Re: Implicit conversion and method call
GeeRay wrote:
Hi all,
how can I create a template class to decorate a type and use it as the
type itself?
For example:
I want to do this:
#include <iostream>
class A
{
public:
A(){};
virtual ~A(){};
void foo(){ std::cout << "foo" << std::endl;};
};
template<class T>
class B
{
public:
B(){};
virtual ~B(){};
operator T(){return instance;};
private:
T instance;
};
int main(int argn, char* argv[])
{
B<A> b();
Drop the parentheses, otherwise you're declaring a function. My answer
assumes that the definition of 'b' is like this:
B<A> b;
b.foo();
}
Is it possible?
No. For the member function calls (like the . you use to access the
'foo' member) the conversions are not considered. You can overload the
member access operator for pointers (pretending that your 'B' class is a
pointer), like so:
template<class T> class B { ...
T* operator->() { return &instance; }
};
, then you could write something like
B<A> b;
b->foo();
which is not necessarily the best syntax, of course...
V
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