I know that a C++ compiler picks a template specialization in preference to the primary template:
template<class T, class Enable = void>
class A {}; // primary template
template<class T>
class A<T, std::enable_if_t<std::is_floating_point_v<T>, void>> {
}; // specialization for floating point types
However, I don't understand why the selection fails when a type other than void is used as argument to enable_if:
template<class T>
class A<T, std::enable_if_t<std::is_floating_point_v<T>, int>> {
}; // specialization for floating point types
The compiler would surely "see" class A<T, int>.
So it certainly can't be SFINAE, nevertheless the primary template is preferred to the specialization.
This question arises from a context where a custom type detection machinery could be used instead of enable_if, e.g. an SFINAE friendly type extractor like common_type.
When you have
A<some_floating_point_type> some_name;
The template parameters are some_floating_point_type and the implicit void. When the compiler instantiates
template<class T>
class A<T, std::enable_if_t<std::is_floating_point_v<T>, int>> {
}; // specialization for floating point types
it would get A<some_floating_point_type, int>, which does not match the A<some_floating_point_type, void> type that some_name has. Because of that, the specialization is ignored and you get the primary template. You can verify this by trying to create a A<some_floating_point_type, int> and you'll get that the specialization was picked.
I find it helpful to think of a specialization as an alternate recipe. First the template arguments are deduced, and then if they match any of the specializations, then we switch to using that alternate recipe. If not, then the original recipe is used.
Specializations are irrelevant until the compiler knows which types it is going to use for the primary template.
When you write A<double>, then the compiler looks only at the primary template and sees that you actually mean A<double,void>.
And only then it is looking for specializations. Now, when your specialization is for A<double,int>, then it is not suitable because you asked for A<double,void>.
In simple terms, this
template<class T, class Enable = void>
class A {};
says "A is a template, it has two arguments, the second has a default that is void". Thats the primary template. When you explicitly provide only one parameter then the second argument is void.
Now the specialization:
template<class T>
class A<T, std::enable_if_t<std::is_floating_point_v<T>, int>> {
};
This doesn't change the above. A is still a template with 2 arguments and when the second one is not specified then it is void. In simple terms it either means "substituting T in std::enable_if_t<std::is_floating_point_v<T> is a failure, because the type alias does not exist" when the condition is false, SFINAE kicks in and the specialization is ignored. When the condition is true: "Whenever A is instantiated with arguments T (some type) and int then use this definition".
When you instantiate A<int> then the second template argument is void. The specialization does not change that. That is A<int> is actually A<int,void>. It does not match the specialiazation.
Related
In the C++14 draft standard, [temp.param]/11 says:
If a template-parameter of a primary class template or alias template
is a template parameter pack, it shall be the last template-parameter.
If you try compiling the following template, then the compiler will complain.
template< typename ...Args, void(*f)(Args...) > // ERROR
struct Bar
{};
But how does it work in this case?
template< typename F, F >
struct Bar;
template< typename ...Args, void(*f)(Args...) > // OK ???
struct Bar< void(*)(Args...), f >
{};
I can see that it has something to do with it being part of the specialization class template, but why?
The rule clearly states that it applies to a primary class template. Does this mean that the rules change for specializations?
I tried to search for this in the standard, but couldn't find anything. Can you please shine some light into this.
The rule clearly states that it applies to a primary class template. Does this mean that the rules change for specializations?
Yes. Quite simply because a specialization is not a primary class template. So if the wording was intended to apply to all template declarations, it would say so. Instead, the rule is very much intended to only apply to the primary class template (... and alias templates, which cannot be specialized). There is no such restriction for specializations.
This is fundamentally because it is not possible to provide any template arguments after a template parameter pack in the primary template, but it is definitely possible to do so in specializations. For instance, here's one way to concatenate two tuple specializations:
template <typename T, typename U>
struct tuple_concat;
template <typename... Ts, typename... Us> // <== parameter pack *after* parameter pack
struct tuple_concat<tuple<Ts...>, tuple<Us...>> {
using type = tuple<Ts..., Us...>;
};
This is fine, it works, it's useful. But there's no benefit† from being able to write stuff like this in a primary class/variable/alias template - so it's forbidden for simplicity.
†As with all things C++, there is of course a footnote. You could have been able to provide a trailing defaulted template parameter that is used to trigger a substitution failure. But there are other ways you can go about solving that problem, and then we'll have Concepts soon anyway.
I've been trying more about multi threaded programming in c++, and i was having difficulty understanding std::promise so i began searching for answers on this website, and low and behold, there is somebody with the same question as me. But reading the answer made me even more confused
this is the code in the answer that presumably is a similar implementation of std::packaged_task
template <typename> class my_task;
template <typename R, typename ...Args>
class my_task<R(Args...)>
{
std::function<R(Args...)> fn;
std::promise<R> pr; // the promise of the result
public:
template <typename ...Ts>
explicit my_task(Ts &&... ts) : fn(std::forward<Ts>(ts)...) { }
template <typename ...Ts>
void operator()(Ts &&... ts)
{
pr.set_value(fn(std::forward<Ts>(ts)...)); // fulfill the promise
}
std::future<R> get_future() { return pr.get_future(); }
// disable copy, default move
};
in this code,
1- what does this syntax mean template <typename R, typename ...Args> class my_task<R(Args...)>, more specifically, what is the purpose of <R(Args...)> ?
2- why is there a foroward decleration for the class?
thanks
There was some brief discussion in the comments how 1 and 2 should be two separate questions, but I believe that they both are just two sides to the same exact question, for the following reasons:
template <typename> class my_task;
template <typename R, typename ...Args>
class my_task<R(Args...)>; ....
The first statement declares a template that takes a typename as its sole template parameter. The second statement declares a specialization for that template class.
In this context:
R(Args...)
Will specialize for any typename that matches a function. This template specialization will match any template instantiation that passes a function signature for a typename. Barring any problems within the template itself, this template specialization will be used for:
my_task<int (const char *)>
or, a function that takes a const char * parameter and returns an int. The template specialization will also match:
my_task<Tptr *(Tptr **, int)>
or, a function that takes two parameters, Tptr ** and an int, and returns a Tptr * (here, Tptr is some other class).
The template specialization will NOT match:
my_task<int>
Or
my_task<char *>
Because they are not function signatures. If you try to instantiate this template using a non-function typename you're going to get a compilation error. Why?
Well, that's because the template is not defined:
template<typename> class my_task;
Don't think of this as just a forward declaration. it's a forward declaration of a template that takes a template parameter, and the template will not be defined anywhere. Rather, the template declaration allows for a subsequent template specialization declaration, that will match only specific types passed as a template parameter.
This is a common programming technique for restricting the kinds of typenames or classes that can be used with a particular template. Instead of allowing a template to be used with just any typename or class, the template can only be used with some subset. In this case, a function typename, or signature.
It also makes it easier for the template itself to explicitly reference -- in this case -- to the template parameter's return type, and the parameter types. If the template has just a bland, single typename as a template parameter, it can't easily access the function's return type, or the function parameter's types.
1: What does this syntax mean template <typename R, typename ...Args> class my_task<R(Args...)>
This is a specialization of the class template my_task. The <R(Args...)> after the name means it is specialized for that type, and that type is a function. R(Args...) is the type of a function taking Args parameters and returning R. So, my_task<void()> mt; for example would make Args be an empty parameter pack, and R would be void.
2: Why is there a forward declaration for the class?
The class is declared, but unlike an ordinary forward declaration, the un-specialized version isn't defined. This class is only intended to work when the type is a function, so if someone tries to use something that isn't a function (like my_task<int>), it will give an error about the type being undefined.
my_task<void*(int, int)> mt1; //R = void*, Args = int, int
my_task<int> mt2; //error: use of undefined class
In the answer to this post "(Partially) specializing a non-type template parameter of dependent type", it states:
The type of a template parameter corresponding to a specialized
non-type argument shall not be dependent on a parameter of the
specialization. [ Example:
template <class T, T t> struct C {};
template <class T> struct C<T, 1>; // error
template< int X, int (*array_ptr)[X] > class A {};
int array[5];
template< int X > class A<X,&array> { }; // error
—end example ]
My question is why is this restriction here? There is at least one use case where I find that this restriction interferes with writing clean code. E.g.
template <typename T, T*>
struct test;
template <typename T>
struct test<T, nullptr> // or struct test<T, (T*)nullptr>
{
};
template <typename R, typename...ARGs, R(*fn)(ARGs...)>
struct test<R(ARGs...), fn>
{
};
Though I'm unsure if there are other cases that stating a constant based on a type is a problem beyond not making any sense.
Anyone have a reason for why this is so?
(IMHO) The most common reasons the standard disallows a specific feature are:
The feature is covered by another mechanism in the language, rendering it superfluous.
It contradicts existing language logic and implementation, making its implementation potentially code breaking.
Legacy: the feature was left out in the first place and now we've built a lot without it that it's almost forgotten (see partial function template specialization).
Difficulty of implementation is rarely a factor, though it may take some time for compiler implementations to catch up with evolution on the "hard" stuff.
You could always wrap your non type template parameter in another type:
template < typename T1, typename T2 >
struct Demo {}; // primary template
template < typename T >
struct Demo<T, integral_constant<T, 0>> {}; // specialization
I doubt this hack falls into case 1. Case 3 is always a possibility so lets examine case 2. To do this, we have to know which are the related rules the standard imposes on class templates partial specializations.
14.5.5 Class template partial specializations
A non-type argument is non-specialized if it is the name of a non-type parameter. All other non-type arguments are specialized. (C1)
Within the argument list of a class template partial specialization, the following restrictions apply:
A partially specialized non-type argument expression shall not involve a template parameter of the partial specialization except when the argument expression is a simple identifier. (C2)
The type of a template parameter corresponding to a specialized non-type argument shall not be dependent on a parameter of the specialization. (C3)
I marked the first three Clauses I found relevant (the third is the one in question). According to C1 in our case we have a specialized non-type argument so C2 should stand, yet this
template <class T, T t> struct C {};
template <class T> struct C<T, 1>;
is actually
template <class T, T t> struct C {};
template <class T> struct C<T, T(1)>; // notice the value initialization
so the partially specialized non type argument T t involves the template parameter of the partial specialization class T in an expression other than an identifier; furthermore such specializations are bound to involve class T in a value initialization which will always be a violation of the rules. Then C3 comes along and clears that out for us so that we won't have to make that deduction every time.
So far we've established that the rules are in sync with themselves but this does NOT prove case 2 (once we remove the initial limitation every other related limitation falls apart). We'd have to dive into matching class template partial specializations rules; partial ordering is considered out of scope here because if we can produce valid candidates it's up to the programmer to put together a well formed program (i.e. not create ambiguous uses of class templates).
Section
Matching of class template partial specializations [temp.class.spec.match]
describes the (give or take) "pattern matching" process involved in template specialization. Rule 1 is the overall workflow of the procedure and the subsequent rules are those that define correctness
A partial specialization matches a given actual template argument list if the template arguments of the partial specialization can be deduced from the actual template argument list
A non-type template argument can also be deduced from the value of an actual template argument of a non-type parameter of the primary template.
In a type name that refers to a class template specialization, (e.g., A) the argument list shall match the template parameter list of the primary template. The template arguments of a specialization are deduced from the arguments of the primary template.
These rules are not violated by allowing the type of a template parameter corresponding to a specialized non-type argument to be dependent on a parameter of the specialization. So IMHO there is no specific reason why we can't have this feature in future revisions of the language: legacy is to blame. Sadly I didn't manage to find any language proposals with the initiative to introduce this feature.
An allocator can optionally have nested types like pointer, const_pointer. But one can always use these interface with std::allocator_traits<Allocator>, which would provide a default version of these types if they are absent in Allocator.
How is std::allocator_traits implemented? How can a template choose a default version of nested type when it's absent?
The solution is to refer to the type T::pointer in a context where it does not cause an error if it is not a valid type, instead it causes template argument deduction to fail. The general form of this is known as SFINAE, which stands for "Substitution Failure Is Not An Error". For a explanation of how it works see my SFINAE Functionality Is Not Arcane Esoterica presentation.
There are various techniques, often involving overloaded function templates, but my current favourite uses the void_t idiom to select a partial specialization of a class template:
template<typename T>
using void_t = void;
template<typename T, typename = void>
struct get_pointer
{
using type = typename T::value_type*;
};
template<typename T>
struct get_pointer<T, void_t<typename T::pointer>>
{
using type = typename T::pointer;
};
Now given an allocator type A you can use typename get_pointer<A>::type to refer to A::pointer if that exists, otherwise A::value_type*
The code above works because when A::pointer is a valid type the partial specialization matches and is more specialized than the primary template, and so gets used. When A::pointer is not a valid type the partial specialization is ill-formed, so the primary template is used.
This is from the C++ Standard Library xutility header that ships with VS2012.
template<class _Elem1,
class _Elem2>
struct _Ptr_cat_helper
{ // determines pointer category, nonscalar by default
typedef _Nonscalar_ptr_iterator_tag type;
};
template<class _Elem>
struct _Ptr_cat_helper<_Elem, _Elem>
{ // determines pointer category, common type
typedef typename _If<is_scalar<_Elem>::value,
_Scalar_ptr_iterator_tag,
_Nonscalar_ptr_iterator_tag>::type type;
};
Specifically what is the nature of the second _Ptr_cat_helper declaration? The angle brackets after the declarator _Ptr_cat_helper make it look like a specialization. But instead of specifying full or partial types by which to specialize the template it instead just repeats the template argument multiple times.
I don't think I've seen that before. What is it?
UPDATE
We are all clear that the specialization applies to an instantiation of the template where both template arguments are of the same type, but I'm not clear on whether this constitutes a full or a partial specialization, or why.
I thought a specialization was a full specialization when all the template arguments are either explicitly supplied or are supplied by default arguments, and are used exactly as supplied to instantiate the template, and that conversely a specialization was partial either, if not all the template parameters were required due to the specialization supplying one or more (but not all) of them, and/or if the template arguments were used in a form that was modified by the specialization pattern. E.g.
A specialization that is partial because the specialization is supplying at least one, but not all, of the template arguments.
template<typename T, typename U>
class G { public: T Foo(T a, U b){ return a + b; }};
template<typename T>
class G<T, bool> { public: T Foo(T a, bool b){ return b ? ++a : a; }};
A specialization that is partial because the specialization is causing the supplied template argument to be used only partially.
template<typename T>
class F { public: T Foo(T a){ return ++a; }};
template<typename T>
class F<T*> { public: T Foo(T* a){ return ++*a; }};
In this second example if the template were instantiated using A<char*> then T within the template would actually be of type char, i.e. the template argument as supplied is used only partially due to the application of the specialization pattern.
If that is correct then wouldn't that make the template in the original question a full specialization rather than a partial specialization, and if that is not so then where is my misunderstanding?
It is a partial class template specialization for the case when the same type is passed for both parameters.
Maybe this will be easier to read:
template<typename T, typename U>
struct is_same : std::false_type {};
template<typename T>
struct is_same<T,T> : std::true_type {};
EDIT:
When in doubt whether a specialization is an explicit (full) specialization or a partial specialization, you can refer to the standard which is pretty clear on this matter:
n3337, 14.7.3./1
An explicit specialization of any of the following:
[...]
can be declared by a declaration introduced by template<>; that is:
explicit-specialization:
template < > declaration
and n3337, 14.5.5/1
A primary class template declaration is one in which the class
template name is an identifier. A template declaration in which the
class template name is a simple-template-id is a partial
specialization of the class template named in the simple-template-id. [...]
Where simple-template-id is defined in the grammar like this:
simple-template-id:
template-name < template-argument-list opt >
template-name
identifier
So, wherever there's template<>, it's a full specialization, anything else is a partial specialization.
You can also think about it this way: Full template specialization specializes for exactly one possible instantiation of the primary template. Anything else is a partial specialization. Example in your question is a partial specialization because while it limits the arguments to be of the same type, it still allows for indifinitely many distinct arguments the template can be instantiated with.
A specialization like this, for example
template<>
vector<bool> { /* ... */ };
is a full specialization because it kicks in when the type is bool and only bool.
Hope that helps.
And just a note I feel it's worth mentioning. I guess you already know - function templates can't be partialy specialized. While this
template<typename T>
void foo(T);
template<typename T>
void foo(T*);
might looks like a partial specialization of foo for pointers on the first glance, it is not - it's an overload.
You mention specifying "full or partial types" when performing specialization of a template, which suggests that you are aware of such language feature as partial specialization of class templates.
Partial specialization has rather extensive functionality. It is not limited to simply specifying concrete arguments for some of the template parameters. It also allows defining a dedicated version of template for a certain groups of argument types, based on their cv-qualifications or levels of indirection, as in the following example
template <typename A, typename B> struct S {};
// Main template
template <typename A, typename B> struct S<A *, B *> {};
// Specialization for two pointer types
template <typename A, typename B> struct S<const A, volatile B> {};
// Specialization for const-qualified type `A` and volatile-qualified type `B`
And it also covers specializations based on whether some template arguments are identical or different
template <typename A> struct S<A, A> {};
// Specialization for two identical arguments
template <typename A> struct S<A, A *> {};
// Specialization for when the second type is a pointer to the first one
As another, rather curios example, partial specialization of a multi-argument template can be used to fully override the main template
template <typename A, typename B> struct S<B, A> {};
// Specialization for all arguments
Now, returning to your code sample, partial specialization for two identical arguments is exactly what is used in the code you posted.