c++ constructor question

Question

I am still confused with the CTORs:

Question 1:
why line 15 call A:A(int) instead of A:A(double&)?

Question 2:
why line 18 did not call A:A(B&)?

#include <iostream>
using namespace std;

class B{};

class A{
public:
   A(int )    {cout<<"A::A(int)"<<endl;}
   A(double&){cout<<"A::A(double&)"<<endl;} // it will work if it is A(double), without the &
   A(B&){cout<<"A::A(B&)"<<endl;}
};

int main()
{
/*line 15*/   A obj((double)2.1);  // this will call A(int), why?
   B obj2;
   A obj3(obj2);
/*line 18*/   A obj4(B);          // this did not trigger any output why?
}

Most reasonable compilers would at least print out a warning about a double-to-int conversion. — EboMike
– EboMike, Commented Jul 21, 2011 at 18:33
@Kerrek: Yes. Yes, it does. (Though note the warning, and you will have to forgive codepad's -Werror.) — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Jul 21, 2011 at 18:33
@Tomalak:Hah, I didn't know you can omit the parentheses for the default constructor. You can even say A x(int);... something new every day:-) — Kerrek SB
– Kerrek SB, Commented Jul 21, 2011 at 18:38
@Kerrek: Huh? You have to omit parentheses for the default constructor, otherwise you're declaring a function. Which is also what you're doing in A x(int);. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Jul 21, 2011 at 18:40

kennytm · Accepted Answer · 2011-07-21 18:29:16Z

6

Line 15: A(double&) can only take lvalues, i.e. variables that can be assigned to. (double)2.1 is an rvalue. Use A(const double&) if you need to accept rvalues as reference too.

Line 18: B is a type, not a value. A obj4(B); only declares a function with name obj4 taking a B and returning an A.

answered Jul 21, 2011 at 18:29

kennytm

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4 Comments

John Crane Over a year ago

Thanks all for the fast response. I changed line 18 to be A obj4(B()), it is still not working. I then changed it to be A obj4((B())), then it works. Is A obj4(B()) still a function?

Lightness Races in Orbit Over a year ago

@John: Yes. It is a function called obj4 that takes a function with no name that takes nothing and returns a B, and returns an A. :) Read more.

John Crane Over a year ago

Now I start to regret becoming a programmer :)

Lightness Races in Orbit Over a year ago

@John: Just wait until you get to templates! For what it's worth, you absorbed all these responses very quickly and from what I can tell, you understood them. So... good job!

Lightness Races in Orbit · Accepted Answer · 2011-07-21 18:30:52Z

Line 15

Your reference constructors accept only references to non-const objects.

That is, you wrote:

A(int)     { cout<<"A::A(int)"<<endl; }
A(double&) { cout<<"A::A(double&)"<<endl; }
A(B&)      { cout<<"A::A(B&)"<<endl; }

instead of:

A(int)           { cout<<"A::A(int)"<<endl; }
A(double const&) { cout<<"A::A(double const&)"<<endl; }
A(B const&)      { cout<<"A::A(B const&)"<<endl; }

The problem with this is that, in A obj((double)2.1), you invoked the constructors with temporary objects (in this case, the double literal 2.1 (that cast to double is pointless), and temporary objects may not bind to references that are not references to const.

So the A(double&) constructor cannot be called; the only one left that might match is A(int), and the required conversion is performed on your data (2.1 → 2) to make this happen. (Your compiler will normally warn you that this occurring.)

So, use the form of constructor that accepts reference to const.

Line 18

As for your A obj4(B), B is a type so you're declaring a function named obj4 that takes an (unnamed) B and returns an A.

Lightness Races in Orbit · Accepted Answer · 2011-07-21 18:51:50Z

0

Because & means it is a reference type and on line 15 you pass a value, not a reference.

edited Jul 21, 2011 at 18:51

Lightness Races in Orbit

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answered Jul 21, 2011 at 18:28

Filip Ekberg

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15 Comments

Lightness Races in Orbit Over a year ago

You never "pass" a reference.

orlp Over a year ago

How would you go about passing a reference? I would be very interested :)

Filip Ekberg Over a year ago

@Tomalak, Maybe I used the wrong words, but it does expect a reference and not a value. So, is my answer wrong?

Filip Ekberg Over a year ago

@nightcracker, it's just a "phrase" commonly used where I mostly spend my days.. I mean that it expects a reference and he is "passing" hence "using"/"sending" or whatever word you want for it.

Lightness Races in Orbit Over a year ago

@Filip: I think you've confused terminology. You only ever pass "values" into functions; some functions copy them into new objects, others work with the existing objects ("accept by reference") and others just don't match at all... like this one. That's because you cannot bind the temporary/rvalue literal 2.1 to a reference parameter. But there is no way that you can fix that by "passing a reference" from the callsite.

|

Puppy · Accepted Answer · 2011-07-21 18:29:11Z

-1

Line 15, the double is an rvalue (a temporary) and cannot convert to a non-const reference. For Line 18, the most likely explanation is that your compiler simply optimized out the constructor call.

answered Jul 21, 2011 at 18:29

Puppy

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4 Comments

orlp Over a year ago

If the compiler optimized it out then he must file a bug because the compiler is not allowed to optimize anything out that puts out output. I think KennyTM nailed it about line 18.

Puppy Over a year ago

@Nightcracker: Constructors have special leeway in this regard.

Lightness Races in Orbit Over a year ago

@DeadMG: Nonsense. Line 18 doesn't even contain a constructor call.

Lightness Races in Orbit Over a year ago

@nightcracker: Unfortunately, though, constructors do have special semantics and can be optimised out even when they have side effects. It's pretty rubbish.

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