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Overloading 'casting' operator

Hi all,

I have a templated Vector3D class which holds (x,y,z) components as the
specified type. I quite often wish to cast a Vector3D holding ints into
a Vector3D holding floats and vice versa. Like so:

Vector3D<int> intVec(10,20,30 );
Vector3D<float> floatVec = intVec;

Of course this doesn't work. I would be happy if instead the following
worked:
Vector3D<int> intVec(10,20,30 );
Vector3D<float> floatVec = static_cast< Vector3D<float> >(intVec);

but of course that doesn't either. I have read online that it is not
possible to overload the static_cast operator (why, incidently?) but
what is the best approach to this problem? Ideally I would like the
first example to compile but generate a warning and the second example
to be fine (as works for built-in types).

Any thoughts appriciated,

David
Jan 17 '06 #1
8 9164
For the first one, I think you can overload = by
template <class T>
class Vector3D{
.....
template <class N> operator=(const Vector3D<N>& rvalue)
{....}
.....
};
my first reply may not be correct.

I don't know the answer to the second one though.
David Williams wrote:
Hi all,

I have a templated Vector3D class which holds (x,y,z) components as the
specified type. I quite often wish to cast a Vector3D holding ints into
a Vector3D holding floats and vice versa. Like so:

Vector3D<int> intVec(10,20,30 );
Vector3D<float> floatVec = intVec;

Of course this doesn't work. I would be happy if instead the following
worked:
Vector3D<int> intVec(10,20,30 );
Vector3D<float> floatVec = static_cast< Vector3D<float> >(intVec);

but of course that doesn't either. I have read online that it is not
possible to overload the static_cast operator (why, incidently?) but
what is the best approach to this problem? Ideally I would like the
first example to compile but generate a warning and the second example
to be fine (as works for built-in types).

Any thoughts appriciated,

David


Jan 17 '06 #2
David Williams wrote:
I have a templated Vector3D class which holds (x,y,z) components as the
specified type. I quite often wish to cast a Vector3D holding ints into
a Vector3D holding floats and vice versa. Like so:

Vector3D<int> intVec(10,20,30 );
Vector3D<float> floatVec = intVec;

Of course this doesn't work. I would be happy if instead the following
worked:
Vector3D<int> intVec(10,20,30 );
Vector3D<float> floatVec = static_cast< Vector3D<float> >(intVec);

but of course that doesn't either. I have read online that it is not
possible to overload the static_cast operator (why, incidently?) but
what is the best approach to this problem? Ideally I would like the
first example to compile but generate a warning and the second example
to be fine (as works for built-in types).


Without code showing the definition of template<typena me T> Vector3D,
it's difficult to respond. See the FAQ:

http://www.parashift.com/c++-faq-lit...t.html#faq-5.8

As another poster said, a templated operator=() may give you what you
want for the first issue. For the second issue, although you can't
overload the static_cast operator directly, you could make the
static_cast operator work the way you want by providing a templated
copy constructor - something like (VERY untested code):

template<typena me T> class Vector3D {
public:
template<typena me U>
explicit Vector3D(const Vector3D<U>&);
};

Best regards,

Tom

Jan 17 '06 #3
David Williams wrote:
I have a templated Vector3D class which holds (x,y,z) components as the
specified type. I quite often wish to cast a Vector3D holding ints into
a Vector3D holding floats and vice versa. Like so:

Vector3D<int> intVec(10,20,30 );
Vector3D<float> floatVec = intVec;
What you need here is not, as other have suggested, an overloaded
operator=. This is not assignment, but initialization -- i.e.,
construction. For construction, we use constructors. If you define a
single-argument constructor which takes another Vector3D of a different
type, you can do what you want here. It's important to recognize the
difference, though, between the above, assignment, and casting.

Initialization (construction) creates a new instance, based in some way
on the parameter(s) provided to the constructor.

Assignment modifies an existing instance, based in some way on the
parameter provided to operator=.

Casting is more complicated -- it depends on whether you're casting by
value, pointer, or reference. What's more, it is possible to provide
an overloaded "conversion operator" (not casting operator) which allows
implicit (or explicit, if that keyword is used) conversion to another
type. A conversion operator is sort of the inverse of a constructor.
Of course this doesn't work. I would be happy if instead the following
worked:

Vector3D<int> intVec(10,20,30 );
Vector3D<float> floatVec = static_cast< Vector3D<float> >(intVec);


You should not be satisfied with this. Implement an appropriate
single-argument constructor and you'll have a better result.

Luke

Jan 17 '06 #4
Luke Meyers wrote:
David Williams wrote:
I have a templated Vector3D class which holds (x,y,z) components as the
specified type. I quite often wish to cast a Vector3D holding ints into
a Vector3D holding floats and vice versa. Like so:

Vector3D<in t> intVec(10,20,30 );
Vector3D<floa t> floatVec = intVec;

What you need here is not, as other have suggested, an overloaded
operator=. This is not assignment, but initialization -- i.e.,
construction. For construction, we use constructors. If you define a
single-argument constructor which takes another Vector3D of a different
type, you can do what you want here. It's important to recognize the
difference, though, between the above, assignment, and casting.


Ah yes, I can see that now. However, could it be argued (according to
the 'Big Three' rule) that if I have a custom constructor I should have
a custom assignment operator as well?

Initialization (construction) creates a new instance, based in some way
on the parameter(s) provided to the constructor.
Great, this now seems to work:

template <typename Type>
template <typename CastType>
Vector3D<Type>: :Vector3D(const Vector3D<CastTy pe>& vectorToSet)
{
m_tX = vectorToSet.x() ;
m_tY = vectorToSet.y() ;
m_tZ = vectorToSet.z() ;
}

Assignment modifies an existing instance, based in some way on the
parameter provided to operator=.

Casting is more complicated -- it depends on whether you're casting by
value, pointer, or reference. What's more, it is possible to provide
an overloaded "conversion operator" (not casting operator) which allows
implicit (or explicit, if that keyword is used) conversion to another
type. A conversion operator is sort of the inverse of a constructor.
I found the following code and modified it for my class. Is this the
'conversion operator' you are refering to?

template <typename Type>
template <typename CastType>
Vector3D<Type>: :operator Vector3D<CastTy pe> () throw()
{
return Vector3D<CastTy pe>(m_tX,m_tY,m _tZ);
}

I was confused by the fact that the return type (Vector3D<CastT ype>)
appears to come between 'operator' and the empty brackets. But it's ok?

Of course this doesn't work. I would be happy if instead the following
worked:

Vector3D<in t> intVec(10,20,30 );
Vector3D<floa t> floatVec = static_cast< Vector3D<float> >(intVec);

Come to think of it, I could write a vector_cast function to do the
same. However, given what I now know I don't think it's necessary.

You should not be satisfied with this. Implement an appropriate
single-argument constructor and you'll have a better result.

Luke


Thanks for your help!
Jan 17 '06 #5
David Williams wrote:
Ah yes, I can see that now. However, could it be argued (according to
the 'Big Three' rule) that if I have a custom constructor I should have
a custom assignment operator as well?


No. The 'Big Three' in the rule are specifically the copy constructor,
assignment operator and destructor. The functions that the compiler
will generate for you if your don't define them yourself. If you need
to write your own version of any one of these three, you probably need
to write them all yourself because the compiler generated ones probably
will not do what you want. There is nothing in the 'rule' about any
other constructors you may write.

Gavin Deane

Jan 17 '06 #6
Luke Meyers wrote:
What you need here is not, as other have suggested, an overloaded
operator=. This is not assignment, but initialization -- i.e.,
construction. For construction, we use constructors. If you define a
single-argument constructor which takes another Vector3D of a different
type, you can do what you want here. It's important to recognize the
difference, though, between the above, assignment, and casting.


I agree... but didn't I suggest exactly that?

Best regards,

Tom

Jan 17 '06 #7
Gavin Deane wrote:
David Williams wrote:
Ah yes, I can see that now. However, could it be argued (according to
the 'Big Three' rule) that if I have a custom constructor I should have
a custom assignment operator as well?

No. The 'Big Three' in the rule are specifically the copy constructor,
assignment operator and destructor. The functions that the compiler
will generate for you if your don't define them yourself. If you need
to write your own version of any one of these three, you probably need
to write them all yourself because the compiler generated ones probably
will not do what you want. There is nothing in the 'rule' about any
other constructors you may write.

Gavin Deane


Ok, thanks.
Jan 17 '06 #8
Yeah, you're right. It's copy constructor instead of assignment
operator, thanks for pointing out.

Jan 18 '06 #9

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