Ttyjfyk

I'm trying to understand the below example, but I'm a bit confused from the three different template and struct declarations.

Could you please describe what will happen for the below call? which of the templates will be used and when?

Also why does the first template+class declaration is lacking "<S...>" right after the struct declaration?(see what's commented out)? when it's right to add it and when it's not?

#include <iostream>
#include <stdio.h>
using namespace std;


template<typename... S>
struct Example /* <S...> */ ; 

template<typename H, typename... T>
struct Example<H, T...>

 static const size_t value = sizeof(H) + Example<T...>::value;
;

template<>
struct Example<>

 static const size_t value = 0;
;


int main()
 cout << Example<long, int, char>::value << endl;
 return 0;

Output: 13

The first declares the template of a struct named Example, accepting any number of types:

template<typename... S>
struct Example /* <S...> */ ;

If the name of the newly declared template was followed by <>, with or without arguments, it would be a specialization instead!

The second defines a partial specialization for at least one type-argument:

template<typename H, typename... T>
struct Example<H, T...>

 static const size_t value = sizeof(H) + Example<T...>::value;
;

And the last one defines a full specialization for no type-arguments:

template<>
struct Example<>

 static const size_t value = 0;
;

Take note that template is followed by empty <>-brackets.

It doesn't matter that the partial specialization is defined before the full specialization because instantiation must be deferred until the templates type-arguments are known.

The specific instance you use, Example<long,int,char>::value, depends on Example<int, char>::value, which depends on Example<char>, which leads to the base-case:

Example<long, int, char>::value = sizeof(long) + Example<int, char>::value; // sizeof(long) + sizeof(int) + 1 + 0
Example<int, char>::value = sizeof(int) + Example<char>::value; // sizeof(int) + 1 + 0
Example<char>::value = sizeof(char) + Example<>::value; // 1 + 0
Example<>::value = 0;

Of course, the example could be simplified:

template <class... T>
struct Example 
 static const size_t value = 0;
 static_assert(!sizeof...(T), "The base-template only handles no template arguments.");
;
template <class H, class... T>
struct Example 
 static const size_t value = sizeof(H) + Example<T...>::example;
;

Or with C++17 fold-expressions:

template <class... T>
struct Example 
 static const size_t value = 0 + ... + sizeof(T);
;

As an aside, there are good reasons never to use using namespace std;, I wonder why you #include <stdio.h>, and return 0; is redundant for main().

Only answering this part of your question:

Also why does the first template+class declaration is lacking < S...> right after the struct declaration?(see what's commented out)? when it's right to add it and when it's not?

• 
  

  When you declare a templated function/class/struct/type, you only use the angle bracket< > once, before the declaration:

  
  
  

  template <typename T> 
  void foo(T x);
  

  
  


• 
  

  When you declare a specific instantiation of the general template, you use < > twice, once empty before the declaration, then again with the specific template parameters for which you're instantating:

  
  
  

  template <>
  void foo<int>(int& x);
  

  
  


• 
  

  When you declare a specific specialization of the general template, you use < > twice, once empty before the declaration, then again with the specific template parameters for which you're instantating:

  
  
  

  template 
  void foo<int>(int& x);
  

  
  

More on the last two items (and how they differ):

Difference between instantiation and specialization in c++ templates

Also why does the first template+class declaration is lacking "< S...>" right after the struct declaration?(see what's commented out)? when it's right to add it and when it's not?

It seems to me that is better to start from this point.

First of all, the following (removed the <S...> commented) is a declaration (attention: declaration only, not definition) of a template struct Example that receive a variadic list of type template parameters

template<typename... S>
struct Example; 

You can also avoid to use the S and write simply

template <typename...>
struct Example; 

because the name of the variadic list isn't used in this context.

At this point the compiler know that there is a variadic template struct Example but doesn't know how is made.

Next we add the definition of a specialization of Example that receive one or more template parameter (observe that Example is defined to receive zero or more parameter, so a specialization that receive one or more parameter is special case of Example)

//....... one --> V VVVVV <- or more template parameter
template<typename H, typename... T>
struct Example<H, T...>
 // .........^^^^^^^^^ <- this is a specialization
 static const size_t value = sizeof(H) + Example<T...>::value;
;

The <H, T...> part after Example identifies a specialization (as said).

This specialization define a static const size_t variable initialized with the sum of the sizeof(H) (the sizeof() of the first type template parameter) with the value defined in another Example class: Example<T...>.

So you're observing a recursive definition: value is the sum of the sizeof() of the first parameter (a type) with the sum of the sizeof() of the following types.

Suggestion: if you use variadic templates, you can use also constexpr, so better define value as constexpr

 static constexpr std::size_t value = sizeof(H) + Example<T...>::value;

Or better, you can inherit from std::integral_constant

template <typename H, typename... T>
struct Example <H, T...> 
 : public std::integral_constant<std::size_t, sizeof(H) + Example<T...>>
 ;

so you inherit value from std::integral_constant with additional useful facilities (by example: automatic conversion to std::size_t in a context where a std::size_t is required)

Every recursion needs a ground case, so you have

template<>
struct Example<>

 static const size_t value = 0;
;

the declaration of another specialization of Example; this time the case with exactly zero template parameter (Example<>). In this case you have the definition of a value that is zero to terminate the recursion.

As before, you can define value as constexpr or, better IMHO, using again std::integral_constant

template <>
struct Example<> : public std::integral_constant<std::size_t, 0u>
 ;

Now you have defined two specializations for Example: one for the one-or-more parameters cases, one for the zero-parameters case. So you have covered all cases for Example that is declared receiving zero-or-more parameters; there is no needs to declare the generic (not specialized version) of Example.

As observed by Deduplicator, you can define the generic case and only one specialization: if you write

template <typename...>
struct Example : public std::integral_constant<std::size_t, 0u>
 ;

template <typename T, typename ... Ts>
struct Example<T, Ts...>
 : public std::integral_constant<std::size_t, sizeof(T)+Example<Ts...>>
 ;

you first declare Example receiving zero-or-more parameters and define the generic case with a value zero (the ground case), next you define a one-or-more specialization.

Considering that the compiler select the more specialized version (when more version matches), the compiler select the specialization when there is one-or-more parameters (bot versions match but the specialization is more specialized) and the generic version when there are zero parameters (because the specialization doesn't matches).

This way is a little more synthetic but can be less clear.

Could you please describe what will happen for the below call? which of the templates will be used and when?

Now should be simple to understand.

When you write

Example<long, int, char>::value

you ask for the value of Example<long, int, char>.

Three parameters, so the one-or-more specialization is selected, that is

value = sizeof(long) + Example<int, char>::value;

for the same reason, the value in Example<int, char> is

value = sizeof(int) + Example<char>::value;

and the value in Example<char> is

value = sizeof(char) + Example<>::value;

Now, for Example<>::value, the zero-parameters specialization is selected and Example<>::value is zero.

Concluding, we have that value in Example<long, int, char> is initialized with

 value = sizeof(long) + sizeof(int) + sizeof(char) + 0;

You tagged C++11, so it's a pity you can't use C++17 (template folding) where you can avoid recursion at all and define Example as a using

template <typename ... Ts>
using Example = std::integral_constant<std::size_t, (... + sizeof(Ts))>;