Partial specializations that does not specialize any template arguments although are still accepted.

[piotrek...@gmail.com]

Hi All,

Let's consider the following code:

template <typename T, int N1, int N2>
class myclass
{
public:
  void f() { std::cout << "not specialized, N1 = " << N1 << ", N2 = " << N2 << "\n"; }
};

int main()
{
  myclass<double,10,20> mc;
  mc.f();
}

The output of the above program is obviously:

not specialized, N1 = 10, N2 = 20

Adding the following specialization will cause compilation error:

template <typename T, int N1, int N2>
class myclass<T,N1,N2>
{
public:
  void f() { std::cout << "specialized, N1 = " << N1 << ", N2 = " << N2 << "\n"; }
};

the error message says:

partial specialization ‘myclass<T, N1, N2>’ does not specialize any template arguments

So far everything is clear. But when I change the order of the parameters in the parameter list, then unexpectedly the code starts to compile:

template <typename T, int N1, int N2>
class myclass
{
public:
  void f() { std::cout << "not specialized, N1 = " << N1 << ", N2 = " << N2 << "\n"; }
};

template <int N1, int N2, typename T>
class myclass<T,N1,N2>
{
public:
  void f() { std::cout << "specialized, N1 = " << N1 << ", N2 = " << N2 << "\n"; }
};

int main()
{
  myclass<double,10,20> mc;
  mc.f();
}

The above code produces:

specialized, N1 = 10, N2 = 20

The result is the same for gcc 4.8.1 and clang 3.3.

Why the above code compile even though the partial specialization doesn't specialize any template arguments?

[Robert Haberlach]

AFAICS, that code is ill-formed with a diagnostic required by [temp.class.spec]/(8.2). Perhaps implementations don't actually check that the partial
specialization is more specialized, but use a heuristic that checks whether the declared template parameters are simply enumerated in order in the
argument list?

On 10/13/2016 01:10 PM, piotrek...@gmail.com wrote:
> Hi All,
>
> Let's consider the following code:
>
> |
> template<typenameT,intN1,intN2>
> classmyclass
> {
> public:

> voidf(){std::cout <<"not specialized, N1 = "<<N1 <<", N2 = "<<N2 <<"\n";}

> };
>
> intmain()
> {
> myclass<double,10,20>mc;
> mc.f();
> }
> |
>
> The output of the above program is obviously:
>
> not specialized, N1 = 10, N2 = 20
>
> Adding the following specialization will cause compilation error:
>
> |
> template<typenameT,intN1,intN2>
> classmyclass<T,N1,N2>
> {
> public:

> voidf(){std::cout <<"specialized, N1 = "<<N1 <<", N2 = "<<N2 <<"\n";}

> };
> |
>
> the error message says:
>
> partial specialization ‘myclass<T, N1, N2>’ does not specialize any template arguments
>
> So far everything is clear. But when I change the order of the parameters in the parameter list, then unexpectedly the code starts to compile:
>
> |
> template<typenameT,intN1,intN2>
> classmyclass
> {
> public:

> voidf(){std::cout <<"not specialized, N1 = "<<N1 <<", N2 = "<<N2 <<"\n";}

> };
>
> template<intN1,intN2,typenameT>
> classmyclass<T,N1,N2>
> {
> public:

> voidf(){std::cout <<"specialized, N1 = "<<N1 <<", N2 = "<<N2 <<"\n";}
> };
>
> intmain()
> {

> myclass<double,10,20>mc;
> mc.f();
> }
> |
>
> The above code produces:
>
> specialized, N1 = 10, N2 = 20
>
> The result is the same for gcc 4.8.1 and clang 3.3.
>
> Why the above code compile even though the partial specialization doesn't specialize any template arguments?
>

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