Below is an example of the problem I am having. I don't understand how
I can get the compiler to see deriv &operator=(cons t T &rhs).
I am sure this is a common problem - any suggestions?
#include <iostream>
#include <vector>
struct base
{
virtual ~base()=0 {};
};
template<class T>
struct deriv : public base
{
deriv(): data_(0) {};
deriv(const T &data): data_(data) {};
deriv &operator=(cons t T &rhs) { data_=rhs; return (*this); }
deriv &operator=(cons t deriv &rhs)
{ data_=rhs.data_ ; return (*this); }
T data_;
};
struct container
{
base &operator[](int i) { return *(data_[i]); };
std::vector<bas e *data_;
};
void fill_cont(conta iner &data);
int main(int argc, char *argv[])
{
container data;
fill_cont(data) ;
std::cout << "data[0]=" << typeid(data[0]).name() << std::endl;
// output: data[0]=struct deriv<int>
data[0]=12445;
std::cout << "data[1]=" << typeid(data[1]).name() << std::endl;
// output: data[1]=struct deriv<double>
data[1]=4.5667;
std::cout << "data[2]=" << typeid(data[2]).name() << std::endl;
// output: data[2]=struct deriv<char *>
data[2]="test";
return 0;
}
void fill_cont(conta iner &data)
{
data.data_.push _back(new deriv<int>());
data.data_.push _back(new deriv<double>() );
data.data_.push _back(new deriv<char *>());
}
--
Adrian
Think you know a language? Post to comp.lang... and find out!
10  1789 
Adrian napsal:
Below is an example of the problem I am having. I don't understand how
I can get the compiler to see deriv &operator=(cons t T &rhs).
I am sure this is a common problem - any suggestions?
#include <iostream>
#include <vector>
struct base
{
virtual ~base()=0 {};
};
template<class T>
struct deriv : public base
{
deriv(): data_(0) {};
deriv(const T &data): data_(data) {};
deriv &operator=(cons t T &rhs) { data_=rhs; return (*this); }
deriv &operator=(cons t deriv &rhs)
{ data_=rhs.data_ ; return (*this); }
T data_;
};
struct container
{
base &operator[](int i) { return *(data_[i]); };
std::vector<bas e *data_;
};
void fill_cont(conta iner &data);
int main(int argc, char *argv[])
{
container data;
fill_cont(data) ;
std::cout << "data[0]=" << typeid(data[0]).name() << std::endl;
// output: data[0]=struct deriv<int>
data[0]=12445;
std::cout << "data[1]=" << typeid(data[1]).name() << std::endl;
// output: data[1]=struct deriv<double>
data[1]=4.5667;
std::cout << "data[2]=" << typeid(data[2]).name() << std::endl;
// output: data[2]=struct deriv<char *>
data[2]="test";
return 0;
}
void fill_cont(conta iner &data)
{
data.data_.push _back(new deriv<int>());
data.data_.push _back(new deriv<double>() );
data.data_.push _back(new deriv<char *>());
}
The code is quite long, but for me is strange for example pure virtual
destructor in base class.
Adrian napsal:
Below is an example of the problem I am having. I don't understand how
I can get the compiler to see deriv &operator=(cons t T &rhs).
I am sure this is a common problem - any suggestions?
#include <iostream>
#include <vector>
struct base
{
virtual ~base()=0 {};
};
template<class T>
struct deriv : public base
{
deriv(): data_(0) {};
deriv(const T &data): data_(data) {};
deriv &operator=(cons t T &rhs) { data_=rhs; return (*this); }
deriv &operator=(cons t deriv &rhs)
{ data_=rhs.data_ ; return (*this); }
T data_;
};
struct container
{
base &operator[](int i) { return *(data_[i]); };
std::vector<bas e *data_;
};
void fill_cont(conta iner &data);
int main(int argc, char *argv[])
{
container data;
fill_cont(data) ;
std::cout << "data[0]=" << typeid(data[0]).name() << std::endl;
// output: data[0]=struct deriv<int>
data[0]=12445;
std::cout << "data[1]=" << typeid(data[1]).name() << std::endl;
// output: data[1]=struct deriv<double>
data[1]=4.5667;
std::cout << "data[2]=" << typeid(data[2]).name() << std::endl;
// output: data[2]=struct deriv<char *>
data[2]="test";
return 0;
}
void fill_cont(conta iner &data)
{
data.data_.push _back(new deriv<int>());
data.data_.push _back(new deriv<double>() );
data.data_.push _back(new deriv<char *>());
}
--
Adrian
Think you know a language? Post to comp.lang... and find out!
In container::oper ator[] you are returning reference to base. But there
is no assignment operator in class 'base', so compiler cannot see it.
Adrian wrote:
>
Below is an example of the problem I am having. I don't understand how
I can get the compiler to see deriv &operator=(cons t T &rhs).
The problem is that this function doesn't exist in base, which is the static
type. The compiler will only search the object's static type.
>
I am sure this is a common problem - any suggestions?
Surely. You can cast to whatever type you really have:
dynamic_cast<de riv<int>&>(data[0]) = 12445;
Another "solution" follows.
#include <iostream>
#include <vector>
struct base
{
template <class Tbase& operator=(const T&);
virtual ~base()=0 {};
of course this don't work, but your compiler will tell you that.
};
template<class T>
struct deriv : public base
{
deriv(): data_(0) {};
deriv(const T &data): data_(data) {};
deriv &operator=(cons t T &rhs) { data_=rhs; return (*this); }
deriv &operator=(cons t deriv &rhs)
{ data_=rhs.data_ ; return (*this); }
T data_;
};
template <class T>
base& base::operator= (const T& rhs)
{
return dynamic_cast<de riv<T>&>(*this) = rhs;
}
>
struct container
{
base &operator[](int i) { return *(data_[i]); };
std::vector<bas e *data_;
};
void fill_cont(conta iner &data);
int main(int argc, char *argv[])
{
container data;
fill_cont(data) ;
std::cout << "data[0]=" << typeid(data[0]).name() << std::endl;
// output: data[0]=struct deriv<int>
data[0]=12445;
std::cout << "data[1]=" << typeid(data[1]).name() << std::endl;
// output: data[1]=struct deriv<double>
data[1]=4.5667;
std::cout << "data[2]=" << typeid(data[2]).name() << std::endl;
// output: data[2]=struct deriv<char *>
data[2]="test";
Of course "test" has type const char[5], not char*, so this will not work.
>
return 0;
}
void fill_cont(conta iner &data)
{
data.data_.push _back(new deriv<int>());
data.data_.push _back(new deriv<double>() );
data.data_.push _back(new deriv<char *>());
}
--
Robert Bauck Hamar
Der er to regler for suksess:
1. Fortell aldri alt du vet.
– Roger H. Lincoln
On 2006-12-04 21:07, Adrian wrote:
Below is an example of the problem I am having. I don't understand how
I can get the compiler to see deriv &operator=(cons t T &rhs).
I am sure this is a common problem - any suggestions?
#include <iostream>
#include <vector>
struct base
{
virtual ~base()=0 {};
};
template<class T>
struct deriv : public base
{
deriv(): data_(0) {};
deriv(const T &data): data_(data) {};
deriv &operator=(cons t T &rhs) { data_=rhs; return (*this); }
deriv &operator=(cons t deriv &rhs)
{ data_=rhs.data_ ; return (*this); }
T data_;
};
struct container
{
base &operator[](int i) { return *(data_[i]); };
std::vector<bas e *data_;
};
void fill_cont(conta iner &data);
int main(int argc, char *argv[])
{
container data;
fill_cont(data) ;
std::cout << "data[0]=" << typeid(data[0]).name() << std::endl;
// output: data[0]=struct deriv<int>
data[0]=12445;
std::cout << "data[1]=" << typeid(data[1]).name() << std::endl;
// output: data[1]=struct deriv<double>
data[1]=4.5667;
std::cout << "data[2]=" << typeid(data[2]).name() << std::endl;
// output: data[2]=struct deriv<char *>
data[2]="test";
return 0;
}
void fill_cont(conta iner &data)
{
data.data_.push _back(new deriv<int>());
data.data_.push _back(new deriv<double>() );
data.data_.push _back(new deriv<char *>());
}
I'm quite sure you can't do that. Besides from what Ondra Holub has said
what you are trying to do is to create a heterogeneous container, which
I'm quite sure you can't do. At least not like that, you might succeed
if you first find out what kind of deriv it is and cast it to that type.
But then there would not be much of a point, would it?
--
Erik Wikström
Ondra Holub wrote:
In container::oper ator[] you are returning reference to base. But there
is no assignment operator in class 'base', so compiler cannot see it.
Should be. The compiler should generate one by default.
--
Adrian
Think you know a language? Post to comp.lang... and find out!
Ondra Holub wrote:
The code is quite long, but for me is strange for example pure virtual
destructor in base class.
This is a useful trick for insuring that the class is abstract.
--
Adrian
Think you know a language? Post to comp.lang... and find out!
Robert Bauck Hamar wrote:
Adrian wrote:
The problem is that this function doesn't exist in base, which is
the static
type. The compiler will only search the object's static type.
>I am sure this is a common problem - any suggestions?
Surely. You can cast to whatever type you really have:
dynamic_cast<de riv<int>&>(data[0]) = 12445;
Another "solution" follows.
>#include <iostream>
#include <vector>
struct base
{
template <class Tbase& operator=(const T&);
> virtual ~base()=0 {};
of course this don't work, but your compiler will tell you that.
What wont work. Nothing wrong with a virtual destructor?
template <class T>
base& base::operator= (const T& rhs)
{
return dynamic_cast<de riv<T>&>(*this) = rhs;
}
Perfect. I didnt try that. I assumed the compiler would never be able
to deduce T.
Thanks for that.
>
Of course "test" has type const char[5], not char*, so this will not work.
Was just throwing things into the example :-)
--
Adrian
Think you know a language? Post to comp.lang... and find out!
Adrian wrote:
Robert Bauck Hamar wrote:
>Adrian wrote:
The problem is that this function doesn't exist in base, which is
the static
>type. The compiler will only search the object's static type.
>>I am sure this is a common problem - any suggestions?
Surely. You can cast to whatever type you really have:
dynamic_cast<d eriv<int>&>(dat a[0]) = 12445;
Another "solution" follows.
>>#include <iostream>
#include <vector>
struct base
{
template <class Tbase& operator=(const T&);
>> virtual ~base()=0 {};
of course this don't work, but your compiler will tell you that.
What wont work. Nothing wrong with a virtual destructor?
The dtor _should_, of course, be virtual. But the above line is a syntax
error. You _can_ do:
class foo {
virtual ~foo() = 0;
};
foo::~foo() {}
but not
class foo {
virtual ~foo() = 0 {}
};
--
Robert Bauck Hamar
Der er to regler for suksess:
1. Fortell aldri alt du vet.
- Roger H. Lincoln
Robert Bauck Hamar wrote:
The dtor _should_, of course, be virtual. But the above line is a syntax
error. You _can_ do:
class foo {
virtual ~foo() = 0;
};
foo::~foo() {}
but not
class foo {
virtual ~foo() = 0 {}
};
Gotya. Actually the compiler allows it. I hate it when they dont
follow standards, makes porting a nightmare.
Just found it in the standard. 10.4.2. Makes sense
--
Adrian
Think you know a language? Post to comp.lang... and find out! This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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