How do I check whether a reference is const?












3















I was writing a test for my iterator types and wanted to check that the reference returned by de-referencing iterators provided by begin() and cbegin() are non-const and const respectively.



I tried doing something similar to the following : -



#include <type_traits>

#include <iostream>

#include <vector>



int main() {

    std::vector<int> vec{0};



    std::cout << std::is_const<decltype(*vec.begin())>::value << std::endl;

    std::cout << std::is_const<decltype(*vec.cbegin())>::value << std::endl;

}


But this prints 0 for both cases.



Is there a way to check if a reference is const?



I can use C++11/14/17 features.






c++ reference iterator const typetraits







share|improve this question

























  • Reference can never be const-qualified. Only the type to which a reference is referred can be const-qualified. std::is_const_v<std::remove_reference_t<T>>.

    – felix
    Nov 23 '18 at 10:19


















3















I was writing a test for my iterator types and wanted to check that the reference returned by de-referencing iterators provided by begin() and cbegin() are non-const and const respectively.



I tried doing something similar to the following : -



#include <type_traits>

#include <iostream>

#include <vector>



int main() {

    std::vector<int> vec{0};



    std::cout << std::is_const<decltype(*vec.begin())>::value << std::endl;

    std::cout << std::is_const<decltype(*vec.cbegin())>::value << std::endl;

}


But this prints 0 for both cases.



Is there a way to check if a reference is const?



I can use C++11/14/17 features.






c++ reference iterator const typetraits







share|improve this question

























  • Reference can never be const-qualified. Only the type to which a reference is referred can be const-qualified. std::is_const_v<std::remove_reference_t<T>>.

    – felix
    Nov 23 '18 at 10:19
















3












3








3


1






I was writing a test for my iterator types and wanted to check that the reference returned by de-referencing iterators provided by begin() and cbegin() are non-const and const respectively.



I tried doing something similar to the following : -



#include <type_traits>

#include <iostream>

#include <vector>



int main() {

    std::vector<int> vec{0};



    std::cout << std::is_const<decltype(*vec.begin())>::value << std::endl;

    std::cout << std::is_const<decltype(*vec.cbegin())>::value << std::endl;

}


But this prints 0 for both cases.



Is there a way to check if a reference is const?



I can use C++11/14/17 features.






c++ reference iterator const typetraits







share|improve this question
















I was writing a test for my iterator types and wanted to check that the reference returned by de-referencing iterators provided by begin() and cbegin() are non-const and const respectively.



I tried doing something similar to the following : -



#include <type_traits>

#include <iostream>

#include <vector>



int main() {

    std::vector<int> vec{0};



    std::cout << std::is_const<decltype(*vec.begin())>::value << std::endl;

    std::cout << std::is_const<decltype(*vec.cbegin())>::value << std::endl;

}


But this prints 0 for both cases.



Is there a way to check if a reference is const?



I can use C++11/14/17 features.






c++ reference iterator const typetraits




c++ reference iterator const typetraits






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 23 '18 at 10:21









einpoklum

36.5k28132261




36.5k28132261










asked Nov 23 '18 at 10:07









FlauFlau

90611329




90611329













  • Reference can never be const-qualified. Only the type to which a reference is referred can be const-qualified. std::is_const_v<std::remove_reference_t<T>>.

    – felix
    Nov 23 '18 at 10:19





















  • Reference can never be const-qualified. Only the type to which a reference is referred can be const-qualified. std::is_const_v<std::remove_reference_t<T>>.

    – felix
    Nov 23 '18 at 10:19



















Reference can never be const-qualified. Only the type to which a reference is referred can be const-qualified. std::is_const_v<std::remove_reference_t<T>>.

– felix
Nov 23 '18 at 10:19







Reference can never be const-qualified. Only the type to which a reference is referred can be const-qualified. std::is_const_v<std::remove_reference_t<T>>.

– felix
Nov 23 '18 at 10:19














4 Answers
4






active

oldest

votes


















4














Remove the reference to get the referenced type to inspect its constness. A reference itself is never const - even though references to const may colloquially be called const references:



std::is_const_v<std::remove_reference_t<decltype(*it)>>





share|improve this answer



















  • 1





    It calls for the definition of template<class T> constexpr bool is_const_ref_v = std::is_const_v<std::remove_reference_t<T>>;.

    – YSC
    Nov 23 '18 at 10:22





















4














*it will be a reference rather than the referenced type (int& or const int& rather than int or const int in your case). So, you need to remove the reference:



#include <iostream>

#include <type_traits>

#include <vector>



int main() {

    std::vector<int> vec{0};



    for(auto it=vec.begin(); it!=vec.end(); ++it) {

        std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;

    }



    for(auto it=vec.cbegin(); it!=vec.cend(); ++it) {

        std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;

    }

}


This produces:



0

1





share|improve this answer































    2














    is_const always returns false for references. Instead, do:



    std::is_const_v<std::remove_reference_t<decltype(*v.begin() )>> // false
    
std::is_const_v<std::remove_reference_t<decltype(*v.cbegin())>> // true





    share|improve this answer
























    • Is is_const_v preferable over is_const ?

      – Flau
      Nov 23 '18 at 10:24






    • 1





      @Flau it's just shorter. Drawback is that it requires C++17.

      – eerorika
      Nov 23 '18 at 10:25











    • @Flau _v variants are a shorthand to avoid writing ::value; the same way _t ones avoid ::type.

      – Acorn
      Nov 23 '18 at 10:25








    • 1





      Ah I see, I missed that difference - thanks.

      – Flau
      Nov 23 '18 at 10:26



















    1














    You can check the notes on document here:
    https://en.cppreference.com/w/cpp/types/is_const




    • Notes



    If T is a reference type then is_const::value is always false. The
    proper way to check a potentially-reference type for const-ness is to
    remove the reference: is_const::type>.




    for(auto it=vec.begin(); it!=vec.end(); ++it) {
    
    std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
    
}
    

    
for(auto it=vec.cbegin(); it!=vec.cend(); ++it) {
    
    std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
    
}





    share|improve this answer


























    • @YSC lol, I edited my sentence, but I think instead of giving an answer, I want to give an advice for future question

      – yelliver
      Nov 23 '18 at 10:30












    Your Answer






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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    Remove the reference to get the referenced type to inspect its constness. A reference itself is never const - even though references to const may colloquially be called const references:



    std::is_const_v<std::remove_reference_t<decltype(*it)>>





    share|improve this answer



















    • 1





      It calls for the definition of template<class T> constexpr bool is_const_ref_v = std::is_const_v<std::remove_reference_t<T>>;.

      – YSC
      Nov 23 '18 at 10:22


















    4














    Remove the reference to get the referenced type to inspect its constness. A reference itself is never const - even though references to const may colloquially be called const references:



    std::is_const_v<std::remove_reference_t<decltype(*it)>>





    share|improve this answer



















    • 1





      It calls for the definition of template<class T> constexpr bool is_const_ref_v = std::is_const_v<std::remove_reference_t<T>>;.

      – YSC
      Nov 23 '18 at 10:22
















    4












    4








    4







    Remove the reference to get the referenced type to inspect its constness. A reference itself is never const - even though references to const may colloquially be called const references:



    std::is_const_v<std::remove_reference_t<decltype(*it)>>





    share|improve this answer













    Remove the reference to get the referenced type to inspect its constness. A reference itself is never const - even though references to const may colloquially be called const references:



    std::is_const_v<std::remove_reference_t<decltype(*it)>>






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 23 '18 at 10:18









    eerorikaeerorika

    88.3k663134




    88.3k663134








    • 1





      It calls for the definition of template<class T> constexpr bool is_const_ref_v = std::is_const_v<std::remove_reference_t<T>>;.

      – YSC
      Nov 23 '18 at 10:22
















    • 1





      It calls for the definition of template<class T> constexpr bool is_const_ref_v = std::is_const_v<std::remove_reference_t<T>>;.

      – YSC
      Nov 23 '18 at 10:22










    1




    1





    It calls for the definition of template<class T> constexpr bool is_const_ref_v = std::is_const_v<std::remove_reference_t<T>>;.

    – YSC
    Nov 23 '18 at 10:22







    It calls for the definition of template<class T> constexpr bool is_const_ref_v = std::is_const_v<std::remove_reference_t<T>>;.

    – YSC
    Nov 23 '18 at 10:22















    4














    *it will be a reference rather than the referenced type (int& or const int& rather than int or const int in your case). So, you need to remove the reference:



    #include <iostream>
    
#include <type_traits>
    
#include <vector>
    

    
int main() {
    
    std::vector<int> vec{0};
    

    
    for(auto it=vec.begin(); it!=vec.end(); ++it) {
    
        std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
    
    }
    

    
    for(auto it=vec.cbegin(); it!=vec.cend(); ++it) {
    
        std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
    
    }
    
}


    This produces:



    0
    
1





    share|improve this answer




























      4














      *it will be a reference rather than the referenced type (int& or const int& rather than int or const int in your case). So, you need to remove the reference:



      #include <iostream>
      
#include <type_traits>
      
#include <vector>
      

      
int main() {
      
    std::vector<int> vec{0};
      

      
    for(auto it=vec.begin(); it!=vec.end(); ++it) {
      
        std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
      
    }
      

      
    for(auto it=vec.cbegin(); it!=vec.cend(); ++it) {
      
        std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
      
    }
      
}


      This produces:



      0
      
1





      share|improve this answer


























        4












        4








        4







        *it will be a reference rather than the referenced type (int& or const int& rather than int or const int in your case). So, you need to remove the reference:



        #include <iostream>
        
#include <type_traits>
        
#include <vector>
        

        
int main() {
        
    std::vector<int> vec{0};
        

        
    for(auto it=vec.begin(); it!=vec.end(); ++it) {
        
        std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
        
    }
        

        
    for(auto it=vec.cbegin(); it!=vec.cend(); ++it) {
        
        std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
        
    }
        
}


        This produces:



        0
        
1





        share|improve this answer













        *it will be a reference rather than the referenced type (int& or const int& rather than int or const int in your case). So, you need to remove the reference:



        #include <iostream>
        
#include <type_traits>
        
#include <vector>
        

        
int main() {
        
    std::vector<int> vec{0};
        

        
    for(auto it=vec.begin(); it!=vec.end(); ++it) {
        
        std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
        
    }
        

        
    for(auto it=vec.cbegin(); it!=vec.cend(); ++it) {
        
        std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
        
    }
        
}


        This produces:



        0
        
1






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 23 '18 at 10:19









        einpoklumeinpoklum

        36.5k28132261




        36.5k28132261























            2














            is_const always returns false for references. Instead, do:



            std::is_const_v<std::remove_reference_t<decltype(*v.begin() )>> // false
            
std::is_const_v<std::remove_reference_t<decltype(*v.cbegin())>> // true





            share|improve this answer
























            • Is is_const_v preferable over is_const ?

              – Flau
              Nov 23 '18 at 10:24






            • 1





              @Flau it's just shorter. Drawback is that it requires C++17.

              – eerorika
              Nov 23 '18 at 10:25











            • @Flau _v variants are a shorthand to avoid writing ::value; the same way _t ones avoid ::type.

              – Acorn
              Nov 23 '18 at 10:25








            • 1





              Ah I see, I missed that difference - thanks.

              – Flau
              Nov 23 '18 at 10:26
















            2














            is_const always returns false for references. Instead, do:



            std::is_const_v<std::remove_reference_t<decltype(*v.begin() )>> // false
            
std::is_const_v<std::remove_reference_t<decltype(*v.cbegin())>> // true





            share|improve this answer
























            • Is is_const_v preferable over is_const ?

              – Flau
              Nov 23 '18 at 10:24






            • 1





              @Flau it's just shorter. Drawback is that it requires C++17.

              – eerorika
              Nov 23 '18 at 10:25











            • @Flau _v variants are a shorthand to avoid writing ::value; the same way _t ones avoid ::type.

              – Acorn
              Nov 23 '18 at 10:25








            • 1





              Ah I see, I missed that difference - thanks.

              – Flau
              Nov 23 '18 at 10:26














            2












            2








            2







            is_const always returns false for references. Instead, do:



            std::is_const_v<std::remove_reference_t<decltype(*v.begin() )>> // false
            
std::is_const_v<std::remove_reference_t<decltype(*v.cbegin())>> // true





            share|improve this answer













            is_const always returns false for references. Instead, do:



            std::is_const_v<std::remove_reference_t<decltype(*v.begin() )>> // false
            
std::is_const_v<std::remove_reference_t<decltype(*v.cbegin())>> // true






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 23 '18 at 10:20









            AcornAcorn

            5,91511339




            5,91511339













            • Is is_const_v preferable over is_const ?

              – Flau
              Nov 23 '18 at 10:24






            • 1





              @Flau it's just shorter. Drawback is that it requires C++17.

              – eerorika
              Nov 23 '18 at 10:25











            • @Flau _v variants are a shorthand to avoid writing ::value; the same way _t ones avoid ::type.

              – Acorn
              Nov 23 '18 at 10:25








            • 1





              Ah I see, I missed that difference - thanks.

              – Flau
              Nov 23 '18 at 10:26



















            • Is is_const_v preferable over is_const ?

              – Flau
              Nov 23 '18 at 10:24






            • 1





              @Flau it's just shorter. Drawback is that it requires C++17.

              – eerorika
              Nov 23 '18 at 10:25











            • @Flau _v variants are a shorthand to avoid writing ::value; the same way _t ones avoid ::type.

              – Acorn
              Nov 23 '18 at 10:25








            • 1





              Ah I see, I missed that difference - thanks.

              – Flau
              Nov 23 '18 at 10:26

















            Is is_const_v preferable over is_const ?

            – Flau
            Nov 23 '18 at 10:24





            Is is_const_v preferable over is_const ?

            – Flau
            Nov 23 '18 at 10:24




            1




            1





            @Flau it's just shorter. Drawback is that it requires C++17.

            – eerorika
            Nov 23 '18 at 10:25





            @Flau it's just shorter. Drawback is that it requires C++17.

            – eerorika
            Nov 23 '18 at 10:25













            @Flau _v variants are a shorthand to avoid writing ::value; the same way _t ones avoid ::type.

            – Acorn
            Nov 23 '18 at 10:25







            @Flau _v variants are a shorthand to avoid writing ::value; the same way _t ones avoid ::type.

            – Acorn
            Nov 23 '18 at 10:25






            1




            1





            Ah I see, I missed that difference - thanks.

            – Flau
            Nov 23 '18 at 10:26





            Ah I see, I missed that difference - thanks.

            – Flau
            Nov 23 '18 at 10:26











            1














            You can check the notes on document here:
            https://en.cppreference.com/w/cpp/types/is_const




            • Notes



            If T is a reference type then is_const::value is always false. The
            proper way to check a potentially-reference type for const-ness is to
            remove the reference: is_const::type>.




            for(auto it=vec.begin(); it!=vec.end(); ++it) {
            
    std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
            
}
            

            
for(auto it=vec.cbegin(); it!=vec.cend(); ++it) {
            
    std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
            
}





            share|improve this answer


























            • @YSC lol, I edited my sentence, but I think instead of giving an answer, I want to give an advice for future question

              – yelliver
              Nov 23 '18 at 10:30
















            1














            You can check the notes on document here:
            https://en.cppreference.com/w/cpp/types/is_const




            • Notes



            If T is a reference type then is_const::value is always false. The
            proper way to check a potentially-reference type for const-ness is to
            remove the reference: is_const::type>.




            for(auto it=vec.begin(); it!=vec.end(); ++it) {
            
    std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
            
}
            

            
for(auto it=vec.cbegin(); it!=vec.cend(); ++it) {
            
    std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
            
}





            share|improve this answer


























            • @YSC lol, I edited my sentence, but I think instead of giving an answer, I want to give an advice for future question

              – yelliver
              Nov 23 '18 at 10:30














            1












            1








            1







            You can check the notes on document here:
            https://en.cppreference.com/w/cpp/types/is_const




            • Notes



            If T is a reference type then is_const::value is always false. The
            proper way to check a potentially-reference type for const-ness is to
            remove the reference: is_const::type>.




            for(auto it=vec.begin(); it!=vec.end(); ++it) {
            
    std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
            
}
            

            
for(auto it=vec.cbegin(); it!=vec.cend(); ++it) {
            
    std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
            
}





            share|improve this answer















            You can check the notes on document here:
            https://en.cppreference.com/w/cpp/types/is_const




            • Notes



            If T is a reference type then is_const::value is always false. The
            proper way to check a potentially-reference type for const-ness is to
            remove the reference: is_const::type>.




            for(auto it=vec.begin(); it!=vec.end(); ++it) {
            
    std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
            
}
            

            
for(auto it=vec.cbegin(); it!=vec.cend(); ++it) {
            
    std::cout << std::is_const<std::remove_reference<decltype(*it)>::type>::value << std::endl;
            
}






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 23 '18 at 10:29

























            answered Nov 23 '18 at 10:26









            yelliveryelliver

            2,59332152




            2,59332152













            • @YSC lol, I edited my sentence, but I think instead of giving an answer, I want to give an advice for future question

              – yelliver
              Nov 23 '18 at 10:30



















            • @YSC lol, I edited my sentence, but I think instead of giving an answer, I want to give an advice for future question

              – yelliver
              Nov 23 '18 at 10:30

















            @YSC lol, I edited my sentence, but I think instead of giving an answer, I want to give an advice for future question

            – yelliver
            Nov 23 '18 at 10:30





            @YSC lol, I edited my sentence, but I think instead of giving an answer, I want to give an advice for future question

            – yelliver
            Nov 23 '18 at 10:30


















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            Tangent Lines Diagram Along Smooth Curve

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            7 1 I am trying to replicate a diagram and have a minimal example. I don't know how to add additional tangents between the specified points and preserve smoothness. documentclass{article} usepackage{tikz} usetikzlibrary{calc,intersections} begin{document} begin{tikzpicture} % Axes draw [->, name path=x] (-1,0) -- (11,0) node [right] {$x$}; draw [->] (0,-1) -- (0,6) node [above] {$y$}; % Origin node at (0,0) [below left] {$0$}; % Points coordinate (start) at (1,-0.8); coordinate (c1) at (3,3); coordinate (c2) at (5.5,1.5); coordinate (c3) at (8,4); coordinate (end) at (10.5,-0.8); % show the points foreach n in {start,c1,c2,c3,end} fill [black] (n) circle (2pt) node [below] {}; % join the coordinates draw [thick,name path=curve] (start) to[out=70,in=180] (c1) to[out=0,in=180]
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            Yusuf al-Mu'taman ibn Hud

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            Main article: Taifa of Zaragoza Al-Mu'taman Ruler of Zaragoza Reign 1081–1085 Predecessor Ahmad al-Muqtadir Successor Al-Mustain II Born Zaragoza Died 1085 Full name Yusuf ibn Ahmad al-Mu'taman ibn Hūd House Banu Hud Yusuf ibn Ahmad al-Mu'taman ibn Hūd (Arabic: المؤتمن بالله يوسف إبن أحمد إبن هود ‎, al-Mutaman bi l-Lah , died c. 1085) was the third king of the Banu Hud dynasty who ruled over the Taifa of Zaragoza from 1081 to 1085. Yusuf al-Mutaman reigned during the height of power of Muslim Zaragoza , following the thriving period of his father Ahmad al-Muqtadir. He continued his father's efforts and created around him a court of intellectuals, living in the beautiful palace of Aljafería, nicknamed as "the palace of joy". Al-Mu'taman was also a scholarly king and a patron of science, philosophy and arts. As an example of a wise king, he knew astrology, philosophy, and especially mathematics, a discipline in which
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            Zucchini

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            This article is about the vegetable. For other uses, see Zucchini (disambiguation). Zucchini (courgette) A striped and a uniform color zucchini Genus Cucurbita Species Cucurbita pepo Origin 19th century northern Italy The zucchini ( / z uː ˈ k iː n i / , American English) or courgette ( / k ʊər ˈ ʒ ɛ t / , British English) is a summer squash which can reach nearly 1 metre (100 cm; 39 in) in length, but is usually harvested when still immature at about 15 to 25 cm (6 to 10 in). [1] A zucchini is a thin-skinned cultivar of what in Britain and Ireland is referred to as a marrow. [2] [3] In South Africa, a zucchini is known as a baby marrow. Along with certain other squashes and pumpkins, the zucchini belongs to the species Cucurbita pepo . It can be dark or light green. A related hybrid, the golden zucchini, is a deep yellow or orange color. [4] In a culinary context, the zucchini is treated as a vegetable; it is usually cooked and presented as a savory dish o
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