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1

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

share|improve this question

asked Mar 27 at 12:33

dinpindinpin

322 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  

  
  
  
  
  
  

  
  
  Check this: stackoverflow.com/a/49904914/11224588
  
  – Diodacus
  Mar 27 at 12:40
  

  
  
  
  

add a comment | 

0

1

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

share|improve this question

asked Mar 27 at 12:33

dinpindinpin

322 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  

  
  
  
  
  
  

  
  
  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

share|improve this question

asked Mar 27 at 12:33

dinpindinpin

322 bronze badges

template<class T>
void func(T var)

 if constexpr (can_be_dereferenced<T>::value)
 
 // do something with *var;
 

share|improve this question

asked Mar 27 at 12:33

dinpindinpin

322 bronze badges

share|improve this question

asked Mar 27 at 12:33

dinpindinpin

322 bronze badges

asked Mar 27 at 12:33

dinpindinpin

322 bronze badges

asked Mar 27 at 12:33

dinpindinpin

322 bronze badges

2 Answers
2

active

oldest

votes

3

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
 

share|improve this answer

answered Mar 27 at 12:57

ArtyerArtyer

7,75111 gold badge1212 silver badges3333 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This is exactly what I needed. Thanks a lot!
  
  – dinpin
  Mar 27 at 13:42
  

  
  
  
  

add a comment | 

0

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.
 [..]

share|improve this answer

answered Mar 27 at 12:56

vahanchovahancho

16.4k33 gold badges2727 silver badges3535 bronze badges

add a comment | 

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3

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
 

share|improve this answer

answered Mar 27 at 12:57

ArtyerArtyer

7,75111 gold badge1212 silver badges3333 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This is exactly what I needed. Thanks a lot!
  
  – dinpin
  Mar 27 at 13:42
  

  
  
  
  

add a comment | 

3

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
 

share|improve this answer

answered Mar 27 at 12:57

ArtyerArtyer

7,75111 gold badge1212 silver badges3333 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
 

share|improve this answer

answered Mar 27 at 12:57

ArtyerArtyer

7,75111 gold badge1212 silver badges3333 bronze badges

#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
 

share|improve this answer

answered Mar 27 at 12:57

ArtyerArtyer

7,75111 gold badge1212 silver badges3333 bronze badges

answered Mar 27 at 12:57

ArtyerArtyer

7,75111 gold badge1212 silver badges3333 bronze badges

answered Mar 27 at 12:57

ArtyerArtyer

7,75111 gold badge1212 silver badges3333 bronze badges

0

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.
 [..]

share|improve this answer

answered Mar 27 at 12:56

vahanchovahancho

16.4k33 gold badges2727 silver badges3535 bronze badges

add a comment | 

0

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.
 [..]

share|improve this answer

answered Mar 27 at 12:56

vahanchovahancho

16.4k33 gold badges2727 silver badges3535 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.
 [..]

share|improve this answer

answered Mar 27 at 12:56

vahanchovahancho

16.4k33 gold badges2727 silver badges3535 bronze badges

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.
 [..]

share|improve this answer

answered Mar 27 at 12:56

vahanchovahancho

16.4k33 gold badges2727 silver badges3535 bronze badges

answered Mar 27 at 12:56

vahanchovahancho

16.4k33 gold badges2727 silver badges3535 bronze badges

answered Mar 27 at 12:56

vahanchovahancho

16.4k33 gold badges2727 silver badges3535 bronze badges