Why does the compiler prefer f(const void*) to f(const std::string &)?











up vote
30
down vote

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5












Consider the following piece of code:



#include <iostream>

#include <string>



// void f(const char *) { std::cout << "const char *"; } // <-- comment on purpose

void f(const std::string &) { std::cout << "const std::string &"; }

void f(const void *) { std::cout << "const void *"; }



int main()

{

    f("hello");

    std::cout << std::endl;

}


I compiled this program using g++ (Ubuntu 6.5.0-1ubuntu1~16.04) 6.5.0 20181026:



$ g++ -std=c++11 strings_1.cpp -Wall

$ ./a.out



const void *


Note that the comment is there on purpose to test, otherwise the compiler uses f(const char *).



So, why does the compiler pick f(const void*) over f(const std::string &)?






c++ stdstring string-literals function-overloading overload-resolution







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  • 3




    Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
    – geza
    2 days ago










  • @geza awesome. I was looking for it, thanks.
    – omar
    2 days ago










  • The overloading resolution rule here is simple and unchanged in the many C++ versions.
    – curiousguy
    2 days ago






  • 3




    Well, a string literal is not an std::string, it's a static array of chars, which decays to a pointer to its first character. This behavior is inherited from C which never had something like std::string, but ample amounts of code handling strings nonetheless.
    – cmaster
    2 days ago






  • 1




    If you specifically want a std::string literal you can achieve that by adding a s behind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
    – henje
    yesterday















up vote
30
down vote

favorite
5












Consider the following piece of code:



#include <iostream>

#include <string>



// void f(const char *) { std::cout << "const char *"; } // <-- comment on purpose

void f(const std::string &) { std::cout << "const std::string &"; }

void f(const void *) { std::cout << "const void *"; }



int main()

{

    f("hello");

    std::cout << std::endl;

}


I compiled this program using g++ (Ubuntu 6.5.0-1ubuntu1~16.04) 6.5.0 20181026:



$ g++ -std=c++11 strings_1.cpp -Wall

$ ./a.out



const void *


Note that the comment is there on purpose to test, otherwise the compiler uses f(const char *).



So, why does the compiler pick f(const void*) over f(const std::string &)?






c++ stdstring string-literals function-overloading overload-resolution







share|improve this question









New contributor




omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 3




    Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
    – geza
    2 days ago










  • @geza awesome. I was looking for it, thanks.
    – omar
    2 days ago










  • The overloading resolution rule here is simple and unchanged in the many C++ versions.
    – curiousguy
    2 days ago






  • 3




    Well, a string literal is not an std::string, it's a static array of chars, which decays to a pointer to its first character. This behavior is inherited from C which never had something like std::string, but ample amounts of code handling strings nonetheless.
    – cmaster
    2 days ago






  • 1




    If you specifically want a std::string literal you can achieve that by adding a s behind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
    – henje
    yesterday













up vote
30
down vote

favorite
5









up vote
30
down vote

favorite
5






5





Consider the following piece of code:



#include <iostream>

#include <string>



// void f(const char *) { std::cout << "const char *"; } // <-- comment on purpose

void f(const std::string &) { std::cout << "const std::string &"; }

void f(const void *) { std::cout << "const void *"; }



int main()

{

    f("hello");

    std::cout << std::endl;

}


I compiled this program using g++ (Ubuntu 6.5.0-1ubuntu1~16.04) 6.5.0 20181026:



$ g++ -std=c++11 strings_1.cpp -Wall

$ ./a.out



const void *


Note that the comment is there on purpose to test, otherwise the compiler uses f(const char *).



So, why does the compiler pick f(const void*) over f(const std::string &)?






c++ stdstring string-literals function-overloading overload-resolution







share|improve this question









New contributor




omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Consider the following piece of code:



#include <iostream>

#include <string>



// void f(const char *) { std::cout << "const char *"; } // <-- comment on purpose

void f(const std::string &) { std::cout << "const std::string &"; }

void f(const void *) { std::cout << "const void *"; }



int main()

{

    f("hello");

    std::cout << std::endl;

}


I compiled this program using g++ (Ubuntu 6.5.0-1ubuntu1~16.04) 6.5.0 20181026:



$ g++ -std=c++11 strings_1.cpp -Wall

$ ./a.out



const void *


Note that the comment is there on purpose to test, otherwise the compiler uses f(const char *).



So, why does the compiler pick f(const void*) over f(const std::string &)?






c++ stdstring string-literals function-overloading overload-resolution




c++ stdstring string-literals function-overloading overload-resolution






share|improve this question









New contributor




omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 2 days ago









curiousguy

4,47722940




4,47722940






New contributor




omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 days ago









omar

20918




20918




New contributor




omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 3




    Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
    – geza
    2 days ago










  • @geza awesome. I was looking for it, thanks.
    – omar
    2 days ago










  • The overloading resolution rule here is simple and unchanged in the many C++ versions.
    – curiousguy
    2 days ago






  • 3




    Well, a string literal is not an std::string, it's a static array of chars, which decays to a pointer to its first character. This behavior is inherited from C which never had something like std::string, but ample amounts of code handling strings nonetheless.
    – cmaster
    2 days ago






  • 1




    If you specifically want a std::string literal you can achieve that by adding a s behind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
    – henje
    yesterday














  • 3




    Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
    – geza
    2 days ago










  • @geza awesome. I was looking for it, thanks.
    – omar
    2 days ago










  • The overloading resolution rule here is simple and unchanged in the many C++ versions.
    – curiousguy
    2 days ago






  • 3




    Well, a string literal is not an std::string, it's a static array of chars, which decays to a pointer to its first character. This behavior is inherited from C which never had something like std::string, but ample amounts of code handling strings nonetheless.
    – cmaster
    2 days ago






  • 1




    If you specifically want a std::string literal you can achieve that by adding a s behind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
    – henje
    yesterday








3




3




Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
– geza
2 days ago




Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
– geza
2 days ago












@geza awesome. I was looking for it, thanks.
– omar
2 days ago




@geza awesome. I was looking for it, thanks.
– omar
2 days ago












The overloading resolution rule here is simple and unchanged in the many C++ versions.
– curiousguy
2 days ago




The overloading resolution rule here is simple and unchanged in the many C++ versions.
– curiousguy
2 days ago




3




3




Well, a string literal is not an std::string, it's a static array of chars, which decays to a pointer to its first character. This behavior is inherited from C which never had something like std::string, but ample amounts of code handling strings nonetheless.
– cmaster
2 days ago




Well, a string literal is not an std::string, it's a static array of chars, which decays to a pointer to its first character. This behavior is inherited from C which never had something like std::string, but ample amounts of code handling strings nonetheless.
– cmaster
2 days ago




1




1




If you specifically want a std::string literal you can achieve that by adding a s behind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
– henje
yesterday




If you specifically want a std::string literal you can achieve that by adding a s behind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
– henje
yesterday












1 Answer
1






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up vote
36
down vote



accepted










Converting to a std::string requires a "user defined conversion".



Converting to void const* does not.



User defined conversions are ordered behind built in ones.






share|improve this answer





















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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    36
    down vote



    accepted










    Converting to a std::string requires a "user defined conversion".



    Converting to void const* does not.



    User defined conversions are ordered behind built in ones.






    share|improve this answer

























      up vote
      36
      down vote



      accepted










      Converting to a std::string requires a "user defined conversion".



      Converting to void const* does not.



      User defined conversions are ordered behind built in ones.






      share|improve this answer























        up vote
        36
        down vote



        accepted







        up vote
        36
        down vote



        accepted






        Converting to a std::string requires a "user defined conversion".



        Converting to void const* does not.



        User defined conversions are ordered behind built in ones.






        share|improve this answer












        Converting to a std::string requires a "user defined conversion".



        Converting to void const* does not.



        User defined conversions are ordered behind built in ones.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 2 days ago









        Yakk - Adam Nevraumont

        178k19184364




        178k19184364






















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