alignas/alignof syntax won't compile Visual Studio 2017









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-1
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The section of code works. But if I instead use the commented out version,



using StorageType = alignas(alignof(T)) char[sizeof(T)];


I get errors.



template <typename T> struct minipool 
union minipool_item
private:
//using StorageType = alignas(alignof(T)) char[sizeof(T)];
using StorageType = char[sizeof(T)];

// Points to the next freely available item.
minipool_item *next;
// Storage of the item. Note that this is a union
// so it is shared with the pointer "next" above.
StorageType datum;
....
;
;


What is the correct syntax?










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  • 2




    Possible duplicate of Where can I use alignas() in C++11?
    – Swordfish
    Nov 10 at 3:45














up vote
-1
down vote

favorite












The section of code works. But if I instead use the commented out version,



using StorageType = alignas(alignof(T)) char[sizeof(T)];


I get errors.



template <typename T> struct minipool 
union minipool_item
private:
//using StorageType = alignas(alignof(T)) char[sizeof(T)];
using StorageType = char[sizeof(T)];

// Points to the next freely available item.
minipool_item *next;
// Storage of the item. Note that this is a union
// so it is shared with the pointer "next" above.
StorageType datum;
....
;
;


What is the correct syntax?










share|improve this question



















  • 2




    Possible duplicate of Where can I use alignas() in C++11?
    – Swordfish
    Nov 10 at 3:45












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











The section of code works. But if I instead use the commented out version,



using StorageType = alignas(alignof(T)) char[sizeof(T)];


I get errors.



template <typename T> struct minipool 
union minipool_item
private:
//using StorageType = alignas(alignof(T)) char[sizeof(T)];
using StorageType = char[sizeof(T)];

// Points to the next freely available item.
minipool_item *next;
// Storage of the item. Note that this is a union
// so it is shared with the pointer "next" above.
StorageType datum;
....
;
;


What is the correct syntax?










share|improve this question















The section of code works. But if I instead use the commented out version,



using StorageType = alignas(alignof(T)) char[sizeof(T)];


I get errors.



template <typename T> struct minipool 
union minipool_item
private:
//using StorageType = alignas(alignof(T)) char[sizeof(T)];
using StorageType = char[sizeof(T)];

// Points to the next freely available item.
minipool_item *next;
// Storage of the item. Note that this is a union
// so it is shared with the pointer "next" above.
StorageType datum;
....
;
;


What is the correct syntax?







c++ c++11






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share|improve this question













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share|improve this question








edited Nov 10 at 3:46









Swordfish

8,86011335




8,86011335










asked Nov 10 at 3:42









Ivan

2,13542754




2,13542754







  • 2




    Possible duplicate of Where can I use alignas() in C++11?
    – Swordfish
    Nov 10 at 3:45












  • 2




    Possible duplicate of Where can I use alignas() in C++11?
    – Swordfish
    Nov 10 at 3:45







2




2




Possible duplicate of Where can I use alignas() in C++11?
– Swordfish
Nov 10 at 3:45




Possible duplicate of Where can I use alignas() in C++11?
– Swordfish
Nov 10 at 3:45












1 Answer
1






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1
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It doesn't work because in C++ there is no mechanism to take an existing type, namely char[sizeof(T)], and create a new type that is identical except for its alignment. If you declare datum to be an array of sizeof(T) chars with the same alignment as T, then the type of datum is still char[sizeof(T)]. The alignment specification can be attached to the member declaration, but not to the type. You can't attach the alignment to the type first, and then use the result of that to declare the member, as you seem to be trying to do.



using StorageType = char[sizeof(T)];
alignas(T) StorageType datum;





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    1 Answer
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    1 Answer
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    active

    oldest

    votes








    up vote
    1
    down vote













    It doesn't work because in C++ there is no mechanism to take an existing type, namely char[sizeof(T)], and create a new type that is identical except for its alignment. If you declare datum to be an array of sizeof(T) chars with the same alignment as T, then the type of datum is still char[sizeof(T)]. The alignment specification can be attached to the member declaration, but not to the type. You can't attach the alignment to the type first, and then use the result of that to declare the member, as you seem to be trying to do.



    using StorageType = char[sizeof(T)];
    alignas(T) StorageType datum;





    share|improve this answer
























      up vote
      1
      down vote













      It doesn't work because in C++ there is no mechanism to take an existing type, namely char[sizeof(T)], and create a new type that is identical except for its alignment. If you declare datum to be an array of sizeof(T) chars with the same alignment as T, then the type of datum is still char[sizeof(T)]. The alignment specification can be attached to the member declaration, but not to the type. You can't attach the alignment to the type first, and then use the result of that to declare the member, as you seem to be trying to do.



      using StorageType = char[sizeof(T)];
      alignas(T) StorageType datum;





      share|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        It doesn't work because in C++ there is no mechanism to take an existing type, namely char[sizeof(T)], and create a new type that is identical except for its alignment. If you declare datum to be an array of sizeof(T) chars with the same alignment as T, then the type of datum is still char[sizeof(T)]. The alignment specification can be attached to the member declaration, but not to the type. You can't attach the alignment to the type first, and then use the result of that to declare the member, as you seem to be trying to do.



        using StorageType = char[sizeof(T)];
        alignas(T) StorageType datum;





        share|improve this answer












        It doesn't work because in C++ there is no mechanism to take an existing type, namely char[sizeof(T)], and create a new type that is identical except for its alignment. If you declare datum to be an array of sizeof(T) chars with the same alignment as T, then the type of datum is still char[sizeof(T)]. The alignment specification can be attached to the member declaration, but not to the type. You can't attach the alignment to the type first, and then use the result of that to declare the member, as you seem to be trying to do.



        using StorageType = char[sizeof(T)];
        alignas(T) StorageType datum;






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 10 at 3:54









        Brian

        63.3k793178




        63.3k793178



























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