tslettebo@chello.no.nospam (Terje Sletteb   ) wrote in message news:<JzBEa.10720$KF1.262725@amstwist00>...

> Well, if a template is a constructor (as is the question in the subject
> line), then it makes sense that it doesn't generate a default
> constructor, and it's consistent, I agree. However, why, then, is a
> template constructor not a copy constructor, as Stefan says? Neither is
> a template assignment operator a copy assignment operator, it seems.
>
> Does it have to do with what I also said, that it prefers (implicitly
> generated) functions over function templates?
>



Well, AFAICS, this is what is happening:
1. It sees that there is a template constructor and if it can be
instantiated, it will, and will prohibit the default form being
generated.
2. It checks to see if the user has a copy ctor, and/or an assignment
operator. If it does, nothing is done. If it doesn't, then one is
generated implicitly.
3. Now if there is a template copy ctor and/or a template assignment
operator defined, and the user wants to do something like this:

Eg A:
//foo is the name of a class.
foo<int> f1, f2;
foo<int> f3 = f1;
f2 = f3;

Here, it seems that it *prefers* the non-templates (overloaded) to the
temlates.

Eg B:
foo<int> f1, f2;
foo<double> f3 = f1;
f2 = f3;

Here, obviously, the templates are used.

This is similar to the overload resolution in this case:



template <class t>
void foo (t a)
{
  cout<<"In template"<<endl;
}

void foo (int a)
{
  cout<<"In normal version"<<endl;
}

int main ( )
{
  foo (3.6); //template.
  foo (3);  //normal function.
  return (0);
}

This is exactly what Terje said.

-Dhruv.

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