alignas/alignof syntax won't compile Visual Studio 2017
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-1
down vote
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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?
c++ c++11
add a comment |
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?
c++ c++11
2
Possible duplicate of Where can I use alignas() in C++11?
– Swordfish
Nov 10 at 3:45
add a comment |
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?
c++ c++11
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
c++ c++11
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
add a comment |
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
add a comment |
1 Answer
1
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)
char
s 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;
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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)
char
s 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;
add a comment |
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)
char
s 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;
add a comment |
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)
char
s 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;
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)
char
s 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;
answered Nov 10 at 3:54
Brian
63.3k793178
63.3k793178
add a comment |
add a comment |
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2
Possible duplicate of Where can I use alignas() in C++11?
– Swordfish
Nov 10 at 3:45