Re: CRTP question
fabio.lombardelli@gmail.com writes:
Hi all,
can somebody explain me why this piece of code doesn't compile?
In particular, why the compiler (on Base<Derived> class) can see _fun,
and why it can't see _DType?
template <typename D>
struct Base
{
typedef typename D::_DType DType;
void fun()
{
static_cast<D*>(this)->_fun();
}
};
struct Derived : public Base<Derived>
{
typedef int _DType;
void _fun() {}
};
Maybe Base<Derived> class knows Derived class just as a forward
declaration.
Exactly.
But if it is so, why it can see the method _fun?
Declarations in a class are instantiated all at once, but definitions
(e.g. the bodies of member functions) are instantiated only when they
are used. You can rewrite the example as:
template <typename D>
struct Base
{
// uncomment for error
// typedef typename D::_DType DType;
void fun();
};
struct Derived : public Base<Derived>
{
typedef int _DType;
void _fun() {}
};
template <typename D>
void Base<D>::fun()
{
static_cast<D*>(this)->_fun();
}
Which should make it pretty clear why fun parses OK. Now you can
imagine transforming fun into a non-member, just to make it extra
clear that it is instantiated later.
template <typename D>
void fun(Base<D>* p)
{
static_cast<D*>(p)->_fun();
}
Regards,
--
Dave Abrahams
Boost Consulting
www.boost-consulting.com
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