Passing class to constructor, when no such constructor exists. Why does it work?

Question

I'm having a class that looks like:

class A {
public:
    A(float v) 
    { 
        A::v = v; 
    }
    float v;
    float set(float v) 
    {
        A::v = v;
        return v;
    }
    float get(float v) 
    {
        return A::v;
    }
};

Then I instantiate 2 objects of class A:

A* a = new A(1.0);
A* b = new A(*a);

Why are there no errors when my Class A doesn't have a constructor which takes a class A?

Aside: `get` takes an argument: `float get(float v) ` -- but that's not a common expectation/idiom. You may want to change that. — Brian Cain, May 10 '19 at 15:08
@NathanOliver Why not dupe-hammer it? I'm sure we had a similar question yesterday too, although I can't find it right now... — Max Langhof, May 10 '19 at 15:17
@BrianCain Wouldn't do that if it was my own code. This was part of a quiz i took, so the passing of arguments was probably made to cause confusion when used in another part of the code (which is not included in the snippet) :) — Zartock, May 10 '19 at 15:46
@MaxLanghof I didn't do it because the answers here talk about how to get the behavior the OP wants, which isn't covered by the dupe. I didn't want to unilaterally close it. — NathanOliver, May 10 '19 at 16:20

Maxim Egorushkin · Accepted Answer · 2019-05-10T15:09:25.617

11

The compiler generates a copy constructor for you:

If no user-defined copy constructors are provided for a class type (struct, class, or union), the compiler will always declare a copy constructor as a non-explicit inline public member of its class.

You can make the copy constructor and assignment deleted and make the compiler not declare move assignment and constructor by declaring one of move constructor or assignment as deleted:

A(A&&) = delete; // Makes the class non-copyable and non-moveable.

edited May 10 '19 at 15:09

answered May 10 '19 at 15:02

Maxim Egorushkin

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Upvoted as it seems that I have plagiarised your snippet! (Feel free to include any bits from my answer too - this looks to me to be the best answer of the current crop). – Bathsheba May 10 '19 at 15:14
Is there really no dupe that covers this? That's a lot of answers for a question I could swear I've seen yesterday... – Max Langhof May 10 '19 at 15:15
1

@Bathsheba Yours is a rather rare beast `A(A const&&)`. – Maxim Egorushkin May 10 '19 at 15:16
@MaximEgorushkin: Last edit and a wiki as promised. It's hardly "my" answer any more! – Bathsheba May 10 '19 at 15:16

Bathsheba · Answer 2 · 2019-05-10T15:16:03.547

8

It does have the copy constructor: the compiler has generated one for you.

If you want to disable that explicitly, then write

A(const A&) = delete;

in the class declaration; and using

A(A&&) = delete;

deletes all the rule of five functions, except the destructor.

edited May 10 '19 at 15:16

answered May 10 '19 at 15:03

Bathsheba

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The compiler still generates the assignment operator though. – Maxim Egorushkin May 10 '19 at 15:12
1

@MaximEgorushkin: Yup indeed, added the idiomatic way of dealing with that. – Bathsheba May 10 '19 at 15:13
1

nitpick: destructor still generated – sp2danny May 10 '19 at 15:15
One more fix and I'll wiki this! – Bathsheba May 10 '19 at 15:16

vlad_tepesch · Answer 3 · 2019-05-10T20:00:51.463

5

An implicit copy constructor is generated by the Compiler, if you do not specify an own.

One further note:

Try

A a = 3.0f;

Conclusion: always mark constructors that take a single basic data type as explicit ... unless you like the implicit conversion

edited May 10 '19 at 20:00

answered May 10 '19 at 15:03

vlad_tepesch

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"always mark constructors that take a single basic data type as explicit" ... unless you like the implicit conversion. – Brian Cain May 10 '19 at 15:09
3

That advice is out of date; you need to mark multi-argument constructors explicit too to obviate initialisation using `{}`. – Bathsheba May 10 '19 at 15:10
@Bathsheba I'd argue you're still less likely to want to `explicit` those, but it's possible – Lightness Races in Orbit May 10 '19 at 15:12
@BrianCain of course you are right. but in my expierence it is wanted rather seldom, maybe if you imeplement some number class that should behave transparently. – vlad_tepesch May 10 '19 at 20:03

Passing class to constructor, when no such constructor exists. Why does it work?

3 Answers3