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Geo
Guest
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 Posted: Tue Jul 11, 2006 4:23 pm Post subject: Template members of template class |
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Having spent several hours trying to get the following code to compile
in GCC and failing
namespace test
{
template<class U, class V=U&> class Formatter__
{
private:
U ss;
public:
Formatter__(V out) : ss(out)
{
}
template<typename T> Formatter__& operator<<(const T &item)
{
ss << item;
return *this;
}
template<> Formatter__& operator<<(const char* s)
{
ss << do_stuff(s);
return *this;
}
};
};
I eventually discovered that it was illegal in standard C++.
My question is, Why ?
I'm not to bothered with the fact that the template specialization
should be at namespace scope, but why can't I specialize operator<<
without specializing Formatter__. What problem does preventing this
solve ?
I can find lots of posts saying it is illegal, but none that explains
why ?
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Greg Herlihy
Guest
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| Posted: Wed Jul 12, 2006 4:18 pm Post subject: Re: Template members of template class |
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|
Geo wrote:
| Quote: | Having spent several hours trying to get the following code to compile
in GCC and failing
namespace test
{
template<class U, class V=U&> class Formatter__
{
private:
U ss;
public:
Formatter__(V out) : ss(out)
{
}
template<typename T> Formatter__& operator<<(const T &item)
{
ss << item;
return *this;
}
template<> Formatter__& operator<<(const char* s)
{
ss << do_stuff(s);
return *this;
}
};
};
I eventually discovered that it was illegal in standard C++.
My question is, Why ?
|
A (full) specialization of a template defines a single, unique
instantation of a general template for a specific set of parameterized
types. So in order to specialize a template member function of a class
template in full, the parameterized types for both the class template
and the member function must be supplied in order to arrive at a
single, unambiguous instantiation of the member function template.
| Quote: | I'm not to bothered with the fact that the template specialization
should be at namespace scope, but why can't I specialize operator
without specializing Formatter__. What problem does preventing this
solve ?
|
A better question would be what problem would fully specialized member
functions for unspecialized class templates solve? It is easy to
mistake specializing a function template with overloading a function -
but because specializations of a template function are considered only
after a function has been already selected when resolving a function
call, they are not the same. In fact the existence of a specialization
for a template function does not improve its chances of being called
for a specific set of parameter types - an outcome that is often as
surprising as it is unexpected. Therefore the change to make to the
example code above (to have it work as expected) - would be to overload
the operator<< member function instead of trying to specialize it:
Formatter__& operator<<(const char* s)
{
ss << do_stuff(s);
return *this;
}
Greg
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[ FAQ: http://www.comeaucomputing.com/csc/faq.html ] |
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