Reducing number of template arguments for class - c++

I have a method and two classes defined like this:
template<template<class X> class T>
void doSomething()
{
T<int> x;
}
template <class T>
class ClassWithOneArg
{
T t;
};
template <class T1, class T2>
class ClassWithTwoArgs
{
T1 t1;
T2 t2;
};
I can now
doSomething<ClassWithOneArg>();
but I cannot
doSomething<ClassWithTwoArgs>();
However, I'd like to pass ClassWithTwoArgs to doSomething, where T2 = double.
The only method I found is to create
template <class T1>
class ClassWithTwoArgs_Child
: public ClassWithTwoArgs<T1, double>
{
};
and then
doSomething<ClassWithTwoArgs_Child>();
This works, but in my concrete case all classes require a constructor argument and thus I have to create a constructor with this argument also in the _Child-class and pass it to the base which I really want to avoid.
Do you have an idea how to do that?
Thanks a lot!

Indirection is a solution. Instead of a template template parameter you pass a "meta function" -- a function that maps one type to another in form of a struct with a nested class template:
struct mf1 {
template<class Arg1>
struct eval {
typedef ClassTemplateWithOneArg<Arg1> type;
};
};
template<class Arg2>
struct mf2 {
template<class Arg1>
struct eval {
typedef ClassTemplateWithTwoArgs<Arg1,Arg2> type;
};
};
template<class MetaFunc>
void do_something()
{
typedef typename MetaFunc::template eval<int>::type clazztype;
clazztype x;
}
void foo() {
do_something<mf1>();
do_something<mf2<double> >();
}
In C++0x this could be reduced to a "template typedef":
template<class Arg1>
using NewClassTemplate = ClassTemplateWithTwoArgs<Arg1,double>;
which allows you to pass NewClassTemplate as a template template argument which also accepts only one template parameter.

There is no generic solution. Your best bet is
template<class T>
void doSomething()
{
T x;
}
template <class T>
class ClassWithOneArg
{
T t;
};
template <class T1, class T2 = double>
class ClassWithTwoArgs
{
T1 t1;
T2 t2;
};
int main(){
doSomething<ClassWithOneArg<int>>();
doSomething<ClassWithTwoArgs<int, double> >();
}

It seems that what you are after is similar to the rebinding of allocators (given an allocator, containers need to be able to produce an allocator for a different type - e.g std::list<int> might need a allocator<list_node<int> > from allocator<int>.
However, the class templates would have to be modified for this.
template<class T>
void doSomething(const T&)
{
typename T::template rebind_1st<int>::type x;
}
template <class T>
class ClassWithOneArg
{
T t;
public:
template <class U>
struct rebind_1st { typedef ClassWithOneArg<U> type; };
};
template <class T1, class T2>
class ClassWithTwoArgs
{
T1 t1;
T2 t2;
public:
template <class U>
struct rebind_1st { typedef ClassWithTwoArgs<U, T2> type; };
};
int main()
{
doSomething(ClassWithOneArg<char>());
doSomething(ClassWithTwoArgs<char, double>() );
}

Assuming you want declare template instantiations of the same class with a different type for the first template parameter, it appears a version of visitor's code is possible that doesn't require modifying original classes.
template <class T, class NewFirstArg>
struct rebind_1st;
template <template <class> class T, class Arg1, class NewFirstArg>
struct rebind_1st<T<Arg1>, NewFirstArg>
{
typedef T<NewFirstArg> type;
};
template <template <class, class> class T, class Arg1, class Arg2, class NewFirstArg>
struct rebind_1st<T<Arg1, Arg2>, NewFirstArg>
{
typedef T<NewFirstArg, Arg2> type;
};
template <class T>
void foo()
{
typename rebind_1st<T, int>::type x;
(void)x;
}
template <class T>
struct One{};
template <class T1, class T2>
struct Two{};
int main()
{
foo<One<char> >();
foo<Two<char, double> >();
}

This works with MSVC:
template<class T>
void doSomething()
{
T x;
}
// class definitions omitted...
void test() {
doSomething<ClassWithOneArg<int> >();
doSomething<ClassWIthTwoArgs<int, double> >();
}
I do not fully understand why you want to define the first parameter of your template template parameter to be int inside of doSomething. Looks like a "template smell" to me, since doSomething has to know a lot about its template template parameter.
Wouldn't it be cleaner to call doSomething the way i proposed? (But obviously i don't know the context of your calls).

Related

c++ templates overloading method depend on class type

I have class like this:
template<typename T>
MyClass{
//myFunc();
}
I want to create myFunc method that return numeric value if class template is numeric and return nothing (void) when class template is not numeric.
For now, I got sth like this:
template<typename T>
MyClass{
template <typename returnT>
returnT myFunc();
}
template <typename T>
template <typename returnT>
typename std::enable_if<std::is_arithmetic<T>::value>
T MyClass<T>::myFunc()
{
return T::value;
}
template <typename T>
template <typename returnT>
typename std::enable_if<!std::is_arithmetic<T>::value>
void MyClass::myFunc()
{
//do sth
}
of course, that doesn't work. Is that a good idea to solve this problem this way? What is "smart" and working solution?
As an alternative to the constexpr if solution already supplied, here is your initial idea in it's working form.
#include <type_traits>
#include <iostream>
template<typename T>
struct MyClass{
template <typename returnT = T, std::enable_if_t<std::is_arithmetic_v<returnT>, bool> = true>
T myFunc();
template <typename returnT = T, std::enable_if_t<!std::is_arithmetic_v<returnT>, bool> = true>
void myFunc();
};
template <typename T>
template <typename returnT, std::enable_if_t<std::is_arithmetic_v<returnT>, bool>>
T MyClass<T>::myFunc()
{
std::cout << "yo\n";
return T{};
}
template <typename T>
template <typename returnT, std::enable_if_t<!std::is_arithmetic_v<returnT>, bool>>
void MyClass<T>::myFunc()
{
std::cout << "yay\n";
}
int main() {
MyClass<int> m;
MyClass<std::string> n;
m.myFunc();
n.myFunc();
}
The simplest way I can think of would be to just use if constexpr:
template <typename T>
class MyClass
{
auto myFunc()
{
if constexpr (std::is_arithmetic_v<T>)
{
return T{};
}
else
{
// do smth
}
}
};
If you can't use C++17, you will have to revert to some SFINAE-based approach. What that would best look like exactly depends a lot on what the actual signatures involved should be. But, for example, you could provide a partial class template specialization for the case of an arithmetic type:
template <typename T, typename = void>
class MyClass
{
void myFunc()
{
// do smth
}
};
template <typename T>
class MyClass<T, std::enable_if_t<std::is_arithmetic<T>::value>>
{
T myFunc()
{
return {};
}
};
Note that an arithmetic type cannot be a class type or enum, so I'm not sure what T::value was trying to achieve in your example code for the case of T being an arithmetic type…
I would create a helper template class to select the return type, and a helper function that uses overloading to perform the right behavior.
template <typename, bool> struct RType;
template <typename T> struct RType<T, false> { typedef void type; };
template <typename T> struct RType<T, true> { typedef T type; };
template<typename T>
class MyClass{
typedef RType<T, std::is_arithmetic<T>::value> R;
void myFuncT(RType<T, false>) {}
T myFuncT(RType<T, true>) { return 0; }
public:
typename R::type myFunc() { return myFuncT(R()); }
};

How do I declare SFINAE class?

Something is not working quite well for me. Is this the way to declare a class, that accepts only floating point template parameter?
template <typename T, swift::enable_if<std::is_floating_point<T>::value> = nullptr>
class my_float;
I fail to define methods outside this class. Doesn't compile, not sure why
Well... not exactly SFINAE... but maybe, using template specialization? Something as follows ?
template <typename T, bool = std::is_floating_point<T>::value>
class my_float;
template <typename T>
class my_float<T, true>
{
// ...
};
If you really want use SFINAE, you can write
template <typename T,
typename = typename std::enable_if<std::is_floating_point<T>::value>::type>
class my_float
{
// ...
};
or also (observe the pointer there isn't in your example)
template <typename T,
typename std::enable_if<std::is_floating_point<T>::value>::type * = nullptr>
class my_float // ------------------------------------------------^
{
};
-- EDIT --
As suggested by Yakk (thanks!), you can mix SFINAE and template specialization to develop different version of your class for different groups of types.
By example, the following my_class
template <typename T, typename = void>
class my_class;
template <typename T>
class my_class<T,
typename std::enable_if<std::is_floating_point<T>::value>::type>
{
// ...
};
template <typename T>
class my_class<T,
typename std::enable_if<std::is_integral<T>::value>::type>
{
// ...
};
is developed for in two versions (two different partial specializations), the first one for floating point types, the second one for integral types. And can be easily extended.
You can also use static_assert to poison invalid types.
template <typename T>
class my_float {
static_assert(std::is_floating_point<T>::value,
"T is not a floating point type");
// . . .
};
It's a little bit more direct, in my opinion.
With either of the other approaches, e.g.
template <typename T, bool = std::is_floating_point<T>::value>
class my_float;
template <typename T> class my_float<T, true> { /* . . . */ };
my_float<int,true> is a valid type. I'm not saying that that's a bad approach, but if you want to avoid this, you'll have to encapsulate
my_float<typename,bool> within another template, to avoid exposing the bool template parameter.
indeed, something like this worked for me (thanks to SU3's answer).
template<typename T, bool B = false>
struct enable_if {};
template<typename T>
struct enable_if<T, true> {
static const bool value = true;
};
template<typename T, bool b = enable_if<T,is_allowed<T>::value>::value >
class Timer{ void start(); };
template<typename T, bool b>
void Timer<T,b>::start()
{ \* *** \*}
I am posting this answer because I did not want to use partial specialization, but only define the behavior of the class outside.
a complete workable example:
typedef std::integral_constant<bool, true> true_type;
typedef std::integral_constant<bool, false> false_type;
struct Time_unit {
};
struct time_unit_seconds : public Time_unit {
using type = std::chrono::seconds;
};
struct time_unit_micro : public Time_unit {
using type = std::chrono::microseconds;
};
template<typename T, bool B = false>
struct enable_if {
};
template<typename T>
struct enable_if<T, true> {
const static bool value = true;
};
template<typename T,
bool b = enable_if<T,
std::is_base_of<Time_unit,
T>::value
>::value>
struct Timer {
int start();
};
template<typename T, bool b>
int Timer<T, b>::start() { return 1; }
int main() {
Timer<time_unit_seconds> t;
Timer<time_unit_micro> t2;
// Timer<double> t3; does not work !
return 0;
}

Partial specialization of specific member functions

#include <iostream>
template <typename T1, typename T2>
class B{
public:
void update(){ std::cerr<<__PRETTY_FUNCTION__<<std::endl; }
void func1(){ std::cerr<<__PRETTY_FUNCTION__<<std::endl; }
void func2(){ std::cerr<<__PRETTY_FUNCTION__<<std::endl; }
};
template <typename T1>
class B<T1, int>{
public:
void update(){ std::cerr<<__PRETTY_FUNCTION__<<"(specialization)"<<std::endl;}
};
int main(){
B<int, double> b1;
b1.update();
b1.func1();
B<int, int> b2;
b2.update();
//b2.func1();//there's no function 'func1' in B<int,int>
}
I want to specialize update function for specific template parameter(data type).
So I tried to specialize template class B but it seems that I have to implement whole member functions again.
Because other members are exactly same between specializations, reimplementing whole members look cumbersome.
Is there any workaround for this case?
Tag-dispatch the call to update:
template <typename> struct tag {};
template <typename T1, typename T2>
class B
{
public:
void update()
{
return update(tag<B>());
}
private:
template <typename U1>
void update(tag<B<U1, int> >)
{
// specialization
}
template <typename U1, typename U2>
void update(tag<B<U1, U2> >)
{
// normal
}
};
DEMO

Applying partially applied template

Having a class like the A, is there a way to apply it to a template like this of B, with T2 set to some type C? But without creating another template class inheriting from A.
template<typename T1, typename T2>
class A
{ };
template<template <typename T1> class T3>
class B
{ };
With C++11 using a template alias works:
template<typename T1, typename T2>
class A
{ };
template<template <typename T1> class T3>
class B
{ };
class C
{ };
template< typename T > using A1 = A< T, C >;
int main()
{
B< A1 > b;
}
without C++11, you are left with what you probably already know:
template< typename T > class A1 : A< T, C > {};
I will propose an alternative solution: do not use template template parameters.
If you write:
template <typename T> struct B {};
Then it can be used with A<int, int> or C<3> or even plain D.
Whilst it is possible to use template template parameters, it is general a bad idea. You should treat the template parameter of a class as an implementation detail and apply the golden rule: do not rely on implementation details.
If you need access to the type, somehow, then use an associated type (aka T::AssociatedType) or a trait (BTraits<T>::AssociatedType).
EDIT: dealing with multiple instantiations of the template template parameter.
Suppose we want to "erase" the template template parameter of such a class:
template <template <typename> class A>
struct Something {
template <typename T>
void doit() { A<T>::doit(); }
};
The C++ standard allocation model is to use an inner rebind structure:
template <typename T>
struct Simple {
template <typename U>
struct rebind { typedef Simple<U> type; };
};
template <typename T0, typename T1>
struct Multi {
template <typename U>
struct rebind { typedef Multi<U, T1> type; };
};
template <typename A>
struct Something {
template <typename T>
void doit() { typedef typename A::rebind<T>::type B; B::doit(); }
};
Note how you can use complex computations in rebind and nothing forces you in blindly passing the type received as parameter.
Whilst another (similar) solution is to ask for a factory (aka, the object passed itself cannot be used but it can build useful objects); for ease of use the C++ containers ask of their allocators that they be both usable in themselves and factories for other types.
Yes, you can do it using C++11's alias template:
template <typename T>
using AA = A<T, C>;
B<AA> b;
Live example

Overloading template classes by template parameter number

Is it possible to have multiple versions of the same class which differ only in the number of template arguments they take?
For instance:
template<typename T>
class Blah {
public:
void operator()(T);
};
template<typename T, typename T2>
class Blah {
public:
void operator()(T, T2);
};
I'm trying to model functor type things which can take a variable number of arguments (up to the number of different templates that were written out).
The simplest answer would be to have just one template, with the maximum number you want to support and use void for a default type on all but the first type. Then you can use a partial specialization as needed:
template<typename T1, typename T2=void>
struct foo {
void operator()(T1, T2);
};
template <typename T1>
struct foo<T1, void> {
void operator()(T1);
};
int main() {
foo<int> test1;
foo<int,int> test2;
test1(0);
test2(1,1);
}
A template can have only one base definition. If you need a variable number of arguments and you don't want to use "null type" constructions as #awoodland suggests, and if you have a C++0x compiler, then you can use variadic templates:
template <typename ...Dummy> struct foo; // base case, never instantiated!
template <typename T> struct foo<T> { /*...*/ }; // partial spec. for one parameter
template <typename T, typename U> struct foo<T, U> { /*...*/ }; // ditto for two
This is untested code, I don't have a version of boost handy, but here goes anyway
#include "boost/tuple.h"
template <class T>
class Blah;
template <class T>
class Blah< boost::tuple<T> >
{
void operator()(T arg);
};
template <class T, class U>
class Blah< boost::tuple<T, U> >
{
void operator()(T arg1, U arg2);
};
etc. etc.