Re: template class instantiate without template parameter, automatic type deduction
Fei Liu wrote:
Victor Bazarov wrote:
Fei Liu wrote:
Hello,
We all know that a template function can automatically deduce its
parameter type and instantiate, e.g.
template <tpyename T>
void func(T a);
func(0.f);
This will cause func<float> to be instantiated. The user does not
have to explicitly call func<float>(0.f);
However this line of thinking is broken when it comes to a template
class constructor function, e.g.
class A{
int x;
};
template <typename T>
class C{
C(const T & t) : t(t) {}
T t;
};
template <>
class C<A> {
C(const A & t) : t(t) {}
A t;
};
int main(){
A a;
C c(a);
}
The above code can't be successfully compiled. One has to name the
type returned from the constructor call to pick up the object. But
what really distinguishes from normal function call is that even
C(a) fails, compiler comlains missing template argument.
Correct. The syntax involves the _type_ name, not a function name.
The problem is
sometimes you want automatic (auto) type deduction that a compiler
can provide but you can't get it for constructor call.
That's not true at all. In your example it's not the constructor
that is a template for which the argument needs to be deduced, it's
the type template that needs to be instantiated. When constructors
are concerned, this is the correct (and working) example:
struct C {
template<class T> C(const T& t) {}
};
Well you see this is not the complete code I presented intially. You
threw away class member t with your convenient modification.
It's not "my convenient modification". It's essential to prove the
point. A class template argument _cannot_ be deduced from anything
when trying to define an object of the class instantiated from the
class template. If it were possible, it would be catch-22. Imagine
you have
template<class T> struct C;
template<> struct C<int> {
C(char);
C(double);
};
template<> struct C<char> {
C(bool);
};
Now, how the hell should the compiler decide what 'T' should be when
you declare an object like so
C a(0);
?
If it
needs to be:
template <typenaem U>
struct C{
template<class T> C(const T& t) : t(t) {}
U t;
};
We are back to step 1, once you have a class member whose type depends
on class template parameter.
No, we're not back to step 1. You're confusing two concepts: defining
a type with which to declare an object and deduction of the function
template argument type from the argument the function is called with.
In this particular case you have two template arguments which are NOT
related at all. Making your compiler deduce the 'T' (the template
argument for the constructor) has NO EFFECT on the need to provide
the template argument for 'C' class template when declaring an object
with it -- you need a _type_, not a template, to declare an object.
struct A {};
int main() {
A a;
C c(a); // compiles just fine
}
What's the current best practice to approach such kind of problem,
i.e. automatic type deduction?
Wrap it in a function template.
Indeed I played with it for a bit, but it does not seem to be that
great, for example I could conjure a function like following to
simulate a constructor:
template <typename T, template <typename > class RT>
RT<T> create(const T& t){
return RT<T>(t);
}
It quickly becomes obvious that RT<T> needs to be given to instantiate
it and one does not gain anything.
Care to supply an example to illustrate your point?
Well, wrapping the in a function will not really help you declare or
define a stand-alone object. In order to do that you still need to
instantiate the template, which requires that you _spell it out_:
C<blah> ...
You can, however, use the function wrapper to provide the mechanism
for instantiating a _temporary_ of your type when using it as part of
an expression. Given
template<class T> class C {
T t;
public:
C(T const& t) : t(t) {}
};
template<class T> C<T> makeC(T const& t) {
return C<T>(t);
}
you *cannot* write
... C(42) ...
and expect C<int> to be instantiated. You still would have to write
... C<int>(42) ...
However, you *can* write
... makeC(42) ...
in which case a temporary of type C<int> _will_ be created (after
C<int> is instantiated). That's what I meant.
V
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