In C++14 is it valid to use a double in the dimension of a new expression?












41















In C++14 given the following code:



void foo() {

  double d = 5.0;

  auto p1 = new int[d];

}


clang compiles this without diagnostic while gcc on the other hand produces the following diagnostic (see it live in godbolt):



error: expression in new-declarator must have integral or enumeration type

    7 |     auto p1 = new int[d];

      |                       ^


I specifically labeled this C++14 because in C++11 mode clang treats this as ill-formed and produces the following diagnostic (see it live in godbolt):



error: array size expression must have integral or unscoped enumeration type, not 'double'

    auto p1 = new int[d];

              ^       ~


Is clang correct? If so what changed in C++14 to allow this?






c++ c++14 language-lawyer new-expression







share|improve this question




















  • 2





    Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such as int * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.

    – Thomas Matthews
    Dec 12 '18 at 15:13








  • 5





    @ThomasMatthews no this would end up being a float to integral conversion and would truncate the float

    – Shafik Yaghmour
    Dec 12 '18 at 15:59






  • 1





    @tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)

    – Bob__
    Dec 13 '18 at 9:50






  • 1





    @tomerzeitune Why int? Why not unsigned or long?

    – curiousguy
    Dec 13 '18 at 22:35






  • 1





    @KeithThompson I think the commenters are joking around

    – Shafik Yaghmour
    Dec 14 '18 at 18:11
















41















In C++14 given the following code:



void foo() {

  double d = 5.0;

  auto p1 = new int[d];

}


clang compiles this without diagnostic while gcc on the other hand produces the following diagnostic (see it live in godbolt):



error: expression in new-declarator must have integral or enumeration type

    7 |     auto p1 = new int[d];

      |                       ^


I specifically labeled this C++14 because in C++11 mode clang treats this as ill-formed and produces the following diagnostic (see it live in godbolt):



error: array size expression must have integral or unscoped enumeration type, not 'double'

    auto p1 = new int[d];

              ^       ~


Is clang correct? If so what changed in C++14 to allow this?






c++ c++14 language-lawyer new-expression







share|improve this question




















  • 2





    Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such as int * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.

    – Thomas Matthews
    Dec 12 '18 at 15:13








  • 5





    @ThomasMatthews no this would end up being a float to integral conversion and would truncate the float

    – Shafik Yaghmour
    Dec 12 '18 at 15:59






  • 1





    @tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)

    – Bob__
    Dec 13 '18 at 9:50






  • 1





    @tomerzeitune Why int? Why not unsigned or long?

    – curiousguy
    Dec 13 '18 at 22:35






  • 1





    @KeithThompson I think the commenters are joking around

    – Shafik Yaghmour
    Dec 14 '18 at 18:11














41












41








41


1






In C++14 given the following code:



void foo() {

  double d = 5.0;

  auto p1 = new int[d];

}


clang compiles this without diagnostic while gcc on the other hand produces the following diagnostic (see it live in godbolt):



error: expression in new-declarator must have integral or enumeration type

    7 |     auto p1 = new int[d];

      |                       ^


I specifically labeled this C++14 because in C++11 mode clang treats this as ill-formed and produces the following diagnostic (see it live in godbolt):



error: array size expression must have integral or unscoped enumeration type, not 'double'

    auto p1 = new int[d];

              ^       ~


Is clang correct? If so what changed in C++14 to allow this?






c++ c++14 language-lawyer new-expression







share|improve this question
















In C++14 given the following code:



void foo() {

  double d = 5.0;

  auto p1 = new int[d];

}


clang compiles this without diagnostic while gcc on the other hand produces the following diagnostic (see it live in godbolt):



error: expression in new-declarator must have integral or enumeration type

    7 |     auto p1 = new int[d];

      |                       ^


I specifically labeled this C++14 because in C++11 mode clang treats this as ill-formed and produces the following diagnostic (see it live in godbolt):



error: array size expression must have integral or unscoped enumeration type, not 'double'

    auto p1 = new int[d];

              ^       ~


Is clang correct? If so what changed in C++14 to allow this?






c++ c++14 language-lawyer new-expression




c++ c++14 language-lawyer new-expression






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 13 '18 at 21:10







Shafik Yaghmour

















asked Dec 12 '18 at 14:25









Shafik YaghmourShafik Yaghmour

127k23329546




127k23329546








  • 2





    Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such as int * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.

    – Thomas Matthews
    Dec 12 '18 at 15:13








  • 5





    @ThomasMatthews no this would end up being a float to integral conversion and would truncate the float

    – Shafik Yaghmour
    Dec 12 '18 at 15:59






  • 1





    @tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)

    – Bob__
    Dec 13 '18 at 9:50






  • 1





    @tomerzeitune Why int? Why not unsigned or long?

    – curiousguy
    Dec 13 '18 at 22:35






  • 1





    @KeithThompson I think the commenters are joking around

    – Shafik Yaghmour
    Dec 14 '18 at 18:11














  • 2





    Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such as int * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.

    – Thomas Matthews
    Dec 12 '18 at 15:13








  • 5





    @ThomasMatthews no this would end up being a float to integral conversion and would truncate the float

    – Shafik Yaghmour
    Dec 12 '18 at 15:59






  • 1





    @tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)

    – Bob__
    Dec 13 '18 at 9:50






  • 1





    @tomerzeitune Why int? Why not unsigned or long?

    – curiousguy
    Dec 13 '18 at 22:35






  • 1





    @KeithThompson I think the commenters are joking around

    – Shafik Yaghmour
    Dec 14 '18 at 18:11








2




2





Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such as int * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.

– Thomas Matthews
Dec 12 '18 at 15:13







Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such as int * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.

– Thomas Matthews
Dec 12 '18 at 15:13






5




5





@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float

– Shafik Yaghmour
Dec 12 '18 at 15:59





@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float

– Shafik Yaghmour
Dec 12 '18 at 15:59




1




1





@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)

– Bob__
Dec 13 '18 at 9:50





@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)

– Bob__
Dec 13 '18 at 9:50




1




1





@tomerzeitune Why int? Why not unsigned or long?

– curiousguy
Dec 13 '18 at 22:35





@tomerzeitune Why int? Why not unsigned or long?

– curiousguy
Dec 13 '18 at 22:35




1




1





@KeithThompson I think the commenters are joking around

– Shafik Yaghmour
Dec 14 '18 at 18:11





@KeithThompson I think the commenters are joking around

– Shafik Yaghmour
Dec 14 '18 at 18:11












2 Answers
2






active

oldest

votes


















43














Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:




Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …




to this in the C++14 draft:




Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type std::size_t and shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted to std::size_t. …




In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.



The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.






share|improve this answer





















  • 19





    I am dubious about the usefulness of allowing double to be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(

    – Matthieu M.
    Dec 12 '18 at 15:48






  • 12





    @MatthieuM. I agree, I believe it is a defect and that the intent was really to say contextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(

    – Shafik Yaghmour
    Dec 12 '18 at 15:57








  • 2





    @MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.

    – Shafik Yaghmour
    Dec 14 '18 at 17:56



















1














From c++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:




Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type std::size_t and shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted to std::size_t. [..]




with only this part ([expr.const]) missing from the C++17 draft.






share|improve this answer


























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    2 Answers
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    2 Answers
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    active

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    43














    Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:




    Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …




    to this in the C++14 draft:




    Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type std::size_t and shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted to std::size_t. …




    In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
    single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.



    The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.






    share|improve this answer





















    • 19





      I am dubious about the usefulness of allowing double to be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(

      – Matthieu M.
      Dec 12 '18 at 15:48






    • 12





      @MatthieuM. I agree, I believe it is a defect and that the intent was really to say contextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(

      – Shafik Yaghmour
      Dec 12 '18 at 15:57








    • 2





      @MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.

      – Shafik Yaghmour
      Dec 14 '18 at 17:56
















    43














    Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:




    Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …




    to this in the C++14 draft:




    Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type std::size_t and shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted to std::size_t. …




    In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
    single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.



    The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.






    share|improve this answer





















    • 19





      I am dubious about the usefulness of allowing double to be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(

      – Matthieu M.
      Dec 12 '18 at 15:48






    • 12





      @MatthieuM. I agree, I believe it is a defect and that the intent was really to say contextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(

      – Shafik Yaghmour
      Dec 12 '18 at 15:57








    • 2





      @MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.

      – Shafik Yaghmour
      Dec 14 '18 at 17:56














    43












    43








    43







    Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:




    Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …




    to this in the C++14 draft:




    Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type std::size_t and shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted to std::size_t. …




    In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
    single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.



    The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.






    share|improve this answer















    Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:




    Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …




    to this in the C++14 draft:




    Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type std::size_t and shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted to std::size_t. …




    In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
    single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.



    The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Dec 15 '18 at 15:00

























    answered Dec 12 '18 at 14:25









    Shafik YaghmourShafik Yaghmour

    127k23329546




    127k23329546








    • 19





      I am dubious about the usefulness of allowing double to be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(

      – Matthieu M.
      Dec 12 '18 at 15:48






    • 12





      @MatthieuM. I agree, I believe it is a defect and that the intent was really to say contextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(

      – Shafik Yaghmour
      Dec 12 '18 at 15:57








    • 2





      @MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.

      – Shafik Yaghmour
      Dec 14 '18 at 17:56














    • 19





      I am dubious about the usefulness of allowing double to be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(

      – Matthieu M.
      Dec 12 '18 at 15:48






    • 12





      @MatthieuM. I agree, I believe it is a defect and that the intent was really to say contextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(

      – Shafik Yaghmour
      Dec 12 '18 at 15:57








    • 2





      @MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.

      – Shafik Yaghmour
      Dec 14 '18 at 17:56








    19




    19





    I am dubious about the usefulness of allowing double to be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(

    – Matthieu M.
    Dec 12 '18 at 15:48





    I am dubious about the usefulness of allowing double to be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(

    – Matthieu M.
    Dec 12 '18 at 15:48




    12




    12





    @MatthieuM. I agree, I believe it is a defect and that the intent was really to say contextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(

    – Shafik Yaghmour
    Dec 12 '18 at 15:57







    @MatthieuM. I agree, I believe it is a defect and that the intent was really to say contextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(

    – Shafik Yaghmour
    Dec 12 '18 at 15:57






    2




    2





    @MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.

    – Shafik Yaghmour
    Dec 14 '18 at 17:56





    @MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.

    – Shafik Yaghmour
    Dec 14 '18 at 17:56













    1














    From c++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:




    Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type std::size_t and shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted to std::size_t. [..]




    with only this part ([expr.const]) missing from the C++17 draft.






    share|improve this answer






























      1














      From c++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:




      Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type std::size_t and shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted to std::size_t. [..]




      with only this part ([expr.const]) missing from the C++17 draft.






      share|improve this answer




























        1












        1








        1







        From c++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:




        Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type std::size_t and shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted to std::size_t. [..]




        with only this part ([expr.const]) missing from the C++17 draft.






        share|improve this answer















        From c++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:




        Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type std::size_t and shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted to std::size_t. [..]




        with only this part ([expr.const]) missing from the C++17 draft.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Jan 10 at 8:42









        Bakudan

        14k84266




        14k84266










        answered Dec 14 '18 at 14:34









        gsamarasgsamaras

        52.3k25107193




        52.3k25107193






























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            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
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            Can I use Tabulator js library in my java Spring + Thymeleaf project?

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            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
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