Re: odd behaviour of overloaded * operator

"Jonathan Mcdougall" <>
5 Jun 2006 08:32:10 -0700
<> wrote:

benben wrote: wrote:

 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
  C operator*(double d);

void f(C& c)
  c * 1.0;

This works because the statement becomes


However, this

  1.0 * c

doesn't, because 1)


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
  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
  // ...

  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.


  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).



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