Why does std::optional::operator=(U&&) require U to be a non-scalar type?

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For optional's template<class U = T> optional<T>& operator=(U&& v); the standard demands that (see [optional.assign]/3.16):




This function shall not participate in overload resolution unless
...
conjunction_v<is_scalar<T>, is_same<T, decay_t<U>>> is false
...




Why do we have to exclude case when assigning a scalar of type U == T?






c++ language-lawyer c++17 optional







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  • 3




    probably because then another overload is fine. Looking at a single overload doesnt really give you the full picture
    – user463035818
    4 hours ago










  • Types are all special cases..., but fundamental types and void in particular are very boring!
    – Oliv
    3 mins ago














up vote
7
down vote

favorite












For optional's template<class U = T> optional<T>& operator=(U&& v); the standard demands that (see [optional.assign]/3.16):




This function shall not participate in overload resolution unless
...
conjunction_v<is_scalar<T>, is_same<T, decay_t<U>>> is false
...




Why do we have to exclude case when assigning a scalar of type U == T?






c++ language-lawyer c++17 optional







share|improve this question



















  • 3




    probably because then another overload is fine. Looking at a single overload doesnt really give you the full picture
    – user463035818
    4 hours ago










  • Types are all special cases..., but fundamental types and void in particular are very boring!
    – Oliv
    3 mins ago












up vote
7
down vote

favorite









up vote
7
down vote

favorite











For optional's template<class U = T> optional<T>& operator=(U&& v); the standard demands that (see [optional.assign]/3.16):




This function shall not participate in overload resolution unless
...
conjunction_v<is_scalar<T>, is_same<T, decay_t<U>>> is false
...




Why do we have to exclude case when assigning a scalar of type U == T?






c++ language-lawyer c++17 optional







share|improve this question















For optional's template<class U = T> optional<T>& operator=(U&& v); the standard demands that (see [optional.assign]/3.16):




This function shall not participate in overload resolution unless
...
conjunction_v<is_scalar<T>, is_same<T, decay_t<U>>> is false
...




Why do we have to exclude case when assigning a scalar of type U == T?






c++ language-lawyer c++17 optional




c++ language-lawyer c++17 optional






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edited 3 hours ago









Barry

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172k18296540










asked 4 hours ago









Zelta

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  • 3




    probably because then another overload is fine. Looking at a single overload doesnt really give you the full picture
    – user463035818
    4 hours ago










  • Types are all special cases..., but fundamental types and void in particular are very boring!
    – Oliv
    3 mins ago












  • 3




    probably because then another overload is fine. Looking at a single overload doesnt really give you the full picture
    – user463035818
    4 hours ago










  • Types are all special cases..., but fundamental types and void in particular are very boring!
    – Oliv
    3 mins ago







3




3




probably because then another overload is fine. Looking at a single overload doesnt really give you the full picture
– user463035818
4 hours ago




probably because then another overload is fine. Looking at a single overload doesnt really give you the full picture
– user463035818
4 hours ago












Types are all special cases..., but fundamental types and void in particular are very boring!
– Oliv
3 mins ago




Types are all special cases..., but fundamental types and void in particular are very boring!
– Oliv
3 mins ago












1 Answer
1






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oldest

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12
down vote



accepted










This exists to support:



optional<int> o(42);
o = ; // <== we want this to reset o


We have a bunch of assignment overloads, which take:



  1. nullopt_t

  2. optional const&

  3. optional&&

  4. U&&

  5. optional<U> const&

  6. optional<U>&&

For scalars, specifically, #4 would be a standard conversion whereas anything else would be a user-defined conversion - so it would be the best match. However, the result of that would be assigning o to be engaged with a value of 0. That would mean that o = could potentially mean different things depending on the type of T. Hence, we exclude scalars.



For non-scalars, #4 and #3 would be equivalent (both user-defined conversions), and #3 would win by being a non-template. No problem there.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    12
    down vote



    accepted










    This exists to support:



    optional<int> o(42);
    o = ; // <== we want this to reset o


    We have a bunch of assignment overloads, which take:



    1. nullopt_t

    2. optional const&

    3. optional&&

    4. U&&

    5. optional<U> const&

    6. optional<U>&&

    For scalars, specifically, #4 would be a standard conversion whereas anything else would be a user-defined conversion - so it would be the best match. However, the result of that would be assigning o to be engaged with a value of 0. That would mean that o = could potentially mean different things depending on the type of T. Hence, we exclude scalars.



    For non-scalars, #4 and #3 would be equivalent (both user-defined conversions), and #3 would win by being a non-template. No problem there.






    share|improve this answer
























      up vote
      12
      down vote



      accepted










      This exists to support:



      optional<int> o(42);
      o = ; // <== we want this to reset o


      We have a bunch of assignment overloads, which take:



      1. nullopt_t

      2. optional const&

      3. optional&&

      4. U&&

      5. optional<U> const&

      6. optional<U>&&

      For scalars, specifically, #4 would be a standard conversion whereas anything else would be a user-defined conversion - so it would be the best match. However, the result of that would be assigning o to be engaged with a value of 0. That would mean that o = could potentially mean different things depending on the type of T. Hence, we exclude scalars.



      For non-scalars, #4 and #3 would be equivalent (both user-defined conversions), and #3 would win by being a non-template. No problem there.






      share|improve this answer






















        up vote
        12
        down vote



        accepted







        up vote
        12
        down vote



        accepted






        This exists to support:



        optional<int> o(42);
        o = ; // <== we want this to reset o


        We have a bunch of assignment overloads, which take:



        1. nullopt_t

        2. optional const&

        3. optional&&

        4. U&&

        5. optional<U> const&

        6. optional<U>&&

        For scalars, specifically, #4 would be a standard conversion whereas anything else would be a user-defined conversion - so it would be the best match. However, the result of that would be assigning o to be engaged with a value of 0. That would mean that o = could potentially mean different things depending on the type of T. Hence, we exclude scalars.



        For non-scalars, #4 and #3 would be equivalent (both user-defined conversions), and #3 would win by being a non-template. No problem there.






        share|improve this answer












        This exists to support:



        optional<int> o(42);
        o = ; // <== we want this to reset o


        We have a bunch of assignment overloads, which take:



        1. nullopt_t

        2. optional const&

        3. optional&&

        4. U&&

        5. optional<U> const&

        6. optional<U>&&

        For scalars, specifically, #4 would be a standard conversion whereas anything else would be a user-defined conversion - so it would be the best match. However, the result of that would be assigning o to be engaged with a value of 0. That would mean that o = could potentially mean different things depending on the type of T. Hence, we exclude scalars.



        For non-scalars, #4 and #3 would be equivalent (both user-defined conversions), and #3 would win by being a non-template. No problem there.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 3 hours ago









        Barry

        172k18296540




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