Re: [Boost-users] function template argument deduction with std::tuple

On 23/11/2017 01:11, Ireneusz Szcześniak wrote:

However, it bugs me that I'm still not sure whether I've implemented a function-template specialization or a template for function overloading.  You claim that this is an overload:

...

template <typename Graph> auto get_cost(const Label1<Graph> &l) {   return std::get<0>(l); }

Could you please elaborate on how to tell one from the other?  Had it been a specialization, a compiler should accept the explicit template argument:

...

template <typename Graph> auto get_cost(const Label1<Graph> &l) {   return std::get<0>(l); }

but I get (with gcc version 7.2.0):

...

error: non-class, non-variable partial specialization ‘get_cost’ is not allowed get_cost(const Label1<Graph> &l)

So it seems that I was defining overloads (these don't have template arguments), not specializations as I thought.  The C++ Programming Language, 4th edition, the bottom of page 737, says that you can drop the explicit template argument in the definition of a template specialization, if it can be deduced.  So I dropped the explicit template argument, and still considered the definition the specialization.

I have a hard time understanding the above error message, thought.  I understand that "non-class" simply means that it's not a member function, OK.  But what "non-variable partial specialization" could possibly mean?  What is "non-variable" here, and why "partial specialization" when I gave all (i.e., one) template argument.  I consider it a complete specialization albeit dependent on a template parameter Graph.

I suggest reading http://www.gotw.ca/publications/mill17.htm. The short version is that function templates are a bit weird, and if you have anything other than template<> it's an overload, because they don't do partial specialization.