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How to check whether variable of given type can be dereferenced?
Template function for detecting pointer like (dereferencable) types fails for actual pointer typesHow can I profile C++ code running on Linux?Function passed as template argumentStatic constant string (class member)C++ metafunction to determine whether a type is callableImage Processing: Algorithm Improvement for 'Coca-Cola Can' RecognitionHow is “int main()(([]())());” valid C++?Replacing a 32-bit loop counter with 64-bit introduces crazy performance deviationsenum to string in modern C++11 / C++14 / C++17 and future C++20Are == and != mutually dependent?Does the C++ standard require operator != must be provided for a given iterator type?
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Is there way in C++ to determine whether type of variable is pointer or any iterator with overloaded operator*?
There is standard std::is_pointer, but it say nothing about iterators.
I wanna use it in code like this:
template<class T>
void func(T var)
if constexpr (can_be_dereferenced<T>::value)
// do something with *var;
c++ typetraits
asked Mar 27 at 12:33
dinpindinpin
32 bronze badges
add a comment |
Is there way in C++ to determine whether type of variable is pointer or any iterator with overloaded operator*?
There is standard std::is_pointer, but it say nothing about iterators.
I wanna use it in code like this:
template<class T>
void func(T var)
if constexpr (can_be_dereferenced<T>::value)
// do something with *var;
c++ typetraits
asked Mar 27 at 12:33
dinpindinpin
32 bronze badges
All standard library iterators have certain well defined members, likevalue_type. You should be able to come up with some SFINAE-based approach that checks for their existence. Or, make use ofstd::iterator_traits.
– Sam Varshavchik
Mar 27 at 12:34
1
Check this: stackoverflow.com/a/49904914/11224588
– Diodacus
Mar 27 at 12:40
add a comment |
Is there way in C++ to determine whether type of variable is pointer or any iterator with overloaded operator*?
There is standard std::is_pointer, but it say nothing about iterators.
I wanna use it in code like this:
template<class T>
void func(T var)
if constexpr (can_be_dereferenced<T>::value)
// do something with *var;
c++ typetraits
asked Mar 27 at 12:33
dinpindinpin
32 bronze badges
Is there way in C++ to determine whether type of variable is pointer or any iterator with overloaded operator*?
There is standard std::is_pointer, but it say nothing about iterators.
I wanna use it in code like this:
template<class T>
void func(T var)
if constexpr (can_be_dereferenced<T>::value)
// do something with *var;
c++ typetraits
c++ typetraits
asked Mar 27 at 12:33
dinpindinpin
32 bronze badges
asked Mar 27 at 12:33
dinpindinpin
32 bronze badges
asked Mar 27 at 12:33
dinpindinpin
32 bronze badges
asked Mar 27 at 12:33
dinpindinpin
32 bronze badges
asked Mar 27 at 12:33
dinpindinpin
32 bronze badges
32 bronze badges
All standard library iterators have certain well defined members, likevalue_type. You should be able to come up with some SFINAE-based approach that checks for their existence. Or, make use ofstd::iterator_traits.
– Sam Varshavchik
Mar 27 at 12:34
1
Check this: stackoverflow.com/a/49904914/11224588
– Diodacus
Mar 27 at 12:40
add a comment |
All standard library iterators have certain well defined members, likevalue_type. You should be able to come up with some SFINAE-based approach that checks for their existence. Or, make use ofstd::iterator_traits.
– Sam Varshavchik
Mar 27 at 12:34
1
Check this: stackoverflow.com/a/49904914/11224588
– Diodacus
Mar 27 at 12:40
All standard library iterators have certain well defined members, like
value_type. You should be able to come up with some SFINAE-based approach that checks for their existence. Or, make use of std::iterator_traits.– Sam Varshavchik
Mar 27 at 12:34
All standard library iterators have certain well defined members, like
value_type. You should be able to come up with some SFINAE-based approach that checks for their existence. Or, make use of std::iterator_traits.– Sam Varshavchik
Mar 27 at 12:34
1
1
Check this: stackoverflow.com/a/49904914/11224588
– Diodacus
Mar 27 at 12:40
Check this: stackoverflow.com/a/49904914/11224588
– Diodacus
Mar 27 at 12:40
add a comment |
2 Answers
2
active
oldest
votes
You basically want to check if *var is valid. This is the perfect use case for a SFINAE check:
#include <type_traits>
#include <utility>
namespace detail
// If `*(object of type T)` is valid, this is selected and
// the return type is `std::true_type`
template<class T>
decltype(static_cast<void>(*std::declval<T>()), std::true_type)
can_be_dereferenced_impl(int);
// Otherwise the less specific function is selected,
// and the return type is `std::false_type`
template<class>
std::false_type can_be_dereferenced_impl(...);
template<class T>
struct can_be_dereferenced : decltype(detail::can_be_dereferenced_impl<T>(0)) ;
template<class T>
void func(T var)
if constexpr (can_be_dereferenced<T&>::value)
// Or can_be_dereferenced<decltype((var))>::value
auto&& dereferenced = *var;
// Use dereferenced
answered Mar 27 at 12:57
ArtyerArtyer
7,7511 gold badge12 silver badges33 bronze badges
This is exactly what I needed. Thanks a lot!
– dinpin
Mar 27 at 13:42
add a comment |
You might do it in this way using compile time function overload resolution and SFINAE:
template<class T, typename = std::enable_if_t<std::is_pointer<T>::value>>
void func(T var)
// a pointer
template<class T, typename = std::enable_if_t<!std::is_pointer<T>::value>,
typename = T::value_type>
void func(T var)
// possibly an iterator
int main()
{
int *i = new int(11);
func(i); // Calls the first overload
std::vector<int> v;
std::vector<int>::const_iterator it = v.begin();
func(it); // Calls the second overload
func(2); // Fail, as there is no function for this argument.
[..]
answered Mar 27 at 12:56
vahanchovahancho
16.4k3 gold badges27 silver badges35 bronze badges
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You basically want to check if *var is valid. This is the perfect use case for a SFINAE check:
#include <type_traits>
#include <utility>
namespace detail
// If `*(object of type T)` is valid, this is selected and
// the return type is `std::true_type`
template<class T>
decltype(static_cast<void>(*std::declval<T>()), std::true_type)
can_be_dereferenced_impl(int);
// Otherwise the less specific function is selected,
// and the return type is `std::false_type`
template<class>
std::false_type can_be_dereferenced_impl(...);
template<class T>
struct can_be_dereferenced : decltype(detail::can_be_dereferenced_impl<T>(0)) ;
template<class T>
void func(T var)
if constexpr (can_be_dereferenced<T&>::value)
// Or can_be_dereferenced<decltype((var))>::value
auto&& dereferenced = *var;
// Use dereferenced
answered Mar 27 at 12:57
ArtyerArtyer
7,7511 gold badge12 silver badges33 bronze badges
This is exactly what I needed. Thanks a lot!
– dinpin
Mar 27 at 13:42
add a comment |
You basically want to check if *var is valid. This is the perfect use case for a SFINAE check:
#include <type_traits>
#include <utility>
namespace detail
// If `*(object of type T)` is valid, this is selected and
// the return type is `std::true_type`
template<class T>
decltype(static_cast<void>(*std::declval<T>()), std::true_type)
can_be_dereferenced_impl(int);
// Otherwise the less specific function is selected,
// and the return type is `std::false_type`
template<class>
std::false_type can_be_dereferenced_impl(...);
template<class T>
struct can_be_dereferenced : decltype(detail::can_be_dereferenced_impl<T>(0)) ;
template<class T>
void func(T var)
if constexpr (can_be_dereferenced<T&>::value)
// Or can_be_dereferenced<decltype((var))>::value
auto&& dereferenced = *var;
// Use dereferenced
answered Mar 27 at 12:57
ArtyerArtyer
7,7511 gold badge12 silver badges33 bronze badges
This is exactly what I needed. Thanks a lot!
– dinpin
Mar 27 at 13:42
add a comment |
You basically want to check if *var is valid. This is the perfect use case for a SFINAE check:
#include <type_traits>
#include <utility>
namespace detail
// If `*(object of type T)` is valid, this is selected and
// the return type is `std::true_type`
template<class T>
decltype(static_cast<void>(*std::declval<T>()), std::true_type)
can_be_dereferenced_impl(int);
// Otherwise the less specific function is selected,
// and the return type is `std::false_type`
template<class>
std::false_type can_be_dereferenced_impl(...);
template<class T>
struct can_be_dereferenced : decltype(detail::can_be_dereferenced_impl<T>(0)) ;
template<class T>
void func(T var)
if constexpr (can_be_dereferenced<T&>::value)
// Or can_be_dereferenced<decltype((var))>::value
auto&& dereferenced = *var;
// Use dereferenced
answered Mar 27 at 12:57
ArtyerArtyer
7,7511 gold badge12 silver badges33 bronze badges
You basically want to check if *var is valid. This is the perfect use case for a SFINAE check:
#include <type_traits>
#include <utility>
namespace detail
// If `*(object of type T)` is valid, this is selected and
// the return type is `std::true_type`
template<class T>
decltype(static_cast<void>(*std::declval<T>()), std::true_type)
can_be_dereferenced_impl(int);
// Otherwise the less specific function is selected,
// and the return type is `std::false_type`
template<class>
std::false_type can_be_dereferenced_impl(...);
template<class T>
struct can_be_dereferenced : decltype(detail::can_be_dereferenced_impl<T>(0)) ;
template<class T>
void func(T var)
if constexpr (can_be_dereferenced<T&>::value)
// Or can_be_dereferenced<decltype((var))>::value
auto&& dereferenced = *var;
// Use dereferenced
answered Mar 27 at 12:57
ArtyerArtyer
7,7511 gold badge12 silver badges33 bronze badges
answered Mar 27 at 12:57
ArtyerArtyer
7,7511 gold badge12 silver badges33 bronze badges
answered Mar 27 at 12:57
ArtyerArtyer
7,7511 gold badge12 silver badges33 bronze badges
answered Mar 27 at 12:57
ArtyerArtyer
7,7511 gold badge12 silver badges33 bronze badges
7,7511 gold badge12 silver badges33 bronze badges
This is exactly what I needed. Thanks a lot!
– dinpin
Mar 27 at 13:42
add a comment |
This is exactly what I needed. Thanks a lot!
– dinpin
Mar 27 at 13:42
This is exactly what I needed. Thanks a lot!
– dinpin
Mar 27 at 13:42
This is exactly what I needed. Thanks a lot!
– dinpin
Mar 27 at 13:42
add a comment |
You might do it in this way using compile time function overload resolution and SFINAE:
template<class T, typename = std::enable_if_t<std::is_pointer<T>::value>>
void func(T var)
// a pointer
template<class T, typename = std::enable_if_t<!std::is_pointer<T>::value>,
typename = T::value_type>
void func(T var)
// possibly an iterator
int main()
{
int *i = new int(11);
func(i); // Calls the first overload
std::vector<int> v;
std::vector<int>::const_iterator it = v.begin();
func(it); // Calls the second overload
func(2); // Fail, as there is no function for this argument.
[..]
answered Mar 27 at 12:56
vahanchovahancho
16.4k3 gold badges27 silver badges35 bronze badges
add a comment |
You might do it in this way using compile time function overload resolution and SFINAE:
template<class T, typename = std::enable_if_t<std::is_pointer<T>::value>>
void func(T var)
// a pointer
template<class T, typename = std::enable_if_t<!std::is_pointer<T>::value>,
typename = T::value_type>
void func(T var)
// possibly an iterator
int main()
{
int *i = new int(11);
func(i); // Calls the first overload
std::vector<int> v;
std::vector<int>::const_iterator it = v.begin();
func(it); // Calls the second overload
func(2); // Fail, as there is no function for this argument.
[..]
answered Mar 27 at 12:56
vahanchovahancho
16.4k3 gold badges27 silver badges35 bronze badges
add a comment |
You might do it in this way using compile time function overload resolution and SFINAE:
template<class T, typename = std::enable_if_t<std::is_pointer<T>::value>>
void func(T var)
// a pointer
template<class T, typename = std::enable_if_t<!std::is_pointer<T>::value>,
typename = T::value_type>
void func(T var)
// possibly an iterator
int main()
{
int *i = new int(11);
func(i); // Calls the first overload
std::vector<int> v;
std::vector<int>::const_iterator it = v.begin();
func(it); // Calls the second overload
func(2); // Fail, as there is no function for this argument.
[..]
answered Mar 27 at 12:56
vahanchovahancho
16.4k3 gold badges27 silver badges35 bronze badges
You might do it in this way using compile time function overload resolution and SFINAE:
template<class T, typename = std::enable_if_t<std::is_pointer<T>::value>>
void func(T var)
// a pointer
template<class T, typename = std::enable_if_t<!std::is_pointer<T>::value>,
typename = T::value_type>
void func(T var)
// possibly an iterator
int main()
{
int *i = new int(11);
func(i); // Calls the first overload
std::vector<int> v;
std::vector<int>::const_iterator it = v.begin();
func(it); // Calls the second overload
func(2); // Fail, as there is no function for this argument.
[..]
answered Mar 27 at 12:56
vahanchovahancho
16.4k3 gold badges27 silver badges35 bronze badges
answered Mar 27 at 12:56
vahanchovahancho
16.4k3 gold badges27 silver badges35 bronze badges
answered Mar 27 at 12:56
vahanchovahancho
16.4k3 gold badges27 silver badges35 bronze badges
answered Mar 27 at 12:56
vahanchovahancho
16.4k3 gold badges27 silver badges35 bronze badges
16.4k3 gold badges27 silver badges35 bronze badges
add a comment |
add a comment |
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All standard library iterators have certain well defined members, like
value_type. You should be able to come up with some SFINAE-based approach that checks for their existence. Or, make use ofstd::iterator_traits.– Sam Varshavchik
Mar 27 at 12:34
1
Check this: stackoverflow.com/a/49904914/11224588
– Diodacus
Mar 27 at 12:40