doubts on operator

[alessio211734]

I have a class that defines a generic point where p={a1..an}

template<class T,int iclass Point
{
public:
T v[i];

Point<T,i>()
{
for (int j=0;j<i;j++) v[j]=0;
}


inline T & operator [] ( const int j ){
assert(j>=0 && j<i);
return v[j];
}
inline const T & operator [] ( const int j ) const {
assert(j>=0 && j<i);
return v[j];
}


inline Point<T,i>& operator=(const Point<T,i& num)
{
for (int j=0;j<i;j++) v[j]=num.v[j];
return *this;

}


inline Point<T,i>& operator=(const float* pnum)
{
for (int j=0;j<i;j++) v[j]=pnum[j];
return *this;

}


inline Point<T,ioperat or + ( Point<T,iconst & num) const {
Point<T,ivect;
for (int j=0;j<i;j++) vect.v[j]=v[j]+num.v[j];
return vect;
}
inline Point<T,ioperat or - ( Point<T,iconst & num) const {
Point<T,ivect;
for (int j=0;j<i;j++) vect.v[j]=v[j]-num.v[j];
return vect;
}
inline Point<T,ioperat or * ( const T s ) const {
Point<T,ivect;
for (int j=0;j<i;j++) vect.v[j]=v[j]*s;
return vect;
}

inline Point<T,i>& operator *= ( const T s ){
for (int j=0;j<i;j++) v[j]=v[j]*s;
return *this;
}
}


I define a new class Point2

template<class Tclass Point2: public Point<T,2>
{
public:
Point2<T>()
{
v[0]=0;
v[1]=1;
}

Point2<T>(const T& a, const T& b)
{
v[0]=a;
v[1]=b;
}

T x() { return v[0] };
T y() { return v[1] };

};

and in main program I run:

Point2f a;
Point2f b;
Point2f c;
c=a+b;

I get this error:

error C2679: binary '=' : no operator found which takes a right-hand
operand of type 'Point<T,i>' (or there is no acceptable conversion)
with
[
T=float,
i=2
]
could be 'Point2<T&Point 2<T>::operator =(const Point2<T>
&)'
with
[
T=float
]
while trying to match the argument list '(Point2f,
Point<T,i>)'
with
[
T=float,
i=2
]


Why Point2 don't call the operator= defined in Point<i,T>?

Point2 is a Point<i,Tbecaus e derive by Point<i,T>, I don't
understand...


Please explain me.

[Victor Bazarov]

Re: doubts on operator

alessio211734 wrote:

I have a class that defines a generic point where p={a1..an}
>
template<class T,int iclass Point
{
public:
T v[i];

Public? Really?

>
Point<T,i>()

First off, you don't need to repeat "<T,i>" *inside* the template class
definition.

{
for (int j=0;j<i;j++) v[j]=0;
}
>
>
inline T & operator [] ( const int j ){
assert(j>=0 && j<i);
return v[j];
}
inline const T & operator [] ( const int j ) const {
assert(j>=0 && j<i);
return v[j];
}
>
>
inline Point<T,i>& operator=(const Point<T,i& num)
{
for (int j=0;j<i;j++) v[j]=num.v[j];
return *this;
>
}
>
>
inline Point<T,i>& operator=(const float* pnum)
{
for (int j=0;j<i;j++) v[j]=pnum[j];
return *this;
>
}
>
>
inline Point<T,ioperat or + ( Point<T,iconst & num) const {
Point<T,ivect;
for (int j=0;j<i;j++) vect.v[j]=v[j]+num.v[j];
return vect;
}
inline Point<T,ioperat or - ( Point<T,iconst & num) const {
Point<T,ivect;
for (int j=0;j<i;j++) vect.v[j]=v[j]-num.v[j];
return vect;
}
inline Point<T,ioperat or * ( const T s ) const {
Point<T,ivect;
for (int j=0;j<i;j++) vect.v[j]=v[j]*s;
return vect;
}
>
inline Point<T,i>& operator *= ( const T s ){
for (int j=0;j<i;j++) v[j]=v[j]*s;
return *this;
}
}
>
>
I define a new class Point2
>
template<class Tclass Point2: public Point<T,2>
{
public:
Point2<T>()
{
v[0]=0;
v[1]=1;
}
>
Point2<T>(const T& a, const T& b)
{
v[0]=a;
v[1]=b;
}
>
T x() { return v[0] };
T y() { return v[1] };
>
};
>
and in main program I run:
>
Point2f a;

So, my guess would be that 'Point2f' is a typedef, something like

typedef Point2<floatPoi nt2f;

Next time, please, be verbose about those things.

Point2f b;
Point2f c;
c=a+b;
>
I get this error:
>
error C2679: binary '=' : no operator found which takes a right-hand
operand of type 'Point<T,i>' (or there is no acceptable conversion)
[..]
Why Point2 don't call the operator= defined in Point<i,T>?

Because Point2 has its *own* operator=, which *hides* any that it can
inherit from its base class.

>
Point2 is a Point<i,Tbecaus e derive by Point<i,T>, I don't
understand...

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[alessio211734]

Re: doubts on operator

Because Point2 has its *own* operator=, which *hides* any that it can

inherit from its base class.

What's the operator= that Point2 define?
Point2 define a default operator?
What's the advantages of hereditance if I should re-define an operator
for Point2 again?
Can't use operator= and other operators defined in the base class
without re-define in the derived class Point2?

[Victor Bazarov]

Re: doubts on operator

alessio211734 wrote:

>Because Point2 has its *own* operator=, which *hides* any that it can
>inherit from its base class.

>
What's the operator= that Point2 define?

It's called "the default copy-assignment operator", provided by the
compiler. It has the following signature:

Point2& Point2::operato r=(Point2 const&);

Point2 define a default operator?

Yes.

What's the advantages of hereditance if I should re-define an operator
for Point2 again?

"Hereditanc e"? You don't have to redefine the assignment operator, you
just need to make it visible. See your book about the " 'using'
declaration".

Can't use operator= and other operators defined in the base class
without re-define in the derived class Point2?

Can.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Markus Moll]

Re: doubts on operator

Hi

alessio211734 wrote:

>Because Point2 has its *own* operator=, which *hides* any that it can
>inherit from its base class.

>
What's the operator= that Point2 define?
Point2 define a default operator?
What's the advantages of hereditance if I should re-define an operator
for Point2 again?
Can't use operator= and other operators defined in the base class
without re-define in the derived class Point2?

An assignment operator for a base class is rarely useful. Consider:
A a;
B b;
a = b;

Without additional knowledge (i.e. without a user-defined assignment
operator), this only makes sense if b contains all the parts of an A,
which is the case if B is equal to or derived from A. If A is derived from
B, on the other hand, then how would you fill in the missing data?

Markus

[alessio211734]

Re: doubts on operator

On 1 Ago, 16:36, Markus Moll <markus.m...@es at.kuleuven.ac. bewrote:

Hi
>
alessio211734 wrote:

Because Point2 has its *own* operator=, which *hides* any that it can
inherit from its base class.

>

What's the operator= that Point2 define?
Point2 define a default operator?
What's the advantages of hereditance if I should re-define an operator
for Point2 again?
Can't use operator= and other operators defined in the base class
without re-define in the derived class Point2?

>
An assignment operator for a base class is rarely useful. Consider:
A a;
B b;
a = b;
>
Without additional knowledge (i.e. without a user-defined assignment
operator), this only makes sense if b contains all the parts of an A,
which is the case if B is equal to or derived from A. If A is derived from
B, on the other hand, then how would you fill in the missing data?
>
Markus

I would like define constructors for n-dimensional point.
If I include in the base class these constructors my application work,
but I think that this is not safe because
I can call the constructor with three parameters when I use
Point<float,2fo r example.


Point<T,i>(cons t T& a, const T& b)
{
assert(i==2);
v[0]=a;
v[1]=b;
}

Point<T,i>(cons t T& a, const T& b,const T& c)
{
assert(i==3);
v[0]=a;
v[1]=b;
v[2]=c;
}

I think to delete constructors in base classes and extends the base
class with
for example Point2 derived by Point and in this class define the
constructor with 2 parameters.

What's syntax should use in Point2 for utilize Point operators?

[Ian Collins]

Re: doubts on operator

alessio211734 wrote:

I have a class that defines a generic point where p={a1..an}
>
template<class T,int iclass Point
{
public:
T v[i];
>
Point<T,i>()
{
for (int j=0;j<i;j++) v[j]=0;
}
>
>
inline T & operator [] ( const int j ){

In addition to Victor's comments, the 'inline' is superfluous here. All
member functions defined in the class declaration are inline.

--
Ian Collins.