I'm trying to get this to work:
template<class Type>
typename boost::enable_if< boost::mpl::or_<
boost::is_arithmetic<Type>,
is_string<Type> > >::type
get(const std::string &argPath, const Type &argDefault) {
bool caught = false;
std::stringstream ss;
Type value;
try {
value = ptree_.get<Type>(argPath);
} catch(ptree_bad_path &e) {
caught = true;
}
if(caught)
value = argDefault;
ss << value;
parameters_.insert(std::pair<std::string, std::string>(argPath, ss.str()));
return value;
}
I used the following is_string type trait: Type trait for strings
My goal is to restrict my Type to string or arithmetic type so that I can push it to my stringstream.
So this builds, but when I try to use it, it returns the following errors:
error: void value not ignored as it ought to be
In member function ‘typename
boost::enable_if,
is_string, mpl_::bool_, mpl_::bool_,
mpl_::bool_ >, void>::type FooClass::get(const std::string&,
const Type&) [with Type = uint8_t]’
error: return-statement with a value, in function returning 'void'
Here is how I try to use it:
FooClass f;
item_value = f.get("tag1.tag2.item", DEFAULT_ITEM_VALUE);
Any help is appreciated, thanks in advance!
From http://www.boost.org/doc/libs/1_53_0/libs/utility/enable_if.html, enable_if has a second parameter that defaults to void:
template <bool B, class T = void>
struct enable_if_c {
typedef T type;
};
Seems to me you need to include the return type in your enable_if. (It is defaulting to void now.)
template<class Type>
typename boost::enable_if< boost::mpl::or_<
boost::is_arithmetic<Type>,
is_string<Type> >,
Type >::type
get(const std::string &argPath, const Type &argDefault);
Related
Consider a struct that contains two types - FirstObjects and SecondObjects, which are both std::tuple<>, e.g. something like:
struct Objects
{
using FirstObjects = std::tuple<int, int>;
using SecondObjects = std::tuple<int>;
};
For convenience we add the following enum:
enum class ObjectCategory
{
FIRST,
SECOND
};
Now consider the following class that is templated on a type such as Objects as described above (the only contract is that it have FirstObjects and SecondObjects and that they be std::tuple):
template <typename T>
class SomeClass
{
public:
using FirstObjects = typename T::FirstObjects;
using SecondObjects = typename T::SecondObjects;
template <std::size_t Idx>
using FirstObject = typename std::tuple_element<Idx, FirstObjects>::type;
template <std::size_t Idx>
using SecondObject = typename std::tuple_element<Idx, SecondObjects>::type;
template <ObjectCategory Category, std::size_t Idx>
using ObjectType = std::conditional_t<Category == ObjectCategory::FIRST, FirstObject<Idx>, SecondObject<Idx>>;
template <ObjectCategory Category, std::size_t Idx>
std::enable_if_t<Category == ObjectCategory::FIRST, const ObjectType<Category, Idx>>& getObject()
{
return std::get<Idx>(firstObjects_);
}
template <ObjectCategory Category, std::size_t Idx>
std::enable_if_t<Category == ObjectCategory::SECOND, const ObjectType<Category, Idx>>& getObject()
{
return std::get<Idx>(secondObjects_);
}
template <ObjectCategory Category, std::size_t Idx>
void doSomething()
{
const ObjectType<Category, Idx>& obj = getObject<Category, Idx>();
}
private:
FirstObjects firstObjects_;
SecondObjects secondObjects_;
};
In brief, SomeClass hosts two member variables, firstObjects_ of type T::FirstObjects, and secondObjects_ of type T::SecondObjects, and has a doSomething() function which makes use of the SFINAE-overloaded getObject() getter method that returns the i-th object contained in firstObjects_ or in secondObjects_ depending on the ObjectCategory chosen.
The problem I have now is:
int main()
{
struct Objects
{
using FirstObjects = std::tuple<int, int>;
using SecondObjects = std::tuple<int>;
};
SomeClass<Objects> object{};
object.doSomething<ObjectCategory::FIRST, 0>(); // <------ works fine
object.doSomething<ObjectCategory::FIRST, 1>(); // <------ compiler error
}
The last line makes the compiler complain:
/usr/include/c++/4.9/tuple: In instantiation of 'struct std::tuple_element<1ul, std::tuple<int> >':
70:114: required by substitution of 'template<class T> template<ObjectCategory Category, long unsigned int Idx> using ObjectType = std::conditional_t<(Category == FIRST), typename std::tuple_element<Idx, typename T::FirstObjects>::type, typename std::tuple_element<Idx, typename T::SecondObjects>::type> [with ObjectCategory Category = (ObjectCategory)0; long unsigned int Idx = 1ul; T = main()::Objects]'
87:42: required from 'void SomeClass<T>::doSomething() [with ObjectCategory Category = (ObjectCategory)0; long unsigned int Idx = 1ul; T = main()::Objects]'
104:50: required from here
/usr/include/c++/4.9/tuple:682:12: error: invalid use of incomplete type 'struct std::tuple_element<0ul, std::tuple<> >'
struct tuple_element<__i, tuple<_Head, _Tail...> >
^
In file included from /usr/include/c++/4.9/tuple:38:0,
from 4:
/usr/include/c++/4.9/utility:85:11: error: declaration of 'struct std::tuple_element<0ul, std::tuple<> >'
class tuple_element;
^
In substitution of 'template<class T> template<ObjectCategory Category, long unsigned int Idx> using ObjectType = std::conditional_t<(Category == FIRST), typename std::tuple_element<Idx, typename T::FirstObjects>::type, typename std::tuple_element<Idx, typename T::SecondObjects>::type> [with ObjectCategory Category = (ObjectCategory)0; long unsigned int Idx = 1ul; T = main()::Objects]':
87:42: required from 'void SomeClass<T>::doSomething() [with ObjectCategory Category = (ObjectCategory)0; long unsigned int Idx = 1ul; T = main()::Objects]'
104:50: required from here
70:114: error: no type named 'type' in 'struct std::tuple_element<1ul, std::tuple<int> >'
It seems like it is not liking the fact that SecondObjects only has one element and therefore std::get<1>(secondObjects_) is not working? But why is this even happening in the first place since I'm calling doSomething on ObjectCategory::FIRST and not ObjectCategory::SECOND?
How can this be solved?
------ EDIT 1 ------
Solution pointed out by #Taekahn is to change the getObject() method as follows:
template <ObjectCategory Category, std::size_t Idx>
const auto& getObject()
{
return std::get<Idx>(firstObjects_);
}
template <ObjectCategory Category, std::size_t Idx, typename = std::enable_if_t<Category == ObjectCategory::SECOND>>
const auto& getObject()
{
return std::get<Idx>(secondObjects_);
}
It seems like the above works because the compiler does not evaluate / check the validity of the return type of a method that has been SFINAE'd out when one uses the keyword auto instead of the explicit type.
But this is only conjecture - would really appreciate a more knowledgeable C++ practitioner to weigh in here!
------ EDIT 2 ------
The above results in ambiguity when trying to do
object.doSomething<ObjectCategory::SECOND, 0>();
The WAR I've found so far is to modify the first getter function to
template <ObjectCategory Category, std::size_t Idx, typename = std::enable_if_t<Category == ObjectCategory::FIRST>, bool = 0 /* dummy param */>
auto& getObject()
{
return std::get<Idx>(firstObjects_);
}
i.e. to add the SFINAE as well as a dummy template arg at the end to allow overloading.
The following code compiles perfectly if:
I don't include <iostream> or
I name operator== as alp::operator==.
I suppose there is a problem with <iostream> and operator==, but I don't know what.
I compile the code with gcc 7.3.0, clang++-6.0 and goldbolt. Always the same error.
The problem is that the compiler is trying to cast the parameters of operator== to const_iterator, but why? (I suppose the compiler doesn't see my version of operator==, and looks for other versions).
#include <vector>
#include <iostream> // comment and compile
namespace alp{
template <typename It_base>
struct Iterator {
using const_iterator = Iterator<typename It_base::const_iterator>;
operator const_iterator() { return const_iterator{}; }
};
template <typename It_base>
bool operator==(const Iterator<It_base>& x, const Iterator<It_base>& y)
{ return true;}
}// namespace
struct Func{
int& operator()(int& p) const {return p;}
};
template <typename It, typename View>
struct View_iterator_base{
using return_type = decltype(View{}(*It{}));
using const_iterator =
View_iterator_base<std::vector<int>::const_iterator, Func>;
};
using view_it =
alp::Iterator<View_iterator_base<std::vector<int>::iterator, Func>>;
int main()
{
view_it p{};
view_it z{};
bool x = operator==(z, p); // only compiles if you remove <iostream>
bool y = alp::operator==(z,p); // always compile
}
Error message:
yy.cpp: In instantiation of ‘struct View_iterator_base<__gnu_cxx::__normal_iterator<const int*, std::vector<int> >, Func>’:
yy.cpp:9:73: required from ‘struct alp::Iterator<View_iterator_base<__gnu_cxx::__normal_iterator<const int*, std::vector<int> >, Func> >’
yy.cpp:44:29: required from here
yy.cpp:28:42: error: no match for call to ‘(Func) (const int&)’
using return_type = decltype(View{}(*It{}));
~~~~~~^~~~~~~
yy.cpp:22:10: note: candidate: int& Func::operator()(int&) const <near match>
int& operator()(int& p) const {return p;}
^~~~~~~~
yy.cpp:22:10: note: conversion of argument 1 would be ill-formed:
yy.cpp:28:42: error: binding reference of type ‘int&’ to ‘const int’ discards qualifiers
using return_type = decltype(View{}(*It{}));
~~~~~~^~~~~~~
I've made a more minimal test case here: https://godbolt.org/z/QQonMG .
The relevant details are:
A using type alias does not instantiate a template. So for example:
template<bool b>
struct fail_if_true {
static_assert(!b, "template parameter must be false");
};
using fail_if_used = fail_if_true<true>;
will not cause a compile time error (if fail_if_used isn't used)
ADL also inspects template parameter classes. In this case, std::vector<int>::iterator is __gnu_cxx::__normal_iterator<const int*, std::vector<int> >, Func>, which has a std::vector<int> in it's template. So, operator== will check in the global namespace (always), alp (As alp::Iterator is in alp), __gnu_cxx and std.
Your View_iterator_base::const_iterator is invalid. View_iterator_base::const_interator::result_type is defined as decltype(Func{}(*std::vector<int>::const_iterator{})). std::vector<int>::const_iterator{} will be a vectors const iterator, so *std::vector<int>::const_iterator{} is a const int&. Func::operator() takes an int&, so this means that the expression is invalid. But it won't cause a compile time error if not used, for the reasons stated above. This means that your conversion operator is to an invalid type.
Since you don't define it as explicit, the conversion operator (To an invalid type) will be used to try and match it to the function parameters if they don't already match. Obviously this will finally instantiate the invalid type, so it will throw a compile time error.
My guess is that iostream includes string, which defines std::operator== for strings.
Here's an example without the std namespace: https://godbolt.org/z/-wlAmv
// Avoid including headers for testing without std::
template<class T> struct is_const { static constexpr const bool value = false; } template<class T> struct is_const<const T> { static constexpr const bool value = true; }
namespace with_another_equals {
struct T {};
bool operator==(const T&, const T&) {
return true;
}
}
namespace ns {
template<class T>
struct wrapper {
using invalid_wrapper = wrapper<typename T::invalid>;
operator invalid_wrapper() {}
};
template<class T>
bool operator==(const wrapper<T>&, const wrapper<T>&) {
return true;
}
}
template<class T>
struct with_invalid {
static_assert(!is_const<T>::value, "Invalid if const");
using invalid = with_invalid<const T>;
};
template<class T>
void test() {
using wrapped = ns::wrapper<with_invalid<T>>;
wrapped a;
wrapped b;
bool x = operator==(a, b);
bool y = ns::operator==(a, b);
}
template void test<int*>();
// Will compile if this line is commented out
template void test<with_another_equals::T>();
Note that just declaring operator const_iterator() should instantiate the type. But it doesn't because it is within templates. My guess is that it is optimised out (where it does compile because it's unused) before it can be checked to show that it can't compile (It doesn't even warn with -Wall -pedantic that it doesn't have a return statement in my example).
I have the following method declaration on a basic_buffer class:
const_iterator insert(const_iterator position, typename argument<value_type>::type val)
Notice the type of the second argument. I often use this argument traits that basically decides whether the argument should be passed by copy or by reference when receiving template arguments. In this case, value_type is a typedef of the template argument T. For instance, fundamental types should be passed by copy instead of const reference. Here's the implementation:
template <typename T> struct argument
{
typedef std::conditional<std::is_fundamental<T>::value || std::is_pointer<T>::value, const T, const T &> type;
};
Notice how fundamental and pointer types evaluate to const T and other types evaluate to const T &. This has been working great so far.
Now consider the following function:
template <class T>
void foo()
{
typedef basic_buffer<T> _storage_type;
typedef typename _storage_type::value_type _value_type;
_value_type value = 0;
_storage_type _storage;
_storage.insert(_storage.end(), value);
}
Several details are omitted. This is what I get:
error: no matching member function for call to 'insert'
_storage.insert(_storage.end(), value);
~~~~~~~~~^~~~~~
What surprises me is this overload version not being matched:
note: candidate function not viable: no known conversion from '_value_type' (aka 'unsigned char') to 'typename argument<value_type>::type' (aka 'conditional<std::is_fundamental<unsigned
char>::value || std::is_pointer<unsigned char>::value, const unsigned char, const unsigned char &>') for 2nd argument
const_iterator insert(const_iterator position, typename argument<value_type>::type val)
To make matters even more confusing, if I cast value to _value_type (that, notably, is already its type) it works:
_storage.insert(_storage.end(), static_cast<_value_type>(value));
So I can solve this by casting value, but rather not. What is happening here?
You have
typedef std::conditional<std::is_fundamental<T>::value || std::is_pointer<T>::value, const T, const T &> type;
So type is a std::conditional<std::is_fundamental<T>::value || std::is_pointer<T>::value, const T, const T &>
When you call
_storage.insert(_storage.end(), value);
It is trying to convert value to a std::conditional<std::is_fundamental<T>::value || std::is_pointer<T>::value, const T, const T &>
You need to add ::type to the conditional to get the resulting type from the condition.
typedef std::conditional<std::is_fundamental<T>::value || std::is_pointer<T>::value, const T, const T &>::type type;
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.
I am having a problem in some code using type_traits from boost.
It is quite a complex part of the code, but I could isolate the part that gives the compilation error:
template<const size_t maxLen>
class MyString {
public:
typedef boost::conditional<(maxLen > 0), char[maxLen+1], std::string> ObjExternal;
};
template <class T>
class APIBase {
public:
typedef T obj_type;
typedef typename T::ObjExternal return_type;
};
template <class T>
int edit(const T& field, const typename T::return_type& value)
{
return 0;
}
int myFunction()
{
APIBase<MyString<10> > b;
char c[11];
return edit(b, c);
}
This gives the following error:
test.cpp: In function ‘int myFunction()’:
tes.cpp:109: error: no matching function for call to ‘edit(APIBase >&, char [11])’
tes.cpp:100: note: candidates are: int edit(const T&, const typename T::return_type&) [with T = APIBase >]
However, if I change the line with the code
char c[11];
by
MyString<10>::ObjExternal c;
it works. Similarly, if instead I change the line
typedef boost::conditional<(maxLen > 0), char[maxLen+1], std::string> ObjExternal;
by
typedef char ObjExternal[maxLen+1];
it also works. I am thinking that it is a problem with boost::conditional, as it seems it is not being evaluated right. Is there a problem in my code, or there is an alternative that can be used instead of boost::conditional to have this functionality?
I am thinking about using the 2nd option, but then I could not use maxLen as 0.
You need to use the member typedef type provided by conditional and not the conditional type itself.
Change:
typedef boost::conditional<(maxLen > 0),
char[maxLen+1],
std::string> ObjExternal;
to:
typedef typename boost::conditional<(maxLen > 0),
char[maxLen+1],
std::string>::type ObjExternal;