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 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 at 15:59






  • 1




    @tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
    – Bob__
    Dec 13 at 9:50






  • 1




    @tomerzeitune Why int? Why not unsigned or long?
    – curiousguy
    Dec 13 at 22:35






  • 1




    @KeithThompson I think the commenters are joking around
    – Shafik Yaghmour
    Dec 14 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 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 at 15:59






  • 1




    @tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
    – Bob__
    Dec 13 at 9:50






  • 1




    @tomerzeitune Why int? Why not unsigned or long?
    – curiousguy
    Dec 13 at 22:35






  • 1




    @KeithThompson I think the commenters are joking around
    – Shafik Yaghmour
    Dec 14 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 at 21:10

























asked Dec 12 at 14:25









Shafik Yaghmour

125k23322530




125k23322530








  • 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 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 at 15:59






  • 1




    @tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
    – Bob__
    Dec 13 at 9:50






  • 1




    @tomerzeitune Why int? Why not unsigned or long?
    – curiousguy
    Dec 13 at 22:35






  • 1




    @KeithThompson I think the commenters are joking around
    – Shafik Yaghmour
    Dec 14 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 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 at 15:59






  • 1




    @tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
    – Bob__
    Dec 13 at 9:50






  • 1




    @tomerzeitune Why int? Why not unsigned or long?
    – curiousguy
    Dec 13 at 22:35






  • 1




    @KeithThompson I think the commenters are joking around
    – Shafik Yaghmour
    Dec 14 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 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 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 at 15:59




@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float
– Shafik Yaghmour
Dec 12 at 15:59




1




1




@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 at 9:50




@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 at 9:50




1




1




@tomerzeitune Why int? Why not unsigned or long?
– curiousguy
Dec 13 at 22:35




@tomerzeitune Why int? Why not unsigned or long?
– curiousguy
Dec 13 at 22:35




1




1




@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 at 18:11




@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 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 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 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 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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    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 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 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 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 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 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 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 at 15:00

























    answered Dec 12 at 14:25









    Shafik Yaghmour

    125k23322530




    125k23322530








    • 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 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 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 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 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 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 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 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 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 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 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 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 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










        answered Dec 14 at 14:34









        gsamaras

        50.5k2399185




        50.5k2399185






























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            Are they similar matrix

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            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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