How do I avoid copy constructor in the conversion function?

Question

I am trying to convert between two classes and avoid a temporary object.

Here is my declaration for Square:

class CSquare
{
public:
    int dimension;
    CSquare(int dimension);

    // Conversion to Rectangle
    operator CRectangle();

    ~CSquare(void);
};

and here is my declaration for Rectangle:

class CRectangle
{
public:
    int length;
    int width;

    CRectangle(int length, int width);

    //Copy Constructor
    CRectangle(const CRectangle& anotherRectangle); 

    ~CRectangle(void);
};

Why does

CSquare aSquare = CSquare(10);
    CRectangle convertedRect = (CRectangle) aSquare;

invoke the copy constructor?

I have a conversion function:

CSquare::operator CRectangle(){
    return CRectangle(CSquare::dimension,CSquare::dimension);
}

but I'm still getting a temporary object.

How do I avoid the temporary object?

I forgot to praise you for not having Rectangle deriving from Square or the other way around :) — Matthieu M.
– Matthieu M., Commented Feb 1, 2011 at 18:46

Benjamin Lindley · Accepted Answer · 2011-02-01 16:47:41Z

5

You can avoid copy construction by writing a conversion constructor in CRectangle:

CRectangle::CRectangle(const CSquare & sq)
    :length(sq.dimension),
     width(sq.dimension)
{}
...
...
CSquare aSquare(10);
CRectangle convertedRect(aSquare);

answered Feb 1, 2011 at 16:47

Benjamin Lindley

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3 Comments

Edward Strange Over a year ago

And yet another, better way would be to not define Rectangle and Square differently. Squareness is a predicate that operates on rectangles. If a rectangle has all sides equal then it is square.

Benjamin Lindley Over a year ago

@Noah: Yes, you could define a function for Rectangle to test that, but that doesn't imply that you should not define a Square class, if one is warranted.

Matthieu M. Over a year ago

while we are it, perhaps that recommending explicit in front of CRectangle would make sense :) ?

Edward Strange · Accepted Answer · 2011-02-01 16:49:47Z

2

Because you're copying a Rectangle.

I honestly don't know what else to say. Compilers are free to optimize away the call in the case you're showing, but they don't have to. Apparently yours is not.

Edit....

As to how to avoid it, you really can't. You could try to crank up the optimizations but it can be extremely difficult to see if you've succeeded, especially with such a small, trivial construct. If I where you I wouldn't be so worried about it unless I really found, by profiling my code, that an inordinate amount of time was being spent in that function. Then I'd be looking for ways to crunch the algorithms using the code to reduce copies.

edited Feb 1, 2011 at 16:49

answered Feb 1, 2011 at 16:37

Edward Strange

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1 Comment

Matthieu M. Over a year ago

I am not sure whether Visual C++ was optimizing in Debug Mode actually.

mskfisher · Accepted Answer · 2011-11-02 14:54:49Z

1

How do I avoid the temporary object?

You don't. That's how conversions work. The temporary may be elided (constructing directly into the destination), but that is not required.

It sounds like you want move semantics, which are in C++0x. Even then, the temporary "exists", but it can be moved instead of copied, which, for types where it matters, is often much more efficient.

Is this square/rectangle case not really the situation you care about and only used here for an example?

edited Nov 2, 2011 at 14:54

mskfisher

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answered Feb 1, 2011 at 16:53

Fred Nurk

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Comments

James · Accepted Answer · 2011-02-01 16:38:20Z

0

I would assume since operator CRectangle returns an instance of CRectangle, the copy constructor is invoked for the return value.

answered Feb 1, 2011 at 16:38

James

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2 Comments

Edward Strange Over a year ago

No. The constructor is invoked when they use initialization syntax to create a copy of that return in a local object.

Fred Nurk Over a year ago

@NoahRoberts: That is true, but the copy ctor may also be used for the return value. In fact, temporaries in this case can be copied many times, or none with RVO.

Community · Accepted Answer · 2017-05-23 10:33:12Z

0

In general this depends on the compiler. Return value optimization is done by most compiler. So instead of creating a temporary object and assign it to your object the copy Ctor from your object is called.

Also take a look here Copy constructor vs. return value optimization

edited May 23, 2017 at 10:33

CommunityBot

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answered Feb 1, 2011 at 16:41

mkaes

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Comments

nullforce · Accepted Answer · 2011-02-01 16:53:11Z

0

A temporary CRectangle object is being created, then copied to convertedRect.

edited Feb 1, 2011 at 16:53

answered Feb 1, 2011 at 16:38

nullforce

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2 Comments

UncleBens Over a year ago

Not possible in standard C++.

Edward Strange Over a year ago

@UncleBens - what? What is not possible?

Mark B · Accepted Answer · 2011-02-01 17:32:15Z

0

I can't tell which type of object you're saying has a temporary.

If the square, then you need to remove any excuse for the compiler to create one:

CSquare aSquare(10);

If the rectangle, then you shoudl use a converting constructor instead of an operator, as suggested by @PigBen:

Rectangle::CRectangle(const CSquare & sq)
    :length(sq.dimension),
     width(sq.dimension)
{}
// And then:
CRectangle convertedRect(aSquare);

answered Feb 1, 2011 at 17:32

Mark B

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Comments

Zac Howland · Accepted Answer · 2011-02-01 17:49:54Z

0

In this particular case, you do it by not having the conversion function and instead use inheritance. E.g.:

class CRectangle
{
public:
    int length;
    int width;

    CRectangle(int length, int width);
};

class CSquare : public CRectangle
{
public:
    CSquare(int dimension) : length(dimension), width(dimension)
    {}
};

answered Feb 1, 2011 at 17:49

Zac Howland

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