18 '13 at 19:48
@SAM: Because the standard says so. As Kerrek's answer says, the parens are optional. – Xeo Aug 18 '13 at 20:09
@Xeo Common man! Just tell me how do you know every little detail about c++? I'm not asking about just this question but I've seen your other answers too. I mean, just look at`void(), X(answer);` How on earth did you find that out?!?! Also what do you refer for C++? I read TC++PL (3rd & 4th edition). Also how did you learned C++11 in such a short time? – Sam Aug 19 '13 at 06:17
@Xeo And how many years of programming experience did you require to get to this stage? – Sam Aug 19 '13 at 06:25
@SAM: I recommend joining me in the Lounge chat room (check my profile), since this is getting off-topic here. :) – Xeo Aug 19 '13 at 08:19

score 67 · Answer 2 · answered Jul 27 '12 at 15:36

67

The parentheses are optional. What you said is identical to X answer;, and it's a declaration statement.

answered Jul 27 '12 at 15:36

Kerrek SB

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score 9 · Answer 3 · answered Jul 27 '12 at 15:39

9

If you want to declare a variable of the type X, you should do it this way:

X y(answer);

answered Jul 27 '12 at 15:39

huysentruitw

27,376
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He did not ask how to make it call the `X(int)` ctor. – Xeo Jul 27 '12 at 15:41
Yeah but I have a small feeling that it was the thing he ment to do :) – huysentruitw Jul 27 '12 at 15:42
6

@WouterH: Actually, knowing Fred, it's unlikely. He's one of those folks who like to explore the dark corners of the C++ Standard and try and understand it. In a certain RPG he would have lost all his sanity points already ;) – Matthieu M. Jul 27 '12 at 17:40

Why does this call the default constructor?

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