I'm not sure if this is possible, but I would like to count the number of template arguments of any class like:
template <typename T>
class MyTemplateClass { ... };
template <typename T, typename U>
class MyTemplateClass2 { ... };
such that template_size<MyTemplateClass>() == 1 and template_size<MyTemplateClass2>() == 2. I'm a beginner to template templates, so I came up with this function which of course does not work:
template <template <typename... Ts> class T>
constexpr size_t template_size() {
return sizeof...(Ts);
}
because Ts can not be referenced. I also know that it might come to problems when handling variantic templates, but that is not the case, at least for my application.
Thx in advance
There is one...
° Introduction
Like #Yakk pointed out in his comment to my other answer (without saying it explicitly), it is not possible to 'count' the number of parameters declared by a template. It is, on the other hand, possible to 'count' the number of arguments passed to an instantiated template.
Like my other answer shows it, it is rather easy to count these arguments.
So...
If one cannot count parameters...
How would it be possible to instantiate a template without knowing the number of arguments this template is suppose to receive ???
Note
If you wonder why the word instantiate(d) has been stricken throughout this post,
you'll find its explanation in the footnote. So keep reading... ;)
° Searching Process
If one can manage somehow to try to instantiate a template with an increasing number of arguments, and then, detect when it fails using SFINAE (Substitution Failure Is Not An Error), it should be possible to find a generic solution to this problem... Don't you think ?
Obviously, if one wants to be able to also manage non-type parameters, it's dead.
But for templates having only typename parameters...
There is one...
Here are the elements with which one should be able to make it possible:
A template class declared with only typename parameters can receive any type as argument. Indeed, although there can have specializations defined for specific types,
a primary template cannot enforce the type of its arguments.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
The above statement might no longer be true from C++20 concepts.
I cannot try ATM, but #Yakk seems rather confident on the subject. After his comment:
I think concepts breaks this. "a primary template cannot enforce the type of its arguments." is false.
He might be right if constraints are apply before the template instantiation. But...
By doing a quick jump to the introduction to Constraints and concepts, one can read, after the first code example:
"Violations of constraints are detected at compile time, early in the template instantiation process, which leads to easy to follow error messages."
To be confirmed...
It is perfectly possible to create a template having for sole purpose to
be instantiated with any number of arguments. For our use case here, it might contain only ints... (let's call it IPack).
It is possible to define a member template of IPack to define the Next IPack by adding an int to the arguments of the current IPack. So that one can progressively increase its number of arguments...
Here is maybe the missing piece. It is maybe something that most people don't realize.
(I mean, most of us uses it a lot with templates when, for example, the template accesses a member that one of its arguments must have, or when a trait tests for the existence of a specific overload, etc...)
But I think it might help in finding solutions sometimes to view it differently and say:
It is possible to declare an arbitrary type, built by assembling other types, for which the evaluation by the compiler can be delayed until it is effectively used.
Thus, it will be possible to inject the arguments of an IPack into another template...
Lastly, one should be able to detect if the operation succeeded with a testing trait making use of decltype and std::declval. (note: In the end, none of both have been used)
° Building Blocks
Step 1: IPack
template<typename...Ts>
struct IPack {
private:
template<typename U> struct Add1 {};
template<typename...Us> struct Add1<IPack<Us...>> { using Type = IPack<Us..., int>; };
public:
using Next = typename Add1<IPack<Ts...>>::Type;
static constexpr std::size_t Size = sizeof...(Ts);
};
using IPack0 = IPack<>;
using IPack1 = typename IPack0::Next;
using IPack2 = typename IPack1::Next;
using IPack3 = typename IPack2::Next;
constexpr std::size_t tp3Size = IPack3::Size; // 3
Now, one has a means to increase the number of arguments,
with a convenient way to retrieve the size of the IPack.
Next, one needs something to build an arbitrary type
by injecting the arguments of the IPack into another template.
Step 2: IPackInjector
An example on how the arguments of a template can be injected into another template.
It uses a template specialization to extract the arguments of an IPack,
and then, inject them into the Target.
template<typename P, template <typename...> class Target>
struct IPackInjector { using Type = void; };
template<typename...Ts, template <typename...> class Target>
struct IPackInjector<IPack<Ts...>, Target> { using Type = Target<Ts...>; };
template<typename T, typename U>
struct Victim;
template<typename P, template <typename...> class Target>
using IPInj = IPackInjector<P, Target>;
//using V1 = typename IPInj<IPack1, Victim>::Type; // error: "Too few arguments"
using V2 = typename IPInj<IPack2, Victim>::Type; // Victim<int, int>
//using V3 = typename IPInj<IPack3, Victim>::Type; // error: "Too many arguments"
Now, one has a means to inject the arguments of an IPack
into a Victim template, but, as one can see, evaluating Type
directly generates an error if the number of arguments does not
match the declaration of the Victim template...
Note
Have you noticed that the Victim template is not fully defined ?
It is not a complete type. It's only a forward declaration of a template.
The template to be tested will not need to be a complete type
for this solution to work as expected... ;)
If one wants to be able to pass this arbitrary built type to some detection trait one will have to find a way to delay its evaluation.
It turns out that the 'trick' (if one could say) is rather simple.
It is related to dependant names. You know this annoying rule
that enforces you to add ::template everytime you access a member template
of a template... In fact, this rule also enforces the compiler not to
evaluate an expression containing dependant names until it is
effectively used...
Oh I see ! ...
So, one only needs to prepare the IPackInjectors without
accessing its Type member, and then, pass it to our test trait, right ?
It could be done using something like that:
using TPI1 = IPackInjector<IPack1, Victim>; // No error
using TPI2 = IPackInjector<IPack2, Victim>; // No error
using TPI3 = IPackInjector<IPack3, Victim>; // No error
Indeed, the above example does not generate errors, and it confirms
that there is a means to prepare the types to be built and evaluate
them at later time.
Unfortunately, it won't be possible to pass these pre-configured
type builders to our test trait because one wants to use SFINAE
to detect if the arbitrary type can be instantiated or not.
And this is, once again, related to dependent name...
The SFINAE rule can be exploited to make the compiler silently
select another template (or overload) only if the substitution
of a parameter in a template is a dependant name.
In clear: Only for a parameter of the current template instantiation.
Hence, for the detection to work properly without generating
errors, the arbitrary type used for the test will have to be
built within the test trait with, at least, one of its parameters.
The result of the test will be assigned to the Success member...
Step 3: TypeTestor
template<typename T, template <typename...> class C>
struct TypeTestor {};
template<typename...Ts, template <typename...> class C>
struct TypeTestor<IPack<Ts...>, C>
{
private:
template<template <typename...> class D, typename V = D<Ts...>>
static constexpr bool Test(int) { return true; }
template<template <typename...> class D>
static constexpr bool Test(...) { return false; }
public:
static constexpr bool Success = Test<C>(42);
};
Now, and finally, one needs a machinery that will successively try
to instantiate our Victim template with an increasing number of arguments. There are a few things to pay attention to:
A template cannot be declared with no parameters, but it can:
Have only a parameter pack, or,
Have all its parameters defaulted.
If the test procedure begins by a failure, it means that the template must take more arguments. So, the testing must continue until a success, and then, continue until the first failure.
I first thought that it might make the iteration algorithm using template specializations a bit complicated... But after having thought a little about it, it turns out that the start conditions are not relevant.
One only needs to detect when the last test was true and next test will be false.
There must have a limit to the number of tests.
Indeed, a template can have a parameter pack, and such a template can receive an undetermined number of arguments...
Step 4: TemplateArity
template<template <typename...> class C, std::size_t Limit = 32>
struct TemplateArity
{
private:
template<typename P> using TST = TypeTestor<P, C>;
template<std::size_t I, typename P, bool Last, bool Next>
struct CheckNext {
using PN = typename P::Next;
static constexpr std::size_t Count = CheckNext<I - 1, PN, TST<P>::Success, TST<PN>::Success>::Count;
};
template<typename P, bool Last, bool Next>
struct CheckNext<0, P, Last, Next> { static constexpr std::size_t Count = Limit; };
template<std::size_t I, typename P>
struct CheckNext<I, P, true, false> { static constexpr std::size_t Count = (P::Size - 1); };
public:
static constexpr std::size_t Max = Limit;
static constexpr std::size_t Value = CheckNext<Max, IPack<>, false, false>::Count;
};
template<typename T = int, typename U = short, typename V = long>
struct Defaulted;
template<typename T, typename...Ts>
struct ParamPack;
constexpr std::size_t size1 = TemplateArity<Victim>::Value; // 2
constexpr std::size_t size2 = TemplateArity<Defaulted>::Value; // 3
constexpr std::size_t size3 = TemplateArity<ParamPack>::Value; // 32 -> TemplateArity<ParamPack>::Max;
° Conclusion
In the end, the algorithm to solve the problem is not that much complicated...
After having found the 'tools' with which it would be possible to do it, it only was a matter, as very often, of putting the right pieces at the right places... :P
Enjoy !
° Important Footnote
Here is the reason why the word intantiate(d) has been stricken at the places where it was used in relation to the Victim template.
The word instantiate(d) is simply not the right word...
It would have been better to use try to declare, or to alias the type of a future instantiation of the Victim template.
(which would have been extremely boring) :P
Indeed, none of the Victim templates gets ever instantiated within the code of this solution...
As a proof, it should be enough to see that all tests, made in the code above, are made only on forward declarations of templates.
And if you're still in doubt...
using A = Victim<int>; // error: "Too few arguments"
using B = Victim<int, int>; // No errors
template struct Victim<int, int>;
// ^^^^^^^^^^^^^^^^
// Warning: "Explicit instantiation has no definition"
In the end, there's a full sentence of the introduction which might be stricken, because this solution seems to demonstrate that:
It is possible to 'count' the number of parameters declared by a template...
Without instantiation of this template.
#include <utility>
#include <iostream>
template<template<class...>class>
struct ztag_t {};
template <template<class>class T>
constexpr std::size_t template_size_helper(ztag_t<T>) {
return 1;
}
template <template<class, class>class T>
constexpr std::size_t template_size_helper(ztag_t<T>) {
return 2;
}
template <typename T>
class MyTemplateClass { };
template <typename T, typename U>
class MyTemplateClass2 { };
template<template<class...>class T>
struct template_size:
std::integral_constant<
std::size_t,
template_size_helper(ztag_t<T>{})
>
{};
int main() {
std::cout << template_size<MyTemplateClass>::value << "\n";
std::cout << template_size<MyTemplateClass2>::value << "\n";
}
I know of no way without writing out the N overloads to support up to N arguments.
Live example.
Reflection will, of course, make this trivial.
° Before Reading This Post
This post does not answer to "How to get the number of parameters",
it answers to "how to get the number of arguments"...
It is let here for two reasons:
It might help someone who would have mixed up (like I did)
the meaning of parameters and arguments.
The techniques used in this post are closely related to the ones used
to produce the correct answer I've posted as a separate answer.
See my other answer for finding "the number of parameters" of a template.
The answer of Elliott looks more like what one usually does (though the primary template should be fully defined and "do something" IMHO). It uses a template specialization for when a template is passed as argument to the primary template.
Meanwhile, the answer of Elliott vanished...
So I've posted something similar to what he showed below.
(see "Generic Solution")
But, just to show you that you weren't that far from a working solution, and, because I noticed that you have used a function for your try, and, you declared this function constexpr, you could have written it like that:
Note
It is a 'fancy solution', but it works...
template <typename T> class MyTemplateClass {};
template <typename T, typename U> class MyTemplateClass2 {};
template <template <typename...> class T, typename...Ts>
constexpr const size_t template_size(T<Ts...> && v)
{
return sizeof...(Ts);
}
// If the target templates are complete and default constructible.
constexpr size_t size1 = template_size(MyTemplateClass<int>{});
constexpr size_t size2 = template_size(MyTemplateClass2<int, short>{});
// If the target templates are complete but NOT default constructible.
constexpr size_t size3 = template_size(decltype(std::declval<MyTemplateClass<int>>()){});
constexpr size_t size4 = template_size(decltype(std::declval<MyTemplateClass2<int, short>>()){});
Explanation
You said "because Ts can not be referenced", which is true and false, because of the way you made the declaration of template_size.
That is, a template template parameter cannot declare parameters itself (where you placed Ts in the declaration of the function template). It is allowed to do so to give a clue of what the template argument is expected to receive as argument, but it is not a declaration of a parameter name for the current template declaration...
(I hope it's clear enough) ;)
Obviously, it might be a little bit over complicated, but it worth knowing I think that such a construct is possible also... ;)
Generic Solution
template <typename T> class MyTemplateClass {};
template <typename T, typename U> class MyTemplateClass2 {};
template<typename T>
struct NbParams { static constexpr std::size_t Value = 0; };
template<template <typename...> class C, typename...Ts>
struct NbParams<C<Ts...>> { static constexpr std::size_t Value = sizeof...(Ts); };
constexpr size_t size1 = NbParams<MyTemplateClass<int>>::Value;
constexpr size_t size2 = NbParams<MyTemplateClass2<int, short>>::Value;
That is the regular way one does this kind of things... ;)
I want to write a function type that transform a type B in the same type but with the same sign that another type A. I can achieve that using this function:
template <typename A, typename B, bool = std::is_signed_v<B>>
struct same_sign_as{
using type = std::make_signed_t<A>;
};
template <typename A, typename B>
struct same_sign_as<A, B, false>{
using type = std::make_unsigned_t<A>;
};
and then use it as using R = same_sign_as_t<A,B>;.
But I want my code to be readable. When I see a function simliar to same_sign_as I never know if it means the type A with the sign of B or the type of B with the sign of A.
As I don't want to remember this kind of things I'd like to write better something similar to
using R = make_type<A>::same_sign_as<B>
With this I'm not going to make any mistakes.
Is easy to write it:
template <typename A>
struct make_type{
template <typename B>
struct same_sign_as
{ using type = typename __same_sign_as<A, B>::type; };
};
where I changed the name of same_sign_as for __same_sign_as (I want this function to be an implementation detail).
To use make_type write:
using Res = typename make_type<A>::template same_sign_as<B>::type
and that is horrible. You can't read that!!!
Can I define an alias to achieve my goal of writing
using Res = make_type<A>::same_sign_as<B>
?
Lets assume I have numerous atomic structs each having an inner_type:
struct Atomic1{
using inner_type = int;
};
struct Atomic2{
using inner_type = double;
};
struct Atomic3{
using inner_type = bool;
};
...
My client class is a variadic template that can use 1 or more of the above atomic classes:
template<class ...AtomicTypeArgPack>
class MyclassAcceptingAtomicTypes;
I have have a related generic class that accepts Atomic*::inner_type as template parameters:
template<class ...InnerTypeArgPack>
class MyclassAcceptingInnerTypes;
My specific api class is defined but specifying a couple of template types:
using my_first_class_t = MyclassAcceptingAtomicTypes<Atomic1, Atomic2>;
for each specific class, I also have another class of inner types:
using my_first_class_inner_types_t = MyclassAcceptingInnerTypes<Atomic1::inner_type , Atomic2::inner_type >;
Is there is way to automatically generate the second type (i.e. my_first_class_inner_types_t) from the first declaration (my_first_class_t) using template meta programming / meta functions?
Try this:
template <class Atomics>
struct inner_types;
template <template <class...> class T, class... Atomic>
struct inner_types<T<Atomic...>>
{
using type = MyclassAcceptingInnerTypes<typename Atomic::inner_type...>;
};
using atomics = MyclassAcceptingAtomicTypes<Atomic1, Atomic2>;
using inners = MyclassAcceptingInnerTypes<Atomic1::inner_type , Atomic2::inner_type >;
static_assert(std::is_same_v<inner_types<atomics>::type, inners>);
Is there is way to automatically generate the second type (i.e. my_first_class_inner_types_t) from the first declaration (my_first_class_t) using template meta programming / meta functions?
Do you mean something as follows ?
template <typename ... Ts>
constexpr auto foo (MyclassAcceptingAtomicTypes<Ts...> const &)
-> MyclassAcceptingInnerTypes<typename Ts::inner_type...>;
template <typename T>
using bar = decltype(foo(std::declval<T>()));
You can verify that
static_assert( std::is_same<bar<my_first_class_t>,
my_first_class_inner_types_t>{}, "!" );
Consider the following:
template <class...>
struct MyT;
template <class T>
struct MyT<T> {};
template <template <class> class TT = MyT> struct A {}; // fine
using B = A<MyT>; // does not compile
int main() {
return 0;
}
When MyT is used as a default argument of A, the compiler (g++ 5.4.0) is happy. However, when it is used to instantiate A, the story is different:
temp.cpp:19:16: error: type/value mismatch at argument 1 in template parameter list for ‘template<template<class> class TT> struct A’
using B = A<MyT>;
^
temp.cpp:19:16: note: expected a template of type ‘template<class> class TT’, got ‘template<class ...> struct MyT’
I can fix it by introducing an alias:
template <class T>
using MyTT = MyT<T>;
using B = A<MyTT>; // fine
The question: what is the reason for the error and is there a solution without introducing an alias?
EDIT Please note that A is declared to have a template template parameter as shown and that is not given for change.
You cannot do that and you cannot use such a type as a default parameter. The fact that it seems to be accepted as long as you don't rely on it doesn't mean that the default parameter is a valid one.
Consider the following code that explicitly uses the default type:
template <class...>
struct MyT;
template <class T>
struct MyT<T> {};
template <template <class> class TT = MyT> struct A {}; // fine
int main() {
A<> a;
return 0;
}
The error is quite clear:
template template argument has different template parameters than its corresponding template template parameter
Partial specializations are not taken in account in this case, thus the two declarations differ.
You should either declare A as:
template <template <class...> class TT = MyT> struct A;
Or declare somewhere a type that is constrained to a single argument, as an example by means of an using declaration as you did.
First, the default argument doesn't work either.
Second, template template arguements are a strange beast. It would make sense if a template template argument would take anything that could be instantiated with the signature described in the template template argument.
That is not how it works.
Instead it works the other way around.
template<template<class...>class Z> struct foo {};
template<template<class >class Z> struct bar {};
template<class...>struct a{};
template<class >struct b{};
foo will accept a or b.
bar will accept only b.
The correct response to this, once you understand it, is "what the hell?". If you aren't responding "what the hell" back up and see if you can understand it. This basically works backwards from typical typing for arguements in C++; it behaves more like a return type than an argument. (Learn the terms contravariance and covariance if you want to see some of the language that lets you talk about this directly)
This is quite non-intuitive, and why it works this way exactly would involve tracking down the pre-history of C++.
But, as a benefit, a template<class...>class argument is in effect an "any template that only takes type parameters". I find this highly useful.
As a downside, template<class>class arguements are almost completely useless.
Tl;dr: make your template<template parameters be template<template<class...>class, and metaprogram only with templates that only take types. If you have a template that takes values, write a type wrapper that replaces a requirement for a std::size_t X with a std::integral_constant< std::size_t, X >.
Forgetting for a moment the question of "Why would you do this?",
the first version would work if you hadn't done any template specialization.
template <class T>
struct MyT { };
template <template <class> class TT = MyT> struct A
{};
using B = A<MyT>;
With template specialization, the compiler must determine the best match, but since you haven't ever actually provided any template arguments it's ambiguous.
When you introduce MyTT you are using a single template argument, and the compiler is smart enough to see that you have a specialization when there is only one arg:
template <class T>
using MyTT = MyT<T>;
It chooses the specialization instead of the variadic version in this case.
But now we circle back to the grand question... why? Unless within A you're always instantiating MyT with a specific class, it's pointless to use A at all:
template<template<class> class TT = MyT> struct A
{
// use an instance of TT??
TT<int> myInstance; // forced to choose `int`
};
I would like to split your question into 2 parts.
A) Consider the template of your structure is simpler
template <class T>
struct TestStruct {
};
template <
template <class>
class TT = TestStruct
>
struct A
{
int a; // sorry to modify this. This help to show how it works
};
int main() {
A< TestStruct > b;
b.a; // it works!
return 0;
}
It works because of the template class TT only accept the template with < class... > template. The specializated class is not count on this ( because the underlying of it is still template < class ... > )
B) even you update your struct A to the template< class... > one, you still have one more problem. What is the template argument of TT? Please see the example below
template <class...>
struct MyT;
template <class T>
struct MyT<T> {
int a;
};
template <
template <class...>
class TT = MyT
// may be you need to add some more typename here, such as
// typename T1, ... , and then TT<T1> a;
>
struct A
{
TT<int> a;
// Here ! TT is a template only, do not have the template parameters!!!
};
int main() {
A< MyT > b;
b.a; // it works!!
return 0;
}
But, if you really cannot update the signature of those definitions, you can do a proxy class
template< class T >
struct Proxy : MyT<T>
{
};
Say I have a template class like so:
template < typename TParam >
class Test
{
// content
};
I want to pull out the first template parameter of TParam if it's a specialization of a class template. Something like:
template < typename TParam >
class Test
{
using TParamInner = TemplateType<TParam>::Type;
// use TParamInner here
};
Additional info:
I have access to all of C++98.
I have access to a subset of C++11.
I would prefer to avoid the stdlib if possible (assume this is
because I'm using an embedded system for which no stdlib is available and/or because I am heavily memory-constrained)
You can get close with something like:
template <class >
struct first_template_param;
template <template <class...> class Z, class T, class... Ts>
struct first_template_param<Z<T, Ts...>> {
using type = T;
}
It won't handle std::array or any other class templates that take non-type template parameters. But it'll handle all the "normal" class templates. You can always then add extra specializations for all the ones you want:
template <class T, size_t N>
struct first_template_param<std::array<T,N>> {
using type = T;
}
Thanks to #Barry for spurring the solution along.
It's not a complete answer for all template types, but it works for templates where all parameters are types, which is a large number of the most useful templates.
template < typename Head, typename ... Tail >
struct split { using first = Head; };
template <class >
struct cls_template_info; // fails on non-templates
template <template <class...> class Z, class... Ts>
struct cls_template_info<Z<Ts...>>
{
using type = typename split<Ts...>::first; // typename used to disambiguate
};
This can then be used as using T = cls_template_info<std::vector<int>>::first;.
You can't. A template type is never carried up to runtime. You have to instantiate it (this leads to a complete new type), and the compiler then generates the needed code to make it appear as if you have defined specifically for the type parameters you specified. Indeed, in old compilers (this has been solved a lot of time ago) when you instantiate a generic type in several compilation units, that lead to several repetitions of the same code in the final program. But as I've said, this has been solved time ago.