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Re: Operator overloading and copy constructor. Can't find the error.

From:
=?ISO-8859-1?Q?Erik_Wikstr=F6m?= <Erik-wikstrom@telia.com>
Newsgroups:
comp.lang.c++
Date:
Tue, 24 Jul 2007 07:59:08 GMT
Message-ID:
<gjipi.4670$ZA.2128@newsb.telia.net>
On 2007-07-24 02:49, clicwar wrote:

On 23 jul, 06:35, James Kanze <james.ka...@gmail.com> wrote:

On Jul 23, 3:15 am, "Jim Langston" <tazmas...@rocketmail.com> wrote:

"clicwar" <clic...@gmail.com> wrote in message
news:1185144847.041900.64380@q75g2000hsh.googlegroups.com...


    [...]

Here's your program cleaned up a bit and extened a little.


Cleaned up still some more, to make it more idiomatic. (The
changes don't actually make a difference here, but in many
cases, they do, and it's good practice to stick to.)

#include <iostream>
#include <string>

class Vector
{
private:
    float x,y;


Normally, of course, we'd use double everywhere. Float is
reserved for cases where we have to save memory.

public:
    Vector(float u, float v);
    Vector operator+ (const Vector &a) const;


Most of the time, it is preferable to make operator+ a free
function. It only really makes a difference if implicit
conversions are involved, but it's never wrong, so why not be
consistent about it.

Also, the most important single rule about operator overloading
is to do it consistently with what the built in types do. So if
you define +, you also define +=. With the same semantics.
It's fairly frequent to define operator+ in terms of +=, e.g.:

    // Free function...
    Vector operator+( Vector lhs, Vector rhs )
    {
        lhs += rhs ;
        return lhs ;
    }

or
    Vector operator+( Vector const& lhs, Vector const& rhs )
    {
        Vector result( lhs ) ;
        result += rhs ;
        return result ;
    }

(The generation of these free functions can actually be
automated pretty well by means of some clever use of templates,
but I don't think that the original poster is ready for that
yet.)

    Vector operator* (const float a ) const;
    Vector operator* (const Vector &b) const;
    Vector(const Vector &source);
    void Show();
};
void Vector::Show()
{
    std::cout <<"(" << x <<"," <<y <<")";
}
Vector::Vector(float u = 0, float v = 0): x(u), y(v)
{
}


Given that this function will typically be in a separate soruce
file, the default arguments don't make any sense here. They
should be on the function declaration in the class.

And if you use them, you've created a converting constructor,
and you'll definitely want operator+ to be a non-member:

    Vector a, b ;
    a = b + 3.14 ; // Legal, member or not...
    a = 3.14 + b ; // Legal if non-member, illegal if member.

Vector::Vector(const Vector &source)
{
    x = source.x;
    y = source.y;
}


Again, it makes no difference in this particular case, but you
really should prefer initialization lists:

    Vector::Vector(
        Vector const& source )
        : x( source.x )
        , y( source.y )
    {
    }

(And while I'm at it: note the position of the const. In this
case, const Vector and Vector const are both legal, but most of
the time, the const must follow what it modifies, so for
consistency's sake, we always put it after.)

Vector Vector::operator+ (const Vector &a) const
{
    Vector temp;
    temp.x = x + a.x;
    temp.y = y + a.y;
    return temp;
}
Vector Vector::operator* (float a) const
{
    Vector temp;
    temp.x = x * a;
    temp.y = y * a;
    return temp;
}


So you support "v * 3.14", but not "3.14 * v". Not good, IMHO.

--
James Kanze (GABI Software) email:james.ka...@gmail.com
Conseils en informatique orient?e objet/
                   Beratung in objektorientierter Datenverarbeitung
9 place S?mard, 78210 St.-Cyr-l'?cole, France, +33 (0)1 30 23 00 34


James Kanze, thank you for your time.
 After re-re-read your posts, i have to confess that i re-confused
myself again.

I concentrate this confusion in two quick problems. The code below
will help to make the questions make sense:

#include <iostream>
using namespace std;

class Vector {
    private:
        float x,y;
    public:
        Vector(float u, float v);
        Vector();
        Vector operator+ (Vector const &a);
        Vector(Vector &source);
        void Show();
};
void Vector::Show() {
    cout <<"(" << x <<"," <<y <<")";
}
Vector::Vector(float u, float v) {
    x=u; y=v;
}
Vector::Vector() {
    x=0; y=0;
}
Vector::Vector(Vector &source) { //this is the line24
    x = (source.x) ; y=10;
}
Vector Vector::operator+ (Vector const &a) {
    Vector temp;
    temp.x = x + a.x;
    temp.y = y + a.y;
    return (temp);
}

int main() {
    Vector a(3,1), b(5,2), c(a), d;
    d = (a.operator+ (b)); //this is the line36
    cout <<endl<< "Data members of the vector c: ";
    c.Show();
    cout <<endl<< "Data members of the vector d: ";
    d.Show();
    Vector e = b;
    cout <<endl<< "Data members of the vector e: ";
    e.Show();
    Vector f(a);
    cout <<endl<< "Data members of the vector f: ";
    f.Show();

    return 0;
}

And the compiler(g++) says:
teste.cpp: In function `int main()':
teste.cpp:36: error: no matching function for call to
`Vector::Vector(Vector)'
teste.cpp:24: note: candidates are: Vector::Vector(Vector&)


You can't bind a temporary to a reference, you need to make the
parameter to the copy-constructor a const reference:

   Vector::Vector(const Vector& source);

1?)

Exactly. And you have to be careful with copy constructors,
because the compiler is allowed to assume that all they do is
copy; the number of times it actually gets called may vary from
one compiler to the next, or according to the optimization
options.

    So... do you agree that if i want to do any other kind of things
in the instantion of the object , i must provide additional methods
and not use the copy constructor, not even for an innocent y=10 or
y=(source.y)*2 ???


Yes, you should use a normal constructor and pass parameters telling it
what kind of behaviour you expect, perhaps something like

   Vector::Vector(float x, float y);

As the name copy-constructor implies it creates a copy of the object,
even if the compiler did not make any assumptions about it other
programmers will, they expect to get a perfect copy out of it , nothing
else.

2?)

You can't bind a temporary to a non-const reference. With your
original version of the copy constructor, you cannot copy
temporary values, only named variables.
    
I'm not sure if i get this recommendation and i think this is the key
concept that i'm missing, that keeps me confused.
Could please write an example?


I'm not 100% sure on the details, but here's an attempt at explaining:
Your own code above, the on the line 'd = (a.operator+ (b));' operator+
will return a new Vector object, which will then be assigned to d, but
to avoid unnecessary copying the compiler is allowed to optimise away
the assignment and copy-construct the return value directly into d,
instead of first creating a temporary. To do that the copy-constructor
have to be able to create a copy from a temporary object.

--
Erik Wikstr?m

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