Template casting operator

[Makhno]

Hello,
Why does my cast from Vector<class Float> to Vector<float> not work? It
won't compile,

template<class Float> class Vector
{
public:
Vector(Float x1,Float y1,Float z1):x(x1),y(y1) ,z(z1){}
inline Vector<float> operator() () const;

Float x,y,z;
};

template <class Float> inline Vector<float>
Vector<Float>:: operator() () const
{
return Vector<float>(( float)x,(float) y,(float)z);
}
int main()
{
Vector<double> pd(5.6,3.4,2.4) ;

Vector<float> pf=(Vector<floa t>)pd; /* compiler error here */

}

I'd ideally like to be able to cast a Vector<double> to a Vector<float>.

[Makhno]

> You mean with VC7.0? I'm not surprised it doesn't work, but I don't

understand what you mean by 'ignores the cast completely'.

As in 'reacts as though I had never written the cast operator to the Vector
class'.

[Jonathan Turkanis]

"Makhno" <ro**@127.0.0.1 > wrote in message
news:bv******** **@news7.svr.po l.co.uk...

You mean with VC7.0? I'm not surprised it doesn't work, but I don't understand what you mean by 'ignores the cast completely'.
As in 'reacts as though I had never written the cast operator to the

Vector class'.

Yes, that's nasty. Did you try using converting constructors instead?

Jonathan

[Jonathan Turkanis]

"Jonathan Turkanis" <te******@kanga roologic.com> wrote in message:

Yes, that's nasty. Did you try using converting constructors

instead?

You probably need a templated assignment operator too, if you do this.

Jonathan

[Makhno]

> Did you try using converting constructors instead?

Yes, worked straight away. You're right - perhaps I don't need the cast
operator at all.

You probably need a templated assignment operator too, if you do this.

No, as
Vector<float> pf=pd;
and
Vector<float> pf(pd);
are equivalent;

[Jonathan Turkanis]

"Makhno" <ro**@127.0.0.1 > wrote in message
news:bv******** *@newsg4.svr.po l.co.uk...

Did you try using converting constructors instead?
Yes, worked straight away. You're right - perhaps I don't need the

cast operator at all.

You probably need a templated assignment operator too, if you do

this.
No, as
Vector<float> pf=pd;
and
Vector<float> pf(pd);
are equivalent;

But that's just one use of the implicit conversion.

How about:

Vector<double> pd;
Vector<float> pf;
// More stuff here.
pf = pd;

?

I didn't suggest a templated assignement operator first because your
example didn't require it. But if you want a general-purpose
substitute for the conversion operator, you need it.

Jonathan

[Jonathan Turkanis]

"Jonathan Turkanis" <te******@kanga roologic.com> wrote in message
news:bv******** ****@ID-216073.news.uni-berlin.de...

"Makhno" <ro**@127.0.0.1 > wrote in message
news:bv******** *@newsg4.svr.po l.co.uk...

Did you try using converting constructors instead?
Yes, worked straight away. You're right - perhaps I don't need the

cast

operator at all.

You probably need a templated assignment operator too, if you do

this.

No, as
Vector<float> pf=pd;
and
Vector<float> pf(pd);
are equivalent;

I didn't suggest a templated assignement operator first because your
example didn't require it. But if you want a general-purpose
substitute for the conversion operator, you need it.

I spoke too soon. It's non needed here either. But it's not because
of copy initialization.

Jonathan

[Makhno]

> But that's just one use of the implicit conversion.

How about:

Vector<double> pd;
Vector<float> pf;
// More stuff here.
pf = pd;

?

Hmm, this still seems to work, how peculiar. Vector<> doesn't have an
assignment operator.

[Makhno]

> I spoke too soon. It's non needed here either. But it's not because

of copy initialization.

why then?

[Victor Bazarov]

"Jonathan Turkanis" <te******@kanga roologic.com> wrote...

"Makhno" <ro**@127.0.0.1 > wrote in message
news:bv******** *@newsg4.svr.po l.co.uk...

Did you try using converting constructors instead?

Yes, worked straight away. You're right - perhaps I don't need the

cast

operator at all.

You probably need a templated assignment operator too, if you do

this.

No, as
Vector<float> pf=pd;
and
Vector<float> pf(pd);
are equivalent;

But that's just one use of the implicit conversion.

How about:

Vector<double> pd;
Vector<float> pf;
// More stuff here.
pf = pd;

?

I didn't suggest a templated assignement operator first because your
example didn't require it. But if you want a general-purpose
substitute for the conversion operator, you need it.

Why? 'pf = pd' expression will result into the use of compiler-
generated assignment operator and a construction of a temporary
on the right side, no? It should be equivalent to

{ Vector<float> temp(pd); pf = temp; }

No special assignment operator is needed here.

Correct me if I am wrong.

V

[Jonathan Turkanis]

"Victor Bazarov" <v.********@com Acast.net> wrote in message
news:37yUb.1814 15$Rc4.1339653@ attbi_s54...

"Jonathan Turkanis" <te******@kanga roologic.com> wrote...

Why? 'pf = pd' expression will result into the use of compiler-
generated assignment operator and a construction of a temporary
on the right side, no? It should be equivalent to

{ Vector<float> temp(pd); pf = temp; }

No special assignment operator is needed here.

Correct me if I am wrong.

You're not -- I was :(.

Jonathan