How to make a function template a parameter to a test?









up vote
0
down vote

favorite












I want to unit test equivalent templated member function of a class template. (As a background, I want to offer the users of the library "classical" function notation with .bar() and a tacit notation with |, &, etc. But I do not want to duplicate the complete test code.)



#include <utility> 

template <typename T>
struct foo
(U&& u)&&
return bar(std::forward<U>(u));


template <typename U>
auto bar(U&& u) const&
// whatever happens here.
return foo<T>();


template <typename U>
auto operator;

int main()
using void_t = void();
using op_t = foo<void>(foo<void>::*)(void_t)&&;

op_t ops = static_cast<op_t>(&foo<void>::bar<void_t>),
static_cast<op_t>(&foo<void>::operator;

for (const auto& op : ops)
auto sut = (foo<void>.*op)(());
// test the behaviour of sut




clang e.g., reports me that "the address of overloaded function 'bar' cannot be static_cast to type 'op_t'"



Or am I on the wrong track and this is not possible?
(I tried clang 6 and gcc 7)










share|improve this question





















  • What would &foo<void>::bar<void_t> be in the first place, given that bar is not a static member?
    – Henning Koehler
    Nov 10 at 11:15











  • @HenningKoehler A pointer to member.
    – melpomene
    Nov 10 at 11:23










  • But for a non-static member to exist, you need to have an instance first ...
    – Henning Koehler
    Nov 10 at 11:25






  • 1




    I can get this code to compile by changing op_t to using op_t = foo<void>(foo<void>::*)(void_t &&) &&; and sut to auto sut = (foo<void>.*op)(*());. Alternatively, instead of changing sut, you can also change void_t to using void_t = void (*)();.
    – melpomene
    Nov 10 at 11:59







  • 2




    Off Topic: starting from C++17, std::void_t is part of the standard (with a completely different meaning). To avoid clashes, I suggest to give another name to your void_t
    – max66
    Nov 10 at 13:14














up vote
0
down vote

favorite












I want to unit test equivalent templated member function of a class template. (As a background, I want to offer the users of the library "classical" function notation with .bar() and a tacit notation with |, &, etc. But I do not want to duplicate the complete test code.)



#include <utility> 

template <typename T>
struct foo
(U&& u)&&
return bar(std::forward<U>(u));


template <typename U>
auto bar(U&& u) const&
// whatever happens here.
return foo<T>();


template <typename U>
auto operator;

int main()
using void_t = void();
using op_t = foo<void>(foo<void>::*)(void_t)&&;

op_t ops = static_cast<op_t>(&foo<void>::bar<void_t>),
static_cast<op_t>(&foo<void>::operator;

for (const auto& op : ops)
auto sut = (foo<void>.*op)(());
// test the behaviour of sut




clang e.g., reports me that "the address of overloaded function 'bar' cannot be static_cast to type 'op_t'"



Or am I on the wrong track and this is not possible?
(I tried clang 6 and gcc 7)










share|improve this question





















  • What would &foo<void>::bar<void_t> be in the first place, given that bar is not a static member?
    – Henning Koehler
    Nov 10 at 11:15











  • @HenningKoehler A pointer to member.
    – melpomene
    Nov 10 at 11:23










  • But for a non-static member to exist, you need to have an instance first ...
    – Henning Koehler
    Nov 10 at 11:25






  • 1




    I can get this code to compile by changing op_t to using op_t = foo<void>(foo<void>::*)(void_t &&) &&; and sut to auto sut = (foo<void>.*op)(*());. Alternatively, instead of changing sut, you can also change void_t to using void_t = void (*)();.
    – melpomene
    Nov 10 at 11:59







  • 2




    Off Topic: starting from C++17, std::void_t is part of the standard (with a completely different meaning). To avoid clashes, I suggest to give another name to your void_t
    – max66
    Nov 10 at 13:14












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I want to unit test equivalent templated member function of a class template. (As a background, I want to offer the users of the library "classical" function notation with .bar() and a tacit notation with |, &, etc. But I do not want to duplicate the complete test code.)



#include <utility> 

template <typename T>
struct foo
(U&& u)&&
return bar(std::forward<U>(u));


template <typename U>
auto bar(U&& u) const&
// whatever happens here.
return foo<T>();


template <typename U>
auto operator;

int main()
using void_t = void();
using op_t = foo<void>(foo<void>::*)(void_t)&&;

op_t ops = static_cast<op_t>(&foo<void>::bar<void_t>),
static_cast<op_t>(&foo<void>::operator;

for (const auto& op : ops)
auto sut = (foo<void>.*op)(());
// test the behaviour of sut




clang e.g., reports me that "the address of overloaded function 'bar' cannot be static_cast to type 'op_t'"



Or am I on the wrong track and this is not possible?
(I tried clang 6 and gcc 7)










share|improve this question













I want to unit test equivalent templated member function of a class template. (As a background, I want to offer the users of the library "classical" function notation with .bar() and a tacit notation with |, &, etc. But I do not want to duplicate the complete test code.)



#include <utility> 

template <typename T>
struct foo
(U&& u)&&
return bar(std::forward<U>(u));


template <typename U>
auto bar(U&& u) const&
// whatever happens here.
return foo<T>();


template <typename U>
auto operator;

int main()
using void_t = void();
using op_t = foo<void>(foo<void>::*)(void_t)&&;

op_t ops = static_cast<op_t>(&foo<void>::bar<void_t>),
static_cast<op_t>(&foo<void>::operator;

for (const auto& op : ops)
auto sut = (foo<void>.*op)(());
// test the behaviour of sut




clang e.g., reports me that "the address of overloaded function 'bar' cannot be static_cast to type 'op_t'"



Or am I on the wrong track and this is not possible?
(I tried clang 6 and gcc 7)







c++ templates






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 10 at 11:04









Felix Petriconi

370210




370210











  • What would &foo<void>::bar<void_t> be in the first place, given that bar is not a static member?
    – Henning Koehler
    Nov 10 at 11:15











  • @HenningKoehler A pointer to member.
    – melpomene
    Nov 10 at 11:23










  • But for a non-static member to exist, you need to have an instance first ...
    – Henning Koehler
    Nov 10 at 11:25






  • 1




    I can get this code to compile by changing op_t to using op_t = foo<void>(foo<void>::*)(void_t &&) &&; and sut to auto sut = (foo<void>.*op)(*());. Alternatively, instead of changing sut, you can also change void_t to using void_t = void (*)();.
    – melpomene
    Nov 10 at 11:59







  • 2




    Off Topic: starting from C++17, std::void_t is part of the standard (with a completely different meaning). To avoid clashes, I suggest to give another name to your void_t
    – max66
    Nov 10 at 13:14
















  • What would &foo<void>::bar<void_t> be in the first place, given that bar is not a static member?
    – Henning Koehler
    Nov 10 at 11:15











  • @HenningKoehler A pointer to member.
    – melpomene
    Nov 10 at 11:23










  • But for a non-static member to exist, you need to have an instance first ...
    – Henning Koehler
    Nov 10 at 11:25






  • 1




    I can get this code to compile by changing op_t to using op_t = foo<void>(foo<void>::*)(void_t &&) &&; and sut to auto sut = (foo<void>.*op)(*());. Alternatively, instead of changing sut, you can also change void_t to using void_t = void (*)();.
    – melpomene
    Nov 10 at 11:59







  • 2




    Off Topic: starting from C++17, std::void_t is part of the standard (with a completely different meaning). To avoid clashes, I suggest to give another name to your void_t
    – max66
    Nov 10 at 13:14















What would &foo<void>::bar<void_t> be in the first place, given that bar is not a static member?
– Henning Koehler
Nov 10 at 11:15





What would &foo<void>::bar<void_t> be in the first place, given that bar is not a static member?
– Henning Koehler
Nov 10 at 11:15













@HenningKoehler A pointer to member.
– melpomene
Nov 10 at 11:23




@HenningKoehler A pointer to member.
– melpomene
Nov 10 at 11:23












But for a non-static member to exist, you need to have an instance first ...
– Henning Koehler
Nov 10 at 11:25




But for a non-static member to exist, you need to have an instance first ...
– Henning Koehler
Nov 10 at 11:25




1




1




I can get this code to compile by changing op_t to using op_t = foo<void>(foo<void>::*)(void_t &&) &&; and sut to auto sut = (foo<void>.*op)(*());. Alternatively, instead of changing sut, you can also change void_t to using void_t = void (*)();.
– melpomene
Nov 10 at 11:59





I can get this code to compile by changing op_t to using op_t = foo<void>(foo<void>::*)(void_t &&) &&; and sut to auto sut = (foo<void>.*op)(*());. Alternatively, instead of changing sut, you can also change void_t to using void_t = void (*)();.
– melpomene
Nov 10 at 11:59





2




2




Off Topic: starting from C++17, std::void_t is part of the standard (with a completely different meaning). To avoid clashes, I suggest to give another name to your void_t
– max66
Nov 10 at 13:14




Off Topic: starting from C++17, std::void_t is part of the standard (with a completely different meaning). To avoid clashes, I suggest to give another name to your void_t
– max66
Nov 10 at 13:14












1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










I can get the code to compile by changing op_t to



using op_t = foo<void>(foo<void>::*)(void_t &&) &&;
// ^^


and sut to



auto sut = (foo<void>.*op)(*());
// ^


(() can be converted to a function pointer, but *op takes a reference to a function, so we have to dereference the pointer).



Alternatively, instead of changing sut, you can also change void_t to



using void_t = void (*)();
// ^^^


Now void_t is already a function pointer, so you don't have to dereference.






share|improve this answer




















    Your Answer






    StackExchange.ifUsing("editor", function ()
    StackExchange.using("externalEditor", function ()
    StackExchange.using("snippets", function ()
    StackExchange.snippets.init();
    );
    );
    , "code-snippets");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "1"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53238301%2fhow-to-make-a-function-template-a-parameter-to-a-test%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    I can get the code to compile by changing op_t to



    using op_t = foo<void>(foo<void>::*)(void_t &&) &&;
    // ^^


    and sut to



    auto sut = (foo<void>.*op)(*());
    // ^


    (() can be converted to a function pointer, but *op takes a reference to a function, so we have to dereference the pointer).



    Alternatively, instead of changing sut, you can also change void_t to



    using void_t = void (*)();
    // ^^^


    Now void_t is already a function pointer, so you don't have to dereference.






    share|improve this answer
























      up vote
      1
      down vote



      accepted










      I can get the code to compile by changing op_t to



      using op_t = foo<void>(foo<void>::*)(void_t &&) &&;
      // ^^


      and sut to



      auto sut = (foo<void>.*op)(*());
      // ^


      (() can be converted to a function pointer, but *op takes a reference to a function, so we have to dereference the pointer).



      Alternatively, instead of changing sut, you can also change void_t to



      using void_t = void (*)();
      // ^^^


      Now void_t is already a function pointer, so you don't have to dereference.






      share|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        I can get the code to compile by changing op_t to



        using op_t = foo<void>(foo<void>::*)(void_t &&) &&;
        // ^^


        and sut to



        auto sut = (foo<void>.*op)(*());
        // ^


        (() can be converted to a function pointer, but *op takes a reference to a function, so we have to dereference the pointer).



        Alternatively, instead of changing sut, you can also change void_t to



        using void_t = void (*)();
        // ^^^


        Now void_t is already a function pointer, so you don't have to dereference.






        share|improve this answer












        I can get the code to compile by changing op_t to



        using op_t = foo<void>(foo<void>::*)(void_t &&) &&;
        // ^^


        and sut to



        auto sut = (foo<void>.*op)(*());
        // ^


        (() can be converted to a function pointer, but *op takes a reference to a function, so we have to dereference the pointer).



        Alternatively, instead of changing sut, you can also change void_t to



        using void_t = void (*)();
        // ^^^


        Now void_t is already a function pointer, so you don't have to dereference.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 10 at 12:33









        melpomene

        57.4k54489




        57.4k54489



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Stack Overflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53238301%2fhow-to-make-a-function-template-a-parameter-to-a-test%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            How to how show current date and time by default on contact form 7 in WordPress without taking input from user in datetimepicker

            Syphilis

            Darth Vader #20