Termination issue on recursive template function in variadic template - c++

I am trying to create a utility class that will call a specific function on all classes in a list. The purpose behind this is to automate an element of reflection within a hierarchy of classes.
I'm using Visual Studio 2015 to compile some C++ code and I'm getting a compile error when the recursive template function is unfolded because the compiler is having trouble distinguishing between the recursive function and the terminating function.
I've extracted out the core of the class to a simple test case:
#include <iostream>
template< typename ... BaseClasses >
class Meta
{
public:
virtual ~Meta() {}
template< typename T >
void call(const T& val)
{
callOnAllClasses<T, BaseClasses...>(val);
}
private:
template< typename T, typename HeadClass >
void callOnAllClasses(const T& val)
{
auto pObj = dynamic_cast<HeadClass*>(this);
if ( pObj )
pObj->HeadClass::doSomething(val);
}
template< typename T, typename HeadClass, typename ... TailClasses >
void callOnAllClasses(const T& val)
{
auto pObj = dynamic_cast<HeadClass*>(this);
if ( pObj )
pObj->HeadClass::doSomething(val);
callOnAllClasses<T, TailClasses...>(val);
}
};
class A
{
public:
void doSomething(int i)
{
std::cout << "A:" << i << std::endl;
}
};
class B
{
public:
void doSomething(int i)
{
std::cout << "B:" << i << std::endl;
}
};
class C : public B, public A, public Meta<C,B,A>
{
public:
void doSomething(int i)
{
std::cout << "C:" << i << std::endl;
}
};
int main()
{
C c;
c.call(5);
}
This results in the following error when compiled in Visual Studio 2015:
error C2668: 'Meta<C,B,A>::callOnAllClasses': ambiguous call to overloaded function
could be 'void Meta<C,B,A>::callOnAllClasses<T,A,>(const T &)'
or 'void Meta<C,B,A>::callOnAllClasses<T,A>(const T &)'
I've never used variadic templates before so I'm at a bit of a loss as to why this might be going wrong. Any help would be much appreciated!

You problem can be minimized as follows:
template< typename ... Bases >
struct Meta
{
template< typename T >
void call(const T& val)
{
callOnAllClasses<T, Bases...>(val);
}
template< typename T, typename HeadClass >
void callOnAllClasses(const T& val)
{
}
template< typename T, typename HeadClass, typename ... TailClasses >
void callOnAllClasses(const T& val)
{
callOnAllClasses<T, TailClasses...>(val);
}
};
struct C : Meta<int, int, int> { };
int main()
{
C{}.call(5);
}
When TailClasses is empty, both these signatures are ambiguous to the compiler:
template< typename T, typename HeadClass >
void callOnAllClasses(const T& val);
template< typename T, typename HeadClass, typename ... TailClasses >
void callOnAllClasses(const T& val);
Adding an extra template argument to the recursive case helps the compiler disambiguate between the variadic and non-variadic overloads when TailClasses is empty.
template< typename T, typename HeadClass, typename T1, typename ... TailClasses >
void callOnAllClasses(const T& val)
{
auto pObj = dynamic_cast<HeadClass*>(this);
if ( pObj )
pObj->HeadClass::doSomething(val);
callOnAllClasses<T, T1, TailClasses...>(val);
}
live example on godbolt

Related

unpack variadic template parameters

I have a template function:
template<typename R, typename... T>
void function(const std::string& id, R (*f)(T...)) {
switch (sizeof...(T)) {
case 0: f0compile<R>(reinterpret_cast<void (*)()>(f)); break;
case 1: f1compile<R, T>(reinterpret_cast<void (*)()>(f)); break;
case 2: f2compile<R, T1, T2>(reinterpret_cast<void (*)()>(f)); break;
}
...
}
How can I call these functions (f0compile, f1compile, f2compile) ? How can I write the "function" ?
template<typename R>
void f0compile(void (*f)()) {
new F0<R>(f):
...
}
template<typename R, typename T>
void f1compile(void (*f)()) {
new F1<R,T>(f);
...
}
template<typename R, typename T1, typename T2>
void f2compile(void (*f)()) {
new F2<R,T1,T2>(f);
...
}
Thank you for help with these variadic template.
I add the implementation of F0 F1 F2:
template <typename R> struct F0 : F {
F0(void (*_fn)()) : F(typeid(R))
, fn(reinterpret_cast<R(*)()>(_fn))
{}
const void* f() { res = fn(); return &res; }
R res; R (*fn)();
void d() { delete this; }
};
template <typename R, typename T> struct F1 : F {
F1(void (*_fn)(), F* _opd) : F(typeid(R))
, fn(reinterpret_cast<R(*)(T)>(_fn))
, opd(autocast<T>(_opd))
{}
const void* f() { res = fn(*(T*) opd->f()); return &res; }
F* opd;
R res; R (*fn)(T);
void d() { opd->d(); delete this; }
};
template <typename R, typename T1, typename T2> struct F2 : F {
F2(void (*_fn)(), F* _opd1, F* _opd2) : F(typeid(R))
, fn(reinterpret_cast<R(*)(T1,T2)>(_fn))
, opd1(autocast<T1>(_opd1))
, opd2(autocast<T2>(_opd2))
{}
const void* f() { res = fn(*(T1*) opd1->f(), *(T2*) opd2->f()); return &res; }
F* opd1; F* opd2;
R res; R (*fn)(T1,T2);
void d() { opd1->d(); opd2->d(); delete this; }
};
Thank you
struct F {
F(const std::type_info& _type) : type(_type) {}
virtual ~F() {}
const std::type_info& type;
virtual const void* f() = 0;
virtual void d() = 0;
};
Added class F . It rapresent each function / operand on the stack
template <typename T> struct Opd : F {
Opd(T _opd) : F(typeid(T)), res(_opd) { }
const void* f() { return &res; }
T res;
void d() { delete this; }
};
Added class Opd . It represent a specific operand on the stack.
The real program is this (simplified):
double foo(double op1, double op2) {
return op1 + op2;
}
#include <functional>
#include <stack>
#include <type_traits>
class Expression {
public:
struct F {
F(const std::type_info& _type) : type(_type) {}
virtual ~F() {}
const std::type_info& type;
virtual const void* f() = 0;
virtual void d() = 0;
};
public:
Expression() : m_cexpr(NULL) {}
~Expression() {
if (m_cexpr) m_cexpr->d();
}
// function
template<typename R, typename... T> void function(R (*f)(T...), void (*compile)(void (*)(), std::stack<F*>&)) {
m_f = std::make_pair(reinterpret_cast<void (*)()>(f), compile);
}
template<typename R, typename T1, typename T2> static void f2compile(void (*f)(), std::stack<F*>& s) {
auto opd2 = s.top();
s.pop();
auto opd1 = s.top();
s.pop();
s.push(new F2<R,T1,T2>(f, opd1, opd2));
}
void compile() {
if (m_cexpr) m_cexpr->d();
std::stack<F*> s;
s.push(new Opd<double>(1));
s.push(new Opd<double>(2));
m_f.second(m_f.first, s);
m_cexpr = s.top();
s.pop();
assert(s.empty());
}
void* execute() {
return const_cast<void*>(m_cexpr->f());
}
const std::type_info& type() {
return m_cexpr->type;
}
private:
F* m_cexpr;
std::pair<void (*)(), void (*)(void (*)(), std::stack<F*>&)> m_f;
template <typename T> struct Opd : F {
Opd(T _opd) : F(typeid(T)), res(_opd) {}
const void* f() { return &res; }
T res;
void d() { delete this; }
};
template <typename R, typename T1, typename T2> struct F2 : F {
F2(void (*_fn)(), F* _opd1, F* _opd2) : F(typeid(R))
, fn(reinterpret_cast<R(*)(T1,T2)>(_fn))
, opd1(_opd1)
, opd2(_opd2)
{}
const void* f() { res = fn(*(T1*) opd1->f(), *(T2*) opd2->f()); return &res; }
F* opd1; F* opd2;
R res; R (*fn)(T1,T2);
void d() { opd1->d(); opd2->d(); delete this; }
};
};
TEST_CASE("expression") {
Expression e;
e.function(foo, e.f2compile<double, double, double>);
e.compile();
e.execute();
REQUIRE(e.type() == typeid(double));
REQUIRE(*static_cast<double*>(e.execute()) == 3);
}
And my problem is how write better code c++11 using variadic template. How write a function "fNcompile" and a function "FN" with variadic template.
I don't think you need the variadic template. Instead:
template<typename R>
void fcompile(void (*f)()) {
new F0<R>(reinterpret_cast<void (*)()>(f));
...
}
template<typename R, typename T>
void fcompile(void (*f)(T)) {
new F1<R,T>(reinterpret_cast<void (*)()>(f));
...
}
template<typename R, typename T1, typename T2>
void fcompile(void (*f)(T1, T2)) {
new F1<R,T1,T2>(reinterpret_cast<void (*)()>(f));
...
}
Now you can call fcompile<some_type>(some_func) for any some_type and any nullary/unary/binary some_func which returns void.
To answer the specific question, below are variadic FN and fNcompile as close as possible to your existing code. First, though, since you said you're working in C++11, we'll need an equivalent of std::make_index_sequence from C++14. Here's a simple one. You can search for others that are smarter about being less likely to hit compiler template limitations...
namespace cxx_compat {
template <typename T, T... Values>
struct integer_sequence {
static constexpr std::size_t size() const
{ return sizeof...(Values); }
};
template <typename T, T Smallest, T... Values>
struct make_integer_sequence_helper {
static_assert(Smallest > 0,
"make_integer_sequence argument must not be negative");
using type = typename make_integer_sequence_helper<
T, Smallest-1, Smallest-1, Values...>::type;
};
template <typename T, T... Values>
struct make_integer_sequence_helper<T, 0, Values...> {
using type = integer_sequence<T, Values...>;
};
template <typename T, T N>
using make_integer_sequence =
typename make_integer_sequence_helper<T, N>::type;
template <std::size_t... Values>
using index_sequence = integer_sequence<std::size_t, Values...>;
template <std::size_t N>
using make_index_sequence = make_integer_sequence<std::size_t, N>;
template <typename... T>
using index_sequence_for = make_index_sequence<sizeof...(T)>;
} // end namespace cxx_compat
And now, the actual FN and fNcompile:
template <typename R, typename ...T> struct FN : F {
private:
template <typename T>
using any_to_Fstar = F*;
public:
FN(void (*_fn)(), any_to_Fstar<T> ... _opd) : F(typeid(R))
, fn(reinterpret_cast<R(*)(T...)>(_fn))
, opd{_opd...}
{}
FN(R (*_fn)(T...)) : F(typeid(R)), fn(_fn), opd() {}
const void* f() {
f_helper(cxx_compat::index_sequence_for<T...>{});
return &res;
}
std::array<F*, sizeof...(T)> opd;
R res; R (*fn)(T...);
void d() {
for (F* o : opd)
o->d();
delete this;
}
private:
template <std::size_t... Inds>
void f_helper(cxx_compat::index_sequence<Inds...>)
{ res = fn(*(T*) opd[Inds]->f() ...); }
};
template<typename R, typename... T>
static void fNcompile(void (*f)(), std::stack<F*>& s) {
auto* f_obj = new FN<R, T...>(f);
for (std::size_t ind = sizeof...(T); ind > 0;) {
f_obj->opd[--ind] = s.top();
s.pop();
}
s.push(f_obj);
}
What's going on:
To actually call the function pointer, we need access to a number of function arguments at the same time, so to replace the named members opd1, opd2 with a number of F* pointers determined by template instantiation, we use a std::array<F*, sizeof...(T)>, since sizeof...(T) is the number of argument types provided to the template.
For compatibility with the F2 constructor you declared, any_to_Fstar<T> ... _opd declares a number of constructor parameters to match the number of T template arguments, all with the same type F*. (But now fNcompile uses the additional constructor taking just the function pointer instead, and sets the array members afterward.)
To get at these pointers and pass them all to fn in one expression, we need to expand some sort of variadic pack. Here's where index_sequence comes in:
index_sequence_for<T...> is a type alias for index_sequence with a sequence of numbers counting up from zero as template arguments. For example, if sizeof...(T) is 4, then index_sequence_for<T...> is index_sequence<0, 1, 2, 3>.
f just calls a private function f_helper, passing it an object of that index_sequence_for<T...> type.
The compiler can deduce the template argument list for f_helper from matching the index_sequence types: Inds... must be that same sequence of numbers counting up from zero.
In the f_helper body, the expression fn(*(T*) opd[Inds]->f() ...) is instantiated by expanding both the template parameter packs T and Inds to get one list of function arguments for calling fn.
However, use of void pointers and reinterpret_cast is dangerous and rarely actually necessary in C++. There's almost always a safer way using templates. So I'd redesign this to be something more like:
#include <type_traits>
#include <typeinfo>
#include <stdexcept>
#include <memory>
#include <stack>
namespace cxx_compat {
// Define integer_sequence and related templates as above.
template <typename T, typename... Args>
std::unique_ptr<T> make_unique(Args&& ... args)
{
return std::unique_ptr<T>(new T(std::forward<Args>(args)...));
}
} // end namespace cxx_compat
class bad_expression_type : public std::logic_error
{
public:
bad_expression_type(const std::type_info& required,
const std::type_info& passed)
: logic_error("bad_argument_type"),
required_type(required),
passed_type(passed) {}
const std::type_info& required_type;
const std::type_info& passed_type;
};
class Expression
{
public:
class F
{
public:
F() noexcept = default;
F(const F&) = delete;
F& operator=(const F&) = delete;
virtual ~F() = default;
virtual const std::type_info& type() const noexcept = 0;
virtual void compile(std::stack<std::unique_ptr<F>>&) = 0;
template <typename R>
R call_R() const;
};
using F_ptr = std::unique_ptr<F>;
using F_stack = std::stack<F_ptr>;
template <typename R>
class Typed_F : public F
{
public:
const std::type_info& type() const noexcept override
{ return typeid(R); }
virtual R call() const = 0;
};
// Accepts any callable: function pointer, lambda, std::function,
// other class with operator().
template <typename R, typename... T, typename Func,
typename = typename std::enable_if<std::is_convertible<
decltype(std::declval<const Func&>()(std::declval<T>()...)),
R>::value>::type>
void function(Func func)
{
store_func<R, T...>(std::move(func));
}
// Overload for function pointer that does not need explicit
// template arguments:
template <typename R, typename... T>
void function(R (*fptr)(T...))
{
store_func<R, T...>(fptr);
}
template <typename T>
void constant(const T& value)
{
store_func<T>([value](){ return value; });
}
void compile(F_stack& stack)
{
m_cexpr->compile(stack);
}
private:
template <typename Func, typename R, typename... T>
class F_Impl : public Typed_F<R>
{
public:
F_Impl(Func func) : m_func(std::move(func)) {}
void compile(F_stack& stack) override {
take_args_helper(stack, cxx_compat::index_sequence_for<T...>{});
}
R call() const override {
return call_helper(cxx_compat::index_sequence_for<T...>{});
}
private:
template <typename Arg>
int take_one_arg(std::unique_ptr<Typed_F<Arg>>& arg, F_stack& stack)
{
auto* fptr = dynamic_cast<Typed_F<Arg>*>(stack.top().get());
if (!fptr)
throw bad_expression_type(
typeid(Arg), stack.top()->type());
arg.reset(fptr);
stack.top().release();
stack.pop();
return 0;
}
template <std::size_t... Inds>
void take_args_helper(F_stack& stack, cxx_compat::index_sequence<Inds...>)
{
using int_array = int[];
(void) int_array{ take_one_arg(std::get<Inds>(m_args), stack) ..., 0 };
}
template <std::size_t... Inds>
R call_helper(cxx_compat::index_sequence<Inds...>) const {
return m_func(std::get<Inds>(m_args)->call()...);
}
Func m_func;
std::tuple<std::unique_ptr<Typed_F<T>>...> m_args;
};
template <typename R, typename... T, typename Func>
void store_func(Func func)
{
m_cexpr = cxx_compat::make_unique<F_Impl<Func, R, T...>>(
std::move(func));
}
F_ptr m_cexpr;
};
template <typename R>
R Expression::F::call_R() const
{
auto* typed_this = dynamic_cast<const Typed_F<R>*>(this);
if (!typed_this)
throw bad_expression_type(typeid(R), type());
return typed_this->call();
}
TEST_CASE("expression") {
Expression a;
a.constant(1.0);
Expression b;
b.constant(2.0);
Expression c;
c.function(+[](double x, double y) { return x+y; });
Expression::F_stack stack;
a.compile(stack);
REQUIRE(stack.size() == 1);
b.compile(stack);
REQUIRE(stack.size() == 2);
c.compile(stack);
REQUIRE(stack.size() == 1);
REQUIRE(stack.top() != nullptr);
REQUIRE(stack.top()->type() == typeid(double));
REQUIRE(stack.top()->call_R<double>() == 3.0);
}
It would also be possible, but a bit tricky, to support reference and const variations of the argument and result types, for example, using a std::string(*)() function as an argument to an unsigned int(*)(const std::string&) function.

How to call conditionally B::f only if derived from B in C++11?

In case when static polymorphism is used, especially in templates (e.g. with policy/strategy pattern), it may be required to call base function member, but you don't know was instantiated class actually derived from this base or not.
This easily can be solved with old good C++ ellipsis overload trick:
#include <iostream>
template <class I>
struct if_derived_from
{
template <void (I::*f)()>
static void call(I& x) { (x.*f)(); }
static void call(...) { }
};
struct A { void reset() { std::cout << "reset A" << std::endl; } };
struct B { void reset() { std::cout << "reset B" << std::endl; } };
struct C { void reset() { std::cout << "reset C" << std::endl; } };
struct E: C { void reset() { std::cout << "reset E" << std::endl; } };
struct D: E {};
struct X: A, D {};
int main()
{
X x;
if_derived_from<A>::call<&A::reset>(x);
if_derived_from<B>::call<&B::reset>(x);
if_derived_from<C>::call<&C::reset>(x);
if_derived_from<E>::call<&E::reset>(x);
return 0;
}
The question is:
Is there any better simple way (e.g. SFINAE doesn't look so) to achieve same result in C++11/C++14?
Would empty call of ellipsis parameter function be elided by optimizing compiler? Hope such case is not special against any "normal" function.
One option is to introduce two overloads of different priorities and to equip the preferred one with an expression SFINAE.
#include <utility>
template <typename T, typename... Args, typename C, typename R, typename... Params>
auto call_impl(int, R(C::*f)(Args...), T&& t, Params&&... params)
-> decltype((std::forward<T>(t).*f)(std::forward<Params>(params)...))
{
return (std::forward<T>(t).*f)(std::forward<Params>(params)...);
}
template <typename T, typename... Args, typename C, typename R, typename... Params>
void call_impl(char, R(C::*)(Args...), T&&, Params&&...)
{
}
template <typename T, typename... Args, typename C, typename R, typename... Params>
auto call(R(C::*f)(Args...), T&& t, Params&&... params)
-> decltype(call_impl(0, f, std::forward<T>(t), std::forward<Params>(params)...))
{
return call_impl(0, f, std::forward<T>(t), std::forward<Params>(params)...);
}
Test:
int main()
{
X x;
call(&B::reset, x);
}
DEMO
The upper function will be selected first by overload resolution (due to an exact match of 0 against int), and possibly excluded from the set of viable candidates if (t.*f)(params...) is not valid. In the latter case, the call to call_impl falls back to the second overload, which is a no-op.
Given that &A::reset may fail for multiple reasons, and you may not necessarily want to explicitly specify the function's signature, and, on top of that, you want the call to fail if the member function exists, but it does not match function call arguments, then you can exploit generic lambdas:
#include <utility>
#include <type_traits>
template <typename B, typename T, typename F
, std::enable_if_t<std::is_base_of<B, std::decay_t<T>>{}, int> = 0>
auto call(T&& t, F&& f)
-> decltype(std::forward<F>(f)(std::forward<T>(t)))
{
return std::forward<F>(f)(std::forward<T>(t));
}
template <typename B, typename T, typename F
, std::enable_if_t<!std::is_base_of<B, std::decay_t<T>>{}, int> = 0>
void call(T&& t, F&& f)
{
}
Test:
int main()
{
X x;
call<A>(x, [&](auto&& p) { return p.A::reset(); });
call<B>(x, [&](auto&& p) { return p.B::reset(); });
}
DEMO 2
what about something like:
#include <iostream>
#include <type_traits>
struct A { void reset() { std::cout << "reset A" << std::endl; } };
struct B { void reset() { std::cout << "reset B" << std::endl; } };
struct X :public A{};
template <typename T, typename R, typename BT>
typename std::enable_if<std::is_base_of<BT, T>::value, R>::type
call_if_possible(T & obj, R(BT::*mf)())
{
return (obj.*mf)();
}
template <typename T, typename R, typename BT>
typename std::enable_if<!std::is_base_of<BT, T>::value, R>::type
call_if_possible(T & obj, R(BT::*mf)()) { }
int main()
{
X x;
call_if_possible(x, &A::reset);
call_if_possible(x, &B::reset);
}
ideone
edit
maybe more readable way:
template <typename T, typename R, typename BT>
R call_if_possible_impl(T & obj, R(BT::*mf)(), std::false_type){}
template <typename T, typename R, typename BT>
R call_if_possible_impl(T & obj, R(BT::*mf)(), std::true_type)
{
return (obj.*mf)();
}
template <typename T, typename R, typename BT>
R call_if_possible(T & obj, R(BT::*mf)())
{
return call_if_possible_impl(obj, mf, typename std::is_base_of<BT, T>::type());
}
ideone
Basing on previously provided answers by #PiotrSkotnicki and #relaxxx I would like to combine the most simple and readable solution, without SFINAE and other blood-from-the-eyes things. It's just for reference, will not be accepted anyway:
#include <iostream>
#include <type_traits>
template <class Base, class Derived>
using check_base = typename std::is_base_of<Base, Derived>::type;
template <class Base, class Derived, typename Func>
void call(Derived& d, Func&& f)
{
call<Base>(d, std::forward<Func>(f), check_base<Base, Derived>());
}
template <class Base, typename Func>
void call(Base& b, Func&& f, std::true_type)
{
f(b);
}
template <class Base, class Derived, typename Func>
void call(Derived&, Func&&, std::false_type)
{
}
struct A { void reset(int i) { std::cout << "reset A: " << i << std::endl;} };
struct B { void reset() { std::cout << "reset B" << std::endl;} };
struct C { void reset() { std::cout << "reset C" << std::endl;} };
struct E: C { void reset() { std::cout << "reset E" << std::endl;} };
struct D: A, E {};
int main()
{
D d;
int i = 42;
call<A>(d, [&](auto& p) { p.reset(i); } );
call<B>(d, [](auto& p) { p.reset(); } );
call<C>(d, [](auto& p) { p.reset(); } );
call<E>(d, [](auto& p) { p.reset(); } );
}
Live at: http://cpp.sh/5tqa

Specialize class for Base and all Derived

I have a Base-class and a derived class. I would like to design a Class C that reacts differently when a base or derived object is handed over like this:
template<typename T>
class Base {};
class Derived: public Base<int>{}
template<typename T>
class C
{
public:
static int f(T&& arg) {return 1;};
};
// Specialize C, not the function
template<typename T>
class C<T>
{
public:
static int f(T&& arg) {return 2;};
};
template<typename T>
int function(T&& arg) {
return C<T>::f(std::forward<T>(arg));
}
int main()
{
std::cout << function(1) << std::endl;
std::cout << function(Base<float>()) << std::endl;
std::cout << function(Derived()) << std::endl;
}
This is supposed to print:
1
2
2
Is there a way that I can do this with specializing the class C and not the function. In reality I just hand over the type C and everything happens out of my reach.
To clearify, the implementation of function may look like that but I can't really tell. All I know is the signature of f: int f(T&&); And that I have to have a class C that has a function of that signature.
Through Specialization of function():
Example code first:
template< class T >
struct Base
{};
struct Child : Base<int>
{ };
template<typename T>
class C1
{
public:
static int f(T&& arg) {return 1;}
};
template<typename T>
class C2
{
public:
static int f(T&& arg) {return 2;}
};
template<typename T>
typename std::enable_if< std::is_base_of<Base<int>, T>::value, int >::type function(T&& arg)
{
return C2<T>::f(std::forward<T>(arg));
}
template<typename T>
typename std::enable_if< !std::is_base_of<Base<int>, T >::value, int >::type function(T&& arg)
{
return C1<T>::f(std::forward<T>(arg));
}
int main()
{
std::cout << function(1) << std::endl;
std::cout << function(Base<int>()) << std::endl;
std::cout << function(Child()) << std::endl;
return 0;
}
Why:
So the split here is that I've written two versions of function(). The return type of function() looks a little hairy but the first version will be compiled if T has Base< int > as a base type. Otherwise the second version will be used.
std::enable_if<> is a neat tool which causes the template expansion to fail if its condition is not met. In this case we cause the template expansion to fail if T does not have Base< int > as a base class. In the case that the template expansion fails the compiler continues to look for another version of function() which will compile.
C1 and C2 are then two different objects with C2 being used if T has Base as a child, otherwise C1 is used.
Through Specialization of C:
template< class BaseArgument >
Base< BaseArgument > get_Base_type_of( Base< BaseArgument >&& );
template< class T >
struct has_Base
{
typedef char yes;
typedef long no;
template< typename T1 >
static yes Check( decltype( get_Base_type_of( declval<T1>() ) )* );
template< typename T1 >
static no Check( ... );
static const bool value = sizeof( decltype( Check<T>(0) ) ) == sizeof( yes ) ;
};
template< typename T, typename Placeholder = void >
class C
{
public:
static int f(T&& arg) {return 1;};
};
template<typename T>
class C< T, typename std::enable_if< has_Base<T>::value >::type >
{
public:
static int f(T&& arg) {return 2;};
};

templated iterator passing <N> to callback function

I am trying to run a compile-time iterator like:
meta::reverse_iterator<2, 9>::iterate(callback());
meta::reverse_iterator<4, 7>::iterate(callback());
std::cout << "-----------------" << std::endl;
meta::iterator<2, 9>::iterate(callback());
meta::iterator<4, 7>::iterate(callback());
struct callback {
template <int i>
void operator()() {
std::cout << "print !!" << i << std::endl;
}
};
and this is how I've written the meta-iterator:
namespace meta {
template <int Begin, int End, bool done = false>
struct reverse_iterator {
template <typename F>
static void iterate(F f) {
f.template operator()<End>();
reverse_iterator<Begin, End-1, Begin == End-1>::iterate(f);
}
};
template <int Begin, int End>
struct reverse_iterator<Begin, End, true> {
template <typename F>
static void iterate(F) {}
};
template <int Begin, int End, bool done = false>
struct iterator {
template <typename F>
static void iterate(F f) {
iterator<Begin, End - 1, Begin == End - 1>::iterate(f);
f.template operator()<End - 1>();
}
};
template <int Begin, int End>
struct iterator<Begin, End, true> {
template <typename F>
static void iterate(F) {}
};
}
Right now I iterator calls operator()<N> But I want it to be able to call any arbitrary function supplied by user with template parameter <N> (not as run-time argument) How can that be achieved ?
also boost::bind doesn't work with it as it calls the function object of bind instead of the real function. So there should be some way to carry default parameters to the supplied functions.
AFAIK you can't do it directly, but if you really want it you can use a traits class that call member function for you, and for different member functions write different traits:
struct function_operator_call_trait {
template< int N, class T >
void call( T& t ) {t.template operator()<N>();}
};
template< class Arg1, class Arg2 >
struct foo_call_trait {
foo_call_trait( Arg1&& arg1 ) : a1( std::move(arg1) ) {}
foo_call_trait( Arg1&& arg1, Arg2&& arg2 )
: a1( std::move(arg1) ), a2( std::move(arg2) ) {}
template< class T, int N >
void call( T& t ) {t.template foo<N>(a1, a2);}
Arg1 a1;
Arg2 a2;
};
template <int Begin, int End, class traits = function_operator_call_trait, bool done = false>
struct iterator{
iterator() {}
iterator( traits const& t ) : t_( t ) {}
iterator( traits&& t ) : t_( std::move(t) ) {}
template<typename F>
static void iterate(F f){
iterator<Begin, End-1, traits, Begin == End-1>::iterate(f);
t_.call<End-1>( f );
}
traits t_;
};
typedef iterator<2, 9> fc_iterator;
typedef foo_call_trait<int, float> foo_traits;
typedef iterator<2, 9, foo_traits> foo_iterator( foo_traits(1, 2) );

Recursively unhide base class members

I try to write a class that takes a tuple of functions as its argument and overloads operator() for all argument_types of the function. Right now this looks like this:
template<typename T>
struct holder {
T t;
};
template<typename T, std::size_t i>
struct overload_helper : public holder<T>, public overload_helper<T, i - 1> {
overload_helper(T t) : holder<T>({t}) {};
typedef typename std::tuple_element<i - 1, T>::type inner_type;
typename inner_type::result_type operator()(typename inner_type::argument_type x) {
return std::get<i - 1>(holder<T>::t)(x); }
};
template<typename T>
struct overload_helper<T, 1> {
typedef typename std::tuple_element<0 ,T>::type inner_type;
typename inner_type::result_type operator()(typename inner_type::argument_type x) {
return std::get<0>(holder<T>::t)(x); }
};
template<typename T>
struct overload : public overload_helper<T, std::tuple_size<T>::value>
{
typedef void result_type;
overload(const T& t) : overload_helper<T, std::tuple_size<T>::value>(t) { }
};
int main() {
auto all = make_overload(std::make_tuple(
std::function<void(double)>([](double i) {
std::cout << "double" << std::endl;
}), std::function<void(int)>([](int i) {
std::cout << "int" << std::endl; })));
all(1); //fails
all(1.0);
}
The problem is that the base class is hiding each recursive definition of operator(). Is it possible to recursively unhide all definitions with using or is the only way to have a templated operator() and pick the right overload with boost::mpl?
using overload_helper<T, i - 1>::operator() in each overload_helper should do the job, so long they aren't ambiguous.