asked on re-implementing virtual assignment operator of a templated class
Dear all,
I have an issue with re-implementing the assignment operator of
a base class, of which the derived class is templated.
the code:
#include <string>
#include <iostream>
#include <iomanip>
#include <sstream>
#include <exception>
using namespace std;
class AbstractProperty {
public:
virtual string getValue() = 0;
virtual void setValue( string value ) = 0;
virtual AbstractProperty & operator = ( AbstractProperty & rhs ) = 0;
};
template< class T >
class Property : public AbstractProperty {
public:
Property( T value ):
m_value( value ){
}
string getValue(){
stringstream ss;
ss << m_value;
return( ss.str() );
}
void setValue( string value ){
stringstream ss;
ss << value;
ss >> m_value;
}
AbstractProperty & operator = ( AbstractProperty & rhs ){
try{
Property & p = dynamic_cast<Property & >( rhs );
this->m_value = p.m_value;
}
catch( exception & e ){
cout << "eek ! a " << e.what( ) << endl;
}
return( * this );
}
private:
T m_value;
};
void main( void ){
int i1 = 1;
int i2 = 2;
float f = 0.5;
Property<int> Pi1( i1 );
Property<int> Pi2( i2 );
Property<float> Pf( f );
Pi2.setValue( "3" );
Pf.setValue( "0.25" );
cout << "Pi1: " << Pi1.getValue() << endl;
cout << "Pi2: " << Pi2.getValue() << endl;
cout << "Pf: " << Pf.getValue() << endl;
Pi1 = Pf; // works, but with a Bad Dynamic (as expected)
Pi1 = Pi2; // doesn't compile, because
// unresolved external symbol "public: virtual class AbstractProperty
// & __thiscall AbstractProperty::operator
cout << "Pi1: " << Pi1.getValue() << endl;
}
What I *really* don't understand, is why the two Property<int> instances
being assigned results in an unresolved function, shouldnt the two instances
compare as two Property<int> classes, and thus be implemented by the template ?
it's not a matter of the argument to the assignment operator not being constant,
Ive checked.
any input is appreciated,
Jonathan
I have an issue with re-implementing the assignment operator of
a base class, of which the derived class is templated.
the code:
#include <string>
#include <iostream>
#include <iomanip>
#include <sstream>
#include <exception>
using namespace std;
class AbstractProperty {
public:
virtual string getValue() = 0;
virtual void setValue( string value ) = 0;
virtual AbstractProperty & operator = ( AbstractProperty & rhs ) = 0;
};
template< class T >
class Property : public AbstractProperty {
public:
Property( T value ):
m_value( value ){
}
string getValue(){
stringstream ss;
ss << m_value;
return( ss.str() );
}
void setValue( string value ){
stringstream ss;
ss << value;
ss >> m_value;
}
AbstractProperty & operator = ( AbstractProperty & rhs ){
try{
Property & p = dynamic_cast<Property & >( rhs );
this->m_value = p.m_value;
}
catch( exception & e ){
cout << "eek ! a " << e.what( ) << endl;
}
return( * this );
}
private:
T m_value;
};
void main( void ){
int i1 = 1;
int i2 = 2;
float f = 0.5;
Property<int> Pi1( i1 );
Property<int> Pi2( i2 );
Property<float> Pf( f );
Pi2.setValue( "3" );
Pf.setValue( "0.25" );
cout << "Pi1: " << Pi1.getValue() << endl;
cout << "Pi2: " << Pi2.getValue() << endl;
cout << "Pf: " << Pf.getValue() << endl;
Pi1 = Pf; // works, but with a Bad Dynamic (as expected)
Pi1 = Pi2; // doesn't compile, because
// unresolved external symbol "public: virtual class AbstractProperty
// & __thiscall AbstractProperty::operator
cout << "Pi1: " << Pi1.getValue() << endl;
}
What I *really* don't understand, is why the two Property<int> instances
being assigned results in an unresolved function, shouldnt the two instances
compare as two Property<int> classes, and thus be implemented by the template ?
it's not a matter of the argument to the assignment operator not being constant,
Ive checked.
any input is appreciated,
Jonathan
Editors IDEsC++
Last Comment
phoffric
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ASKER
ah. I think I got it, I'm confusing the situation where :
class A{
virtual A & operator = ( A & a ) = 0;
};
class B : public A {
virtual A & operator = ( A & a ){
cout << " here ";
}
}
B * b1 = new B();
B * b2 = new B();
* b1 = * b2;
in this situation, the assignment operator in B is called, even when
b1 and b2 are cast to A *..
J
class A{
virtual A & operator = ( A & a ) = 0;
};
class B : public A {
virtual A & operator = ( A & a ){
cout << " here ";
}
}
B * b1 = new B();
B * b2 = new B();
* b1 = * b2;
in this situation, the assignment operator in B is called, even when
b1 and b2 are cast to A *..
J
>> in this situation, the assignment operator in B is called
In function `B::operator=(B&)':
x.cpp:53: undefined reference to `A::operator=(A&)'
collect2: ld returned 1 exit status
make: *** [xprog] Error 1
In function `B::operator=(B&)':
x.cpp:53: undefined reference to `A::operator=(A&)'
collect2: ld returned 1 exit status
make: *** [xprog] Error 1
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ASKER
@evilrix:
sorry, that was from the top of my mind,
here's what i meant, which builds & runs..
class A{
public:
virtual A & operator = ( A & a ) = 0;
};
class B : public A {
public:
virtual A & operator = ( A & a ){
B & b = static_cast< B & >( a );
cout << " do something b'ish here ";
return( * this );
}
};
void main( void ){
B * b1 = new B();
B * b2 = new B();
* ( A * )b1 = * ( A * )b2;
}
this works, but is it legal ?
@phoffric: thanks, how do you reckon does 2) relate to my example ?
J
sorry, that was from the top of my mind,
here's what i meant, which builds & runs..
class A{
public:
virtual A & operator = ( A & a ) = 0;
};
class B : public A {
public:
virtual A & operator = ( A & a ){
B & b = static_cast< B & >( a );
cout << " do something b'ish here ";
return( * this );
}
};
void main( void ){
B * b1 = new B();
B * b2 = new B();
* ( A * )b1 = * ( A * )b2;
}
this works, but is it legal ?
@phoffric: thanks, how do you reckon does 2) relate to my example ?
J
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@evilrix - LOL - What makes it heavier is that usually when something doesn't compile or build, the examples are there to show just that, and there are comments that indicate the problem. If I were just reading it and not trying the example stated, I would have thought that this would build. Maybe I'm doing something wrong. (I usually like to try out new things to help let it sink in. For some reason, I have never come across this item about trying to inherit the = operator. (Or maybe I saw it and thought "no problem".)
@evilrix - does "just like me" apply to "a former member of the C++ standards council" as well as "a Brit"?
>> does "just like me" apply to "a former member of the C++ standards council" as well as "a Brit"?
Muhahaha -- I wish. No just the Brit part :)
The reason the example code doesn't build is because the function bodies are missing. You'll not thought that even with the function bodies in D it still would build unless you also implement them in B.
Muhahaha -- I wish. No just the Brit part :)
The reason the example code doesn't build is because the function bodies are missing. You'll not thought that even with the function bodies in D it still would build unless you also implement them in B.
struct B {
virtual int operator= (int); // { return 0; }
virtual B& operator= (const B&); // { return *this; };
};
struct D : B {
virtual int operator= (int) { return 0; }
virtual D& operator= (const B&) { return *this; };
};
D dobj1;
D dobj2;
B* bptr = &dobj1;
void f() {
bptr->operator=(99); // calls D::operator=(int)
*bptr = 99; // ditto
bptr->operator=(dobj2); // calls D::operator=(const B&)
*bptr = dobj2; // ditto
dobj1 = dobj2; // calls implicitly-declared
// D::operator=(const D&)
}
// I add main in order to build
int main()
{
}
Ahh so! So that's how you work this C++ Language Spec thingy :)
I was so blind-sided by the question that I didn't even consider this intuitive, oh so clear, obvious answer (thanks to you).
Sorry jochemspek, I didn't mean to go off on a tangent that doesn't help you.
I was so blind-sided by the question that I didn't even consider this intuitive, oh so clear, obvious answer (thanks to you).
Sorry jochemspek, I didn't mean to go off on a tangent that doesn't help you.
why doesnt the compiler resolve the assignment operator to be that of the Property<int> class ?
J