# Re: odd behaviour of overloaded * operator

From:
"Jonathan Mcdougall" <jonathanmcdougall@gmail.com>
Newsgroups:
comp.lang.c++
Date:
5 Jun 2006 08:32:10 -0700
Message-ID:
Cleverbum@hotmail.com wrote:

benben wrote:

Cleverbum@hotmail.com wrote:

Hi,
I've created a class and defined how it should multiply with the other
number classes, but I keep getting errors when using it. Recently I
spotted that the errors came when I used:
double * myClass
as opposed to:
myClass * double
Can someone tell me how I should be overloading my operators to avoid
the above problem?

We don't really know! And how could we? Post some code, buddy!

Sorry, this is how I overloaded it:

Vector operator*(double scalar)
{
Vector result;
result.x = x * scalar;
result.y = y * scalar;
result.z = z * scalar;
return result;
}

That's illegal, but I suspect this is a member function. Next time,
post complete, compilable code.

class C
{
public:
C operator*(double d);
};

void f(C& c)
{
c * 1.0;

This works because the statement becomes

c.operator*(1.0);

However, this

1.0 * c

doesn't, because 1)

1.0.operator*(c);

makes no sense and 2)

operator*(double, const C&);

doesn't exist (remember: a binary operator @ as in x@y can be applied
as a member function x.operator@(y) or as a namespace scope function
operator@(x, y)).

There are three solutions. First, define two operator* at namespace
scope (outside the class):

C operator*(double d, const C& c); // 1.0 * c
C operator*(const C& c, double d); // c * 1.0

These operators may be friends of C if they need to.

The second and third solutions only work if you can [implictly]
construct a C from a double:

class C
{
public:
C(double d);
};

In this case, either define one non member operator* that takes two Cs:

C operator*(const C& c1, const C& c2);

or one member operator* that takes one C:

class C
{
public:
// ...

C operator*(const C& c);
};

The choice between the three solution is not only a matter of taste,
but also depends on the interpretation of the concept of
"encapsulation" or "information hiding". Have fun.

Now,

1.0 * c;

creates a temporary C initialized with 1.0 and both Cs are passed to
operator*. Note that operator*'s parameters must be const references
(or by value, which is "innefficient" however), because that's the only
way you will be able to pass the temporary value (technical: this
temporary is an rvalue and cannot be bound to a non-const reference).

}

Jonathan

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