Explicit member function specialisation - c++11

I am unable to specialize the template member function below. I have looked at the solution given to answer similar question on SOF but the solution that is proposed is same as the code I have below but it does not seem to work. I am missing something for sure.
enum EStep
{
eStep1, eStep2, eStep3
};
template<int16_t iDevice>
struct Device
{
template<EStep step>
static constexpr bool isType() { return false; }
};
template<> template<>
constexpr bool Device<int16_t>::isType<eStep1>()
{
return true;
}

template<> template<>
constexpr bool Device<int16_t>::isType<eStep1>()
{
return true;
}
Device is a template of int16_t, so to specialize it, you'd need to provide an int16_t value as the template argument. e.g.
template<> template<>
constexpr bool Device<999>::isType<eStep1>()

Related

refactor - remove CRPT

I have custom List class like this:
template<class T>
struct iList{
constexpr bool empty() const{
return static_cast<const T *>(this)->size() == 0;
}
// several methods like empty(), some are non const.
};
// there are several classes like this one with different implementations
struct MyList : iList<MyList>{
constexpr unsigned size() const{
return 5;
}
};
int main(){
constexpr MyList list;
static_assert( ! list.empty() );
}
code is constexpr and simplified so I can use static_assert.
I want to get rid of CRTP pattern, because templates become way too ugly. However I do not want to use virtual functions.
One way I am thinking is following:
struct iList2{
protected:
template<class LIST>
constexpr static bool empty_(const LIST &list){
return list.size() == 0;
}
};
struct MyList2 : iList2{
constexpr unsigned size() const{
return 5;
}
constexpr bool empty() const{
return empty_(*this);
}
};
int main(){
constexpr MyList2 list2;
static_assert( ! list2.empty() );
}
In this case I need to call empty_ static member function in each list.
Another way is to use mixin, e.g. to inherit MyList and add empty() method. If I do this as templated class, I can even make typedef. However I do not like this approach as well.
Is there any possibility I am missing?

Checking for method parameter at runtime with SFINAE

I know one could check the existence of a particular method using expression SFINAE in C++11 as follows.
What I can't find though, is an example to do the same, checking method arguments as well. In particular I would like to match a method that takes a const parameter.
#include <iostream>
struct A
{
void method() const
{
return;
}
};
template <typename T, typename = std::string>
struct hasMethod
: std::false_type
{
};
template <typename T>
struct hasMethod<T, decltype(std::declval<T>().method())>
: std::true_type
{ };
int main() {
std::cout << hasMethod<A>::value << std::endl;
}
In reality I would like the hasMethod:: to match
void method(const Type& t) const
{
return;
}
What is the syntax to pass to decltype?
I have tried:
struct hasMethod<T, decltype(std::declval<T>().method(const int&))>
: std::true_type
but it obviously doesn't work.

VS2015 internal compiler error when calling base class constexpr method

the following code produces an internal compiler error (VS2015)
struct A
{
constexpr A(){}
constexpr int bar()
{
return 3;
}
};
struct B : A
{
constexpr B(){}
constexpr int foo()
{
return A::bar();
}
};
int main()
{
constexpr B b;
constexpr int dummy = b.foo();
return 1;
}
However, if i'd remove the A:: qualifier:
constexpr int foo()
{
return bar();
}
it will be compiled.
problem arises when these methods have the same name, and I need to invoke the base class method. (e.g. when using recursive template inheritence)
any workarounds?
The actual problem is b is declared as const (constexpr implies const on objects) and you are trying to call non-const (since C++14, constexpr doesn't imply const on methods, see here) method with the const object...
According to the standard, you should not be able to solve the problem by simply removing A:: nor by the static_cast the way you did. Pre-RTM version of Visual Studio 2015 allows you to do this only because its support for constexpr is preliminary and very buggy. C++11 constexpr (but unfortunately not C++14 extended constexpr) expected to be fully supported in the RTM version of VS 2015 (see here).
The correct version of your code is:
struct A
{
constexpr A(){}
constexpr int bar() const
{
return 3;
}
};
struct B : A
{
constexpr B(){}
constexpr int foo() const
{
return A::bar();
}
};
int main()
{
constexpr B b;
constexpr int dummy = b.foo();
return 1;
}
Found a solution.
"this" should be casted to const A*:
struct B : A
{
constexpr B(){}
constexpr int foo()
{
return static_cast<const A*>(this)->bar();
}
};
Also works when the methods have the same name.

Is it possible to have a branch of code conditionally be compiled in C++?

Title says it all really. Sample code which illustrates the spirit of the affaire:
if( std::is_constructible<T, unsigned long>::value )
{
unsigned long identity = collection.rbegin()->first + 1;
std::shared_ptr<T> newObject(new T(identity));
collection.insert( identity , newObject );
return true;
}
else
{
return false;
}
Tag dispatch.
template<class T>
bool foo_impl(std::true_type){
unsigned long identity = collection.rbegin()->first + 1;
std::shared_ptr<T> newObject(new T(identity));
collection.insert( identity , newObject );
return true;
}
template<class T>
bool foo_impl(std::false_type){
return false;
}
template<class T>
bool foo(){
return foo_impl<T>(std::is_constructible<T, unsigned long>());
}
Since the if statement can be determined at compile time, I'd expect the compiler to be smart and optimize it directly, just as if you had something like
if ( true ) {
// Some code
}
else {
// Anything here any decent compiler will ignore.
}
Another option is to wrap the behaviour you want in a function, and use std::enable_if:
template <typename T, typename = typename enable_if<is_constructible<T, int>::value>::type>
bool foo() {
return true;
}
template <typename T, typename = typename enable_if<!is_constructible<T, int>::value>::type>
bool foo() {
return false;
}
// ...
return foo<T>();
Example: http://ideone.com/sgNVr5
Yet another option is to specialize on the boolean value:
template <bool b> bool foo();
template <>
bool foo<true>(){
return true;
}
template <>
bool foo<false>() {
return false;
}

Is there a way to print a constexpr string during compiletime?

I'm trying to do the following (only relevant parts of code below):
template<typename ContainerType>
struct IsContainerCheck : is_container<ContainerType>
{
static constexpr char* err_value = "Type is not a container model";
};
namespace _check_concept {
template<typename ResultType>
struct run {
constexpr static int apply() {
static_assert(false, IsContainerCheck<ResultType>::err_value)
return 0;
}
};
template<>
struct run<true_t> {
constexpr static int apply() {
return 0;
}
};
}
This fails because the static_assert allows only literals to be printed. The same is with BOOST_STATIC_ASSERT_MSG macro.
So my question is - is there any way to output a constexpr string during compilation?
If there is a gcc extension providing this functionality that would also be great.
Used compiler gcc 4.8.1
GCC does not provide such a mechanism as you want. However you will not need
it if you are able to refactor your code somewhat as illustrated in the
following program. (I have filled in a few gaps so as to give us a
compilable example):
#include <type_traits>
#include <vector>
template<typename ContainerType>
struct is_container
{
static bool const value = false;
};
template<>
struct is_container<std::vector<int>>
{
static bool const value = true;
};
template<typename ContainerType>
struct IsContainerCheck // : is_container<ContainerType> <- Uneccessary
{
static_assert(is_container<ContainerType>::value,
"Type is not a container model");
};
namespace _check_concept {
template<typename ResultType>
struct run {
constexpr static int apply() {
return (IsContainerCheck<ResultType>(),0);
}
};
// No such specialization is necessary. Delete it.
// template<>
// struct run<true_t> {
// constexpr static int apply() {
// return 0;
// }
//};
}
using namespace _check_concept;
int main(int argc, char **argv)
{
auto verdict0 = run<std::vector<int>>::apply();
(void)verdict0;
// The following line will static_assert: "Type is not a container model"
auto verdict1 = run<float>::apply();
(void)verdict1;
return 0;
}
In your specialization _check_concept::struct run<true_t> I presume that
true_t is not an alias or equivalent of std::true_type, but rather
just a place-holder for some ResultType that is a container type. As
the test program shows, no such specialization is now necessary, because
IsContainerCheck<ResultType>() will static_assert, or not, depending
on ResultType, in the unspecialized run<ResultType>::apply().
I had some time (and a good liqueur to come along with it) to think more about the problem. This is what I came up with:
namespace _details {
struct PassedCheck {
constexpr static int printError () {
return 0; //no error concept check passed
}
};
template<template<typename> class ConceptCheck, typename ...ModelTypes>
struct check_concept_impl;
template<template<typename> class ConceptCheck, typename FirstType, typename ...ModelTypes>
struct check_concept_impl<ConceptCheck, FirstType, ModelTypes...> : mpl::eval_if< typename ConceptCheck<FirstType>::type,
check_concept_impl<ConceptCheck, ModelTypes...>,
mpl::identity<ConceptCheck<FirstType>>>
{ };
template<template<typename> class ConceptCheck, typename LastType>
struct check_concept_impl<ConceptCheck, LastType> : mpl::eval_if<typename ConceptCheck<LastType>::type,
mpl::identity<PassedCheck>,
mpl::identity<ConceptCheck<LastType>>>
{ };
}
template<template<typename> class ConceptCheck, typename ...ModelTypes>
struct check_concept {
private:
typedef typename _details::check_concept_impl<ConceptCheck, ModelTypes...>::type result_type;
public:
// the constexpr method assert produces shorter, fixed depth (2) error messages than a nesting assert in the trait solution
// the error message is not trahsed with the stack of variadic template recursion
constexpr static int apply() {
return result_type::printError();
}
};
template<typename ContainerType>
struct IsContainerCheck : is_container<ContainerType>
{
template<typename BoolType = false_t>
constexpr static int printError () {
static_assert(BoolType::value, "Type is not a container model");
return 0;
}
};
and the usage:
check_concept<IsContainerCheck, std::vector<int>, std::vector<int>, float, int>::apply();
The solution is probably not the most elegant one but I it keeps the assert message short:
In file included from ../main.cpp:4:0:
../constraint.check.hpp: In instantiation of ‘static constexpr int IsContainerCheck::printError() [with BoolType = std::integral_constant; ContainerType = float]’:
../constraint.check.hpp:61:34: required from ‘static constexpr int check_concept::apply() [with ConceptCheck = IsContainerCheck; ModelTypes = {std::vector >, std::vector >, float, int}]’
../main.cpp:25:83: required from here
../constraint.check.hpp:74:3: error: static assertion failed: Type is not a container model
static_assert(BoolType::value, "Type is not a container model");
The assert is issued in a constexpr method after the check_concept template specialization has been done. Embedding the static assert directly into the template class definition would drag the whole check_concept_impl recursion stack into the error message.
So changing the IsContainerCheck trait to something like (rest of the changes omitted for readibility):
template<typename ContainerType>
struct IsContainerCheck
{
static_assert(is_container<ContainerType>::type::value, "Type is not a container model");
};
would yield an error
../constraint.check.hpp: In instantiation of ‘struct IsContainerCheck’:
../constraint.check.hpp:36:9: required from ‘struct _details::check_concept_impl’
/usr/include/boost/mpl/eval_if.hpp:38:31: required from ‘struct boost::mpl::eval_if, _details::check_concept_impl, boost::mpl::identity > > >’
../constraint.check.hpp:36:9: required from ‘struct _details::check_concept_impl >, float, int>’
/usr/include/boost/mpl/eval_if.hpp:38:31: required from ‘struct boost::mpl::eval_if, _details::check_concept_impl >, float, int>, boost::mpl::identity > > >’
../constraint.check.hpp:36:9: required from ‘struct _details::check_concept_impl >, std::vector >, float, int>’
../constraint.check.hpp:53:84: required from ‘struct check_concept >, std::vector >, float, int>’
../main.cpp:25:81: required from here
../constraint.check.hpp:72:2: error: static assertion failed: Type is not a container model
static_assert(is_container::type::value, "Type is not a container model");
As you can see each recursive eval_if call is emended in the error description which is bad because it makes the error message dependent from the amount and type of template parameters.

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